5 THE INTEGRAL 5.1 Approxi Approximatin mating g and Computing Computing Area Preliminary Questions Œ2; 5 is divided into six subintervals? 1. What are the right and left endpoints endpoints if Œ2; If the interval Œ2;5 interval Œ2;5 is divided into six subintervals, the length of each subinterval is of the subintervals are then 52 ; 3; 72 ; 4; 92 ; 5, while the left endpoints are 2; are 2; 52 ; 3; 72 ; 4; 92 . SOLUTION
52 6
D 12 . The right endpoints
2. The interval Œ1 interval Œ1;; 5 is divided into eight subintervals. (a) What is the left endpoint of the last subinterval? subinterval? (b) What are the right endpoints of the first two subintervals? SOLUTION
Note that each each of the 8 subintervals has length
51 8
subinterval is 5 is 5 12 (a) The left endpoint of the last subinterval
D 12 .
D 92 . 1 3 1 C D C 1 1 2 (b) The right endpoints of the first two subintervals are 1 are and 2 2 and 1 2 D 2. 3. Which of the following pairs of sums sums are not equal? equal? 4
(a)
4
i;
i
1
`
(b)
5
j;
j 1
D
i
5
X D X
2
j ;
j 1
` 1
4
(c)
4
X X D D X X
1/
(d)
D2
i
i.i
D1
k2
k 2
D
4
.i
X
C 1/;
5
X
.j
1/j
j 2
D
SOLUTION
(a) Only the name of the index variable has been changed, changed, so these two sums are the same. (b) These two sums are not the the same; the second squares the numbers two through five while the first squares the numbers one through four. (c) These two sums are the same. Note that when i ranges from two through five, the expression i 1 ranges from one through four. (d) These two sums are the same. Both sums are 1 2 2 3 3 4 4 5. 100
4. Explain:
100
P D P j
j 1
D
SOLUTION
j 0
D
C
100
j but
P
C
100
1 is not equal to
j 1
1.
j 0
D
The first term in the sum
P
C
D
100 j 0 j
PD
is equal to zero, so it may be dropped. More specifically, 100
100
100
X D C X D X j
0
j 0
On the other hand, the first term in
j 0
D
100
1
D1
100
X ¤X C 1
j 1
D
1:
j 1
D
D x 2 on Œ3 on Œ3;; 7. On Œ3 On Œ3;; 7, the function f function f .x/ D x 2 is a decreasing function; hence, for any subinterval of Œ3 of Œ3;; 7, the function value at
5. Explain why L why L100 SOLUTION
D
this term cannot be dropped. In particular,
100
X
j:
j 1
D
100 j 0 1 is not zero, so
PD
j
j 1
D
R100 for f for f .x/
the left endpoint i s larger than the function value at the right endpoint. Consequently, L Consequently, L100 must be larger than R than R 100 .
546
CHAPTER 5
THE INTEGRAL
Exercises Œ0; 3 and 1. Figure 1 shows the velocity of an object over a 3-min interval. Determine the distance traveled over the intervals Œ0; Œ1;2:5 (remember to convert from km/h to km/min). km/h 30 20 10 min 1
2
3
FIGURE 1
The distance traveled traveled by the object can be determined by calculating the area underneath the velocity graph over over the specified interval. During the interval Œ0 interval Œ0;; 3, the object travels SOLUTION
10 60
1 2
25 60
C
.1/
C
15 60
1 2
20 60
.1/
1 2
D 12 D 0:5 km 0:5 km::
C
0:96 km D 23 0:96 km:: 24
During the interval Œ1 interval Œ1;; 2:5, it travels 25 60
1 2
15 60
1 2
20 60
C
C
2. An ostrich (Figure 2) runs with velocity 20 km/h for 2 minutes, 12 km/h for 3 minutes, and 40 km/h for another minute. Compute the total distance traveled and indicate with a graph how this quantity can be interpreted as an area.
70 km/h. FIGURE 2 Ostriches can reach speeds speeds as high as 70 km/h. SOLUTION
The total distance distance traveled by by the ostrich is 20 60
.2/
C
12 60
.3/
C
40 60
.1/
D 23 C 35 C 23 D 29 15
km. This distance is the area under t he graph below which shows the ostrich’s velocity as a function of time. y
40 30 20 10 0
0
1
2
3
4
5
6
x
3. A rainstorm hit Portland, Maine, in October 1996, resulting in record rainfall. The rainfall rate R.t/ rate R.t/ on on October 21 i s recorded, in centimeters per hour, in the following table, where t where t is is the number of hours since midnight. Compute the total rainfall during this 24-hour period and indicate on a graph how this quantity can be interpreted as an area.
t (h)
0–2
2–4
4–9
9–12
12–20
20–24
R.t/ (cm)
0.5
0.3
1.0
2.5
1.5
0.6
SECTION
SOLUTION
Approximat Approximating ing and Computing Computing Area
5.1
547
Over each interval, the total rainfall is the time interval in hours times the rainfall in centimeters pe perr hour. Thus R
D 2.0:5/ C 2.0:3/ C 5.1:0/ C 3.2:5/ C 8.1:5/ C 4.0:6/ D 28:5 cm 28:5 cm::
The figure below is a graph of the rainfall as a function of time. The area of the shaded region represents the total rainfall. y
2.5 2.0 1.5 1.0 0.5 x
5
4. The velocity of an object is v.t/ is v.t/ and Œ2 and Œ2;; 5. SOLUTION
10
15
20
25
D 12t m/s. Use Eq. (2) and geometry to find the distance traveled over the time intervals Œ0 intervals Œ0;; 2
Œa; b is By equation Eq. (2), the distance distance traveled over over the time interval interval Œa; b
Z
b
v.t/dt
a
Z D
12t dt
I
a
that is, the dist ance traveled is the area under the graph of the velocity function over the interval Œa interval Œa;; b . The graph below shows the 12t m/s 12t m/s over the intervals Œ0 2; 5. Over the interval Œ0 area under the velocity function v.t/ intervals Œ0;; 2 and Œ and Œ2; interval Œ0;; 2, the area is a triangle of base 2 and height 24; therefore, the distance traveled is
D
1 .2/.24/ 2
D 24 meters 24 meters::
2; 5, the area is a trapezoid of height 3 and base lengths 24 and 60; therefore, the distance traveled is Over the interval Œ interval Œ2; 1 .3/.24 2
C 60/ D 126 meters 126 meters::
y
60 50 40 30 20 10 x
1
2
3
4
5
5. Compute R Compute R 5 and L and L5 over Œ0 over Œ0;; 1 using the fol lowing values.
x
SOLUTION
x
0
0.2
0.4
0.6
0.8
1
f.x/ 50 50
48
46
44
42
40
D 1 5 0 D 0:2. 0:2. Thus,
L5
D 0:2.50 C 48 C 46 C 44 C 42/ D 0:2.230/ D 46;
and R5
D 0:2 .48 C 46 C 44 C 42 C 40/ D 0:2.220/ D 44:
The average is 46
C 44 D 45: 2
This estimate is frequently referred to as the Trapezoidal Approximation . 6. Compute R Compute R 6 , L6 , and M and M 3 to estimate the distance traveled over Œ0 over Œ0;; 3 if the velocity at half-second intervals is as follows:
t (s)
0
0.5
1
1.5
2
2.5
3
v (m/s)
0
12
18
25
20
14
20
548
THE INTEGRAL
CHAPTER 5
SOLUTION
D 3 6 0 D 0:5. 0:5. For M For M 3 , t D 3 3 0 D 1. Then R6 D 0:5 s 0:5 s .12 C 18 C 25 C 20 C 14 C 20/ m/s D 0:5.109/ m 0:5.109/ m D 54:5 m;
For R For R6 and L and L 6 , t
L6
D 0:5 sec .0 C 12 C 18 C 25 C 20 C 14/ m/sec D 0:5.89/ m D 44:5 m;
and M 3 Let f .x/ 7. Let f
D 1 sec .12 C 25 C 14/ m/sec 14/ m/sec D 51 m:
D 2x C 3.
(a) Compute R Compute R6 and L and L 6 over Œ0 over Œ0;; 3. (b) Use geometry to find the exact area A area A and and compute the errors A R6 and A L6 in the approximations.
j
SOLUTION
Let f Let f .x/ .x/
j
j
D 2x C 3 on Œ0 on Œ0;; 3.
j
n
o
(a) We partition Œ0 partition Œ0;; 3 into 6 equally-spaced subintervals. The left endpoints of the subintervals are 0; 12 ; 1; 32 ; 2; 52 whereas the
right endpoints are
Let a Let a
n
o
1 3 5 2 ; 1; 2 ; 2; 2 ; 3
.
D 0, b D 3, n D 6, x D .b a/=n D 12 , and x and xk D a C kx , k D 0 ; 1 ; : : : ; 5 (left 5 (left endpoints). Then 5
L6
With x With xk
D
5
X
f .xk /x
D x
k 0
D
X
f .xk /
k 0
D
D 12 .3 C 4 C 5 C 6 C 7 C 8/ D 16:5:
D a C kx, k D 1 ; 2 ; : : : ; 6 (right 6 (right endpoints), we have 6
R6
D
6
X
f .xk /x
k 1
D
D x
X
f .xk /
k 1
D
D 12 .4 C 5 C 6 C 7 C 8 C 9/ D 19:5:
D 12 .3/.6/ C 32 D 18. 18. Thus, L Thus, L6 underestimates the true area .L area .L6 A D D C1:5/. 1:5/.
(b) Via geometry geometry (see figure below), the exact area is A is A 1:5/, 1:5/, while R while R6 overestimates the true area .R area .R6 A y
9 6 3 x
0.5
8. Repeat Exercise 7 for f for f .x/ .x/ SOLUTION
Let f Let f .x/ .x/
1
1.5
2
2.5
3
D 20 3x over 3x over Œ2 Œ2;; 4.
D 20 3x on 3x on Œ2 Œ2;; 4.
n
o
1 1 (a) We partition Œ2;4 partition Œ2;4 into 6 equally-spaced subintervals. The left endpoints of the subintervals are 2; 73 ; 83 ; 3; 10 whereas 3 ; 3
the right endpoints are
Let a Let a
n
7 8 10 11 3 ; 3 ; 3; 3 ; 3 ; 3
o
.
D 2, b D 4, n D 6, x D .b a/=n D 13 , and x and xk D a C kx , k D 0 ; 1 ; : : : ; 5 (left 5 (left endpoints). Then 5
X D
L6
5
f .xk /x
k 0
D
With x With xk
D x
X
f .xk /
k 0
D
D 13 .14 C 13 C 12 C 11 C 10 C 9/ D 23:
D a C kx, k D 1 ; 2 ; : : : ; 6 (right 6 (right endpoints), we have 6
R6
X D
k 1
D
6
f .xk /x
D x
X
f .xk /
k 1
D
D 13 .13 C 12 C 11 C 10 C 9 C 8/ D 21:
1 (b) Via geometry geometry (see figure below), the exact area is A is A 2 .2/.14 1/, 1/, while R while R6 underestimates the true area .R area .R6 A 1/. 1/.
D
D
C 8/ D 22. 22. Thus, L Thus, L 6 overestimates the true area .L6 A D
y
14 12 10 8 6 4 2 x
1
2
3
4
SECTION
5.1
Approximat Approximating ing and Computing Computing Area
549
9. Calculate R Calculate R3 and L and L3
for f for f .x/ .x/
D x 2 x C 4 over Œ1 over Œ1;; 4
f and the rectangles that make up each approximation. Is the area under the graph larger or smaller than Then sketch the graph of f R3? Is it larger or smaller than L than L 3 ? Let f Let f .x/ x 2 x 4 and set a set a a kx , k 0;1;2;3. 0;1;2;3.
SOLUTION
(a) Let x Let x k
D
D C
C
D
D 1, b D 4, n D 3, x D .b a/=n D .4 1/=3 D 1.
Selecting the left endpoints of the subintervals, subintervals, x x k , k
D 0;1;2, 0;1;2, or f1;2;3g, we have
2
L3
X D
2
f .xk /x
D x
k 0
D
Selecting the right endpoints of the subintervals, subintervals, x x k , k
X D
R3
D .1/.4 C 6 C 10/ D 20:
D 1;2;3, 1;2;3, or f2;3;4g, we have
3
X D
f .xk /
k 0
3
f .xk /x
D x
k 1
D
X
f .xk /
k 1
D
D .1/ .1/ .6 C 10 C 16/ D 32:
(b) Here are figures of the three rectangles that approximate the area under the curve f .x/ .x/ over the interval Œ1 interval Œ1;; 4. Clearly, the area under the graph is larger than L3 but smaller than R than R3 . y
y
14
14 L 3
12
10. Let f Let f .x/ .x/ 6
by the sum
D
X
i
D1
p 2 x
f .1
10 8 6 4
x
1.0
1.5
R3
12
10 8 6 4 2.0
2.5
3.0
3.5
x
1.0
1.5
2.0
2.5
3.0
3.5
C 1 and x D 13 . Sketch the graph of f f .x/ .x/ and draw the right-endpoint rectangles whose area is represented
C i x/ x .
1 Because x evaluates f at 1 at 1 i x for i for i from from 1 through 6, i t follows that the interval over which 3 and the sum evaluates f we are considering f considering f is Œ1 is Œ1;; 3. The sketch of f f together with the six rectangles is shown below. below. SOLUTION
D
C
y
3.0 2.5 2.0 1.5 1.0 0.5 x
0.5
1.0
1.5
2.0
2.5
3.0
11. Estimate R Estimate R3 , M 3 , and L and L 6 over Œ0;1:5 over Œ0;1:5 for the function in Figure 3. y
5 4 3 2 1 x
0.5
1
1.5
FIGURE 3 SOLUTION
Let f Let f .x/ on x/ on Œ0; Œ0; 32 be given by Figure 3. For n R3
M 3
D
1 2
D
1 2
3
X D X
f .xk /
k 1 6
k 1
D
f xk
D 3, x D . 32 0/=3 D 12 , fxk gk3 D0 D
D 12 .2 C 1 C 2/ D 2:5;
1 x 2
D 12 .3:25 C 1:25 C 1:25/ D 2:875:
n
o
0; 12 ; 1; 32 . Therefore
550
CHAPTER 5
THE INTEGRAL
For n For n
D 6, x D . 32 0/=6 D 14 , fxk gk6D0 D L6
D 14
5
X
n
f .xk /
k 0
D
o
0; 14 ; 12 ; 34 ; 1; 54 ; 32 . Therefore
D 14 .5 C 3:25 C 2 C 1:25 C 1 C 1:25/ D 3:4375:
12. Calculate the area of the shaded rectangles in Figure 4. Which approximation do these rectangles represent? represent? y y =
4 ! x 1 + x 2
x
3
2
!
1
!
1
!
2
3
FIGURE 4
Each rectangle in Figure 4 has a width of 1 and and the height is taken as the value of the function at the midpoint of the interval. Thus, the area of the shaded rectangles is SOLUTION
1
26 29
22 13
C C
18 5
C
14 5
10 13
6 29
C C
D 18784 9:965: 1885
Because there are six rectangles and the height of each rectangle is taken as the value of the function at the midpoint of the interval, the shaded rectangles represent the approximation M approximation M 6 to the area under the curve. In Exercises Exercises 13–20, 13–20, calculate calculate the appro approximation ximation for for the given given function function and interval. 13. R3 , f.x/ SOLUTION
D 7 x , Œ3; Œ3; 5 .x/ D 7 x on Œ3 Let f Let f .x/ on Œ3;; 5. For n For n D 3, x D .5 3/=3 D 23 , and fxk gk3 D0 D R3
D 23 D 23
3
X
n
o
13 3; 11 3 ; 3 ; 5 . Therefore
.7 xk /
k 1
D
10 3
C 83 C 2 D 23 .8/ D 163 :
D p 6x C 2, Œ1; Œ1; 3 p Let f Let f .x/ .x/ D 6x C 2 on Œ1 on Œ1;; 3. For n For n D 6, x D .3 1/=6 D 13 , and fxk gk6 D0 D
14. L6 , f.x/ SOLUTION
L6
D 13 D 13
15. M 6 , f.x/ SOLUTION
6xk
k 0
Dp C p 8
C2
10
C
p
12
D 4x C 3, Œ5;8 Œ5; 8. For n D Let f.x/ D 4x C 3 on Œ5; M 6
D
1 2
o
1; 43 ; 53 ; 2; 73 ; 83 ; 3 . Therefore
5
X p
C
p
14
6, x
D
7:25; 7:25; 7:75 7:75 . Therefore,
g
n
C4C
.8
p
18
5/=6
D
7:146368:
1 2,
f gk5 D0 D f5:25; 5:25; 5:75; 5:75; 6:25; 6:25; 6:75; 6:75;
and xk
5
X
k 0
D
C 3
4xk
D 12 .24 C 26 C 28 C 30 C 32 C 34/ D 12 .174/ D 87: D x 2 C x , Œ1; 1 Let f.x/ D x 2 C x on Œ1; 1. For n D
16. R5 , f.x/ SOLUTION
o
3 5;1
5, x
D
.1
. 1//=5
D
2 5,
f gk5D0 D
and xk
. Therefore
R5
D
2 5
D 25
5
.xk2 k 1
X
D 14 5
C xk / D
D 28 : 25
2 5
9 25
3 5
C
1 25
1 5
C
1 25
C
1 5
C
9 25
C
3 5
C2
n
3
1 1
1; 5 ; 5 ; 5 ;
SECTION
D x2 C 3jx j, Œ2; 1 SOLUTION Let f.x/ Let f.x/ D x 2 C 3 jx j on Œ2; 1. For n D 0:5; 0:5; 0;0:5; 0;0:5; 1g. Therefore
Approximat Approximating ing and Computing Computing Area
5.1
551
17. L6 , f.x/
L6
D
1 2
5
X
.xk2
k 0
D
6, x
D
.1
. 2//=6
D
1 2,
f gk6 D0 D f2; 1:5; 1;
and xk
C 3 jxk j/ D 12 .10 C 6:75 C 4 C 1:75 C 0 C 1:75/ D 12:125:
p Œ3; 5 18. M 4 , f.x/ D x , Œ3; p on Œ3;; 5. For n 15 17 19 SOLUTION Let f Let f .x/ D x on Œ3 For n D 4, x D .5 3/=4 D 12 , and fxk gk3 D0 D f 13 4 ; 4 ; 4 ; 4 g. Therefore,
M 4
D
1 2
3
X q Dp p
xk
k 0
13 2
15 2
C
C
p
p
!
fxk gk4 D0 D
17 2
C
19 2
3:990135:
D cos2 x , 6 ; 2 Let f Let f .x/ D cos2 x on Πon Π6 ; 2 . For n For n D 4,
19. L4 , f.x/ SOLUTION
D
1 2
D .=2 4 =6/ D 12
x
and
5 ; ; ; ; : 6 4 3 12 2
Therefore L4
3
D 12
0:410236:
k 0
D
D ln x, Œ1; Œ1; 3 14 Let f Let f .x/ D ln x on Œ1 on Œ1;; 3. For n For n D 5, x D .3 1/=5 D 25 , and fxk gk4 D0 D f 65 ; 85 ; 2; 12 5 ; 5 g. Therefore,
20. M 5 , f.x/ SOLUTION
cos2 xk
X
M 5
D
2 5
D
2 5
4
X
ln xk
k 0
D
6 ln 5
C
8 ln 5
C ln 2 C
12 ln 5
C
14 ln 5
1:300224:
In Exercises Exercises 21–26, 21–26, write the sum in summation notation. notation. 21. 47
C 57 C 67 C 77 C 87
SOLUTION
The first term is 4 is 47 , and the last term is 8 is 87 , so it seems the k the kth th term is k is k 7 . Therefore, the sum is: 8
X
k7:
k 4
D
22. .22
C 2/ C .32 C 3/ C .42 C 4/ C .52 C 5/ SOLUTION The first term is 2 is 2 2 C 2, and the last term is 5 is 5 2 C 5, so it seems that the sum limits are 2 are 2 and and 5 5,, and the k the kth th term is 2 k C k. Therefore, the sum is: 5
X
.k 2
k 2 2
3
23. .2
4
5
D
C k/:
C 2/ C .2 C 2/ C .2 C 2/ C .2 C 2/ SOLUTION The first term is 2 is 22 C 2, and the last term is 2 is 25 C 2, so it seems the sum limits are 2 and 2 and 5 5,, and the k the kth th term is 2k C 2.
Therefore, the sum is:
5
X
.2k
k 2
D
24.
p C C p C 1
13
2
23
C
p C C n
n3
C 2/:
552
THE INTEGRAL
CHAPTER 5
p
p
The first term is 1 13 and the last term is k -th term is k k 3 . Therefore, the sum is SOLUTION
p
n
C
C
C n3, so it seems the summation limits are 1 through n, n , and the
n
X p C k
k3:
k 1
25.
D
1
2 n C C C 23 34 .n C 1/.n C 2/
1 . .1 1/.1 2/
The first summand summand is
SOLUTION
C
C
This shows us
1
2 n C C D C 23 34 .n C 1/.n C 2/ 26. e
Ce
=2
Ce
=3
C C e
=n
n
X
i
D1
.i
C
i 1/.i
C 2/ :
The first term is e is e =1 and the last term is e =n , so it seems the sum limits are 1 and n and the k the k th term is e is e =k . Therefore, the sum is SOLUTION
n
e =k :
X
k 1
D
27. Calculate the sums: 5
(a)
5
X
i
9
(b)
D1
4
X
i
4
(c)
D0
X
k3
k 2
D
SOLUTION
5
(a)
5
XDCC D XDCC D X D C 9
i
9
9
9
1
C 9 C 9 D 45. 45. Alternatively,
9
i
5
(b)
4
i
(c)
4
4
23
9
1
1
i
4
33
k 2
D
C 4 C 4 C 4 D 24. 24. Alternatively,
D 45. 45.
5
4
i
.9/.5/
1
5
0 4
k3
5
XDXD D D X D XD 4
D0
i
.4/.6/
D0
D 24. 24.
C 43 D 99. 99. Alternatively,
X D 0@ X 1A 0@ X 1A D D D D X 4
4
k
3
k 2
1
k
3
k
k 1
3
k 1
44 4
C
4 3 2
C
4 2 4
!
14 4
C
1 3 2
C
!
1 2 4
D 99:
28. Calculate the sums: 4
(a)
X
sin j
j 3
D
2
5
1
(b)
k 3
D
k1
2
(c)
X
3j
j 0
D
SOLUTION
4
(a)
X D X XD
sin
j 3 5
(b)
3j
1
D
29. Let b Let b 1 4
X
i
C sin
4 2
D 1 C 0 D 1.
1 1 1 13 D C C D : k1 2 3 4 12
j 0
(a)
D sin
3 2
1
k 3 2
(c)
j 2
D 13 C 1 C 3 D 133 :
D 4, b2 D 1, b3 D 2, and b and b 4 D 4. Calculate: 2
bi
(b)
D2
X
SOLUTION
4
(a)
X
i
D2
.2
j 1
D
bi
D b2 C b3 C b4 D 1 C 2 C .4/ D 1:
bj
bj /
3
(c)
X
k 1
D
1
kb k
SECTION
2
(b)
X D X
2bj
bj
j 1
D
.24 4/
Approximat Approximating ing and Computing Computing Area
5.1
C .21 1/ D 13:
3
(c)
kb k
k 1
D
D 1.4/ C 2.1/ C 3.2/ D 12: 10
30. Assume that a that a1
D 5,
10
(a)
X
i
.4a i
D1
X
i
10
ai
D1
D 20, 20, and
C 3/
X D D X bi
i
7. Calculate:
1
10
10
D2
X
ai
(b)
i
(c)
i
.2ai
3bi /
D1
SOLUTION
10
(a)
.4a i
i
(b)
3/
1 10
10
ai
i
ai
2 10
i
3
1
1
i
4.20/
1
a1
20 .5/
C 3.10/ D 110. 110.
25. 25.
1
10
.2a i
i
4
10
ai
i
(c)
10
X C D X C X D XD D X D D D D XD D D X X D D D D X 3bi /
10
2
1
ai
i
3
bi
1
i
2.20/ 3.7/
1
D 19. 19.
200
j . Hint: Write as a difference of two sums and use f ormula (3).
31. Calculate
j 101
D
SOLUTION
200
X
200
j
j 101
D
100
X X D D j
j
j 1
j 1
D
D
2002 2
C
200 2
!
1002 2
C
100 2
!
D 20100 5050 D 15050:
30
X
32. Calculate
C 1/2 . Hint: Expand and use formulas (3)–(4).
.2j
j 1
D
SOLUTION
30
30
X
C 1/2 D 4
.2j
j 1
D
D4
30
X D
j2
j 1
303 3
C4
C
30
X C X D !D j
j 1
30 2 2
C
1
j 1
30 6
C4
302 2
C
30 2
!
C 30
D 39;710: In Exercises Exercises 33–40, 33–40, use linearity linearity and and formulas formulas (3)–(5) to rewrite rewrite and evaluate evaluate the sums. 20
33.
X
8j 3
j 1
D
20 SOLUTION
X
j 1
D
20
8j
3
D8
X
j
3
j 1
D
D8
204 4
C
20 3 2
C
20 2 4
!
D 8.44;100/ D 352;800. 352;800.
30
34.
X
.4k
3/
k 1
D
SOLUTION
30
X
30
.4k
k 1
D
3/
D4 D4
30
X X D D! k3
k 1
302 2
1
k 1
C 302
3.30/
D 4.465/ 90 D 1770:
553
554
THE INTEGRAL
CHAPTER 5
150
35.
n2
X
n 51
D
SOLUTION
150
150
n2
X
n 51
D
n 1
C
150 2 2
C
150 6
!
503 3
C
50 2 2
C
100 3 2
C
50 6
!
D 1;136;275 42;925 D 1;093;350:
200
36.
n2
n 1
1503 3
D
X
50
n2
X X D D
D
k3
k 101
D
SOLUTION
200
X
200
k3
k 101
D
k 1
C
200 3 2
C
200 2 4
!
1004 4
C
100 2 4
!
D 404;010;000 25;502;500 D 378;507;500:
50
X
k3
k 1
2004 4
D 37.
100
k3
X X D D
D
j.j
1/
j 0
D
SOLUTION
50
50
X
j.j
1/
j 0
50
X D D
.j
2
j/
j 0
D
D
50 2
X X D D !D ! j
503 3
2
j 0
C 502 C 506
X
1/
D
30
38.
2
C 4j3
6j
j 2
D
3
C 502 D 503
50 3
D 124;950 D 41;650: 3
50
j.j
j 0
X
502 2
50
The power sum formula is usable because
j
j 0
D
X
j.j
1/. 1/.
j 1
D
!
SOLUTION
30
X
6j
j 2
D
C
4j 2 3
30
! X D D 6
j
j 2
302 2
D6
C
4 3
0X X 1 0X X 1 A A D @ C @ D ! D D ! D D
30
30
X
j
2
1
6
j
j 2
j 1
C 302 1 C 43
303 3
j
j 1
4 3
2
C 302 C 306 1
46;168 D 6.464/ C 43 .9454/ D 2784 C 37;816 D : 3 3 30
39.
X
.4 m/3
m 1
D
SOLUTION
30
X
30
.4 m/3
m 1
D
X D X D D
.64 48m
m 1
30
64
m 1
D
C 12m2 m3 / 30
1 48
X
m 1
D
30
m
C 12
X
m 1
D
30
m2
X
m 1
D
m3
30
1
j
j 1
2
j2
j 1
SECTION
D 20
40.
X
5
m 1
D
C 3m2
.30/.31/ 64.30/ 48 2
C 12
303 3
C
Approximat Approximating ing and Computing Computing Area
5.1
30 2 2
C
30 6
!
304 4
C
30 3 2
C
D 1920 22;320 C 113;460 216;225 D 123;165:
2
SOLUTION
20
X C 5
m 1
D
3m 2
20
2
D 25
20
X
1
m 1
D
X
C 15
D 25.20/ C 15
m
m 1
D
202 2
C
C 94
20 2
20
X
m2
m 1
! D C
9 4
203 3
C
20 2 2
C
20 6
!
D 500 C 15.210/ C 94 .2870/ D 10107:5: In Exercises Exercises 41–44, 41–44, use formulas formulas (3)–(5) (3)–(5) to evaluate evaluate the the limit. N
41.
lim
N
X
!1 i D1
i N 2 N
Let s Let sN
D
SOLUTION
X
i
D1
i
. Then, N 2 N
i
X D
sN
i
D1
N
1
N 2
D N 2
j3
1 N 4
X D N 2
N 2 2
N 3 2
N 2 4
1
i
i
D1
C
N 2
!
1 D 12 C 2N :
D 12 .
Theref Therefore ore,, lim sN N
!1
N
42.
lim
N
X
!1 j D1
j3 N 4 N
Let s Let sN
D
SOLUTION
X
j 1
D
j3 . Then N 4 1 N 4
sN
D
N
X
j 1
D
D
N 4 4
C
C
!
1 D 14 C 2N C 4N 1 2 :
D 14 .
Theref Therefore ore,, lim sN N
!1
N
43.
lim
N
X
!1 i D1
i2 i N 3
C1 N
Let s Let sN
D
SOLUTION
X
i
D1
i2 i N 3
C 1 . Then N
X D
sN
i
D D 13 .
Theref Therefore ore,, lim sN N
!1
i2 i
N 3
D1
1 N 3
C1 D
"
N 3 3
2
1 N 3
20X 1 4@ D A !
C N 2 C N 6
N
i
i
2
1
N 2 2
0 X 1 0 X 13 @ D A C @ D A5 ! # N
N
i
i
1
1
i
1
C N 2 C N D 13 C 3N 2 2 :
30 2 4
!
555
556
N
44.
THE INTEGRAL
CHAPTER 5
lim
N
X
!1 i D1
! ! X D D X X D
i3 N 4
20 N
N
i3 N 4
Let s Let sN
SOLUTION
i
1
N
1 N 4
sN
20 . Then N
i
i
3
D1
20 N
N
i
1
D1
D
1 N 4
N 4 4
C
N 3 2
C
!
N 2 4
1 D 14 C 2N C 4N 1 2 20:
20
D 14 20 D 794 .
Theref Therefore ore,, lim sN N
!1
In Exercises Exercises 45–50, 45–50, calculate calculate the limit limit for the the given function function and interval. interval. Verify your your answer by by using geometry. geometry.
D 9x, 9x , Œ0; Œ0; 2 D .2 0/=N D D SOLUTION Let f Let f .x/ .x/ D 9x on 9x on Œ0 Œ0;; 2. Let N Let N be be a positive integer and set a set a D 0, b D 2, and x D .b a/=N D 2=N . 2=N . Also, let x let x k D a C kx D 2k=N , 2k=N , k D 1 ; 2 ; : : : ; N be the right endpoints of the N the N subintervals subintervals of Œ0; Œ0; 2. Then
45.
lim RN , f.x/
N
!1
N
RN
D x
X
f .xk /
k 1
D
D
2 N
N
X
9
k 1
D
2k N
N
36
D N 2
X
k
k 1
D
36
D N 2
N 2 2
C
N 2
!
D 18 C 18 : N
The area under the graph is lim RN
D N lim !1
N
!1
18
C
18 N
D 18:
The region under the graph is a tri angle with base 2 and height 18. The area of the region is then the value obtained from the limit of the right-endpoint approximations.
1 2 .2/.18/
D 18, 18, which agrees with
D 3x C 6, Œ1; Œ1; 4 D .4 1/=N D D SOLUTION Let f.x/ D 3x C 6 on Œ1; Œ1; 4. Let N be a positive positive integer integer and and set a D 1, b D 4, and x D .b a/=N D 3=N . 3=N . Also, let x let x k D a C kx D 1 C 3k=N , 3k=N , k D 1 ; 2 ; : : : ; N be be the right endpoints of the N the N subintervals subintervals of Œ1; Œ1; 4. Then
46.
lim RN , f.x/
N
!1
N
RN
D x D
27 N
3 N
N
X D X C D D X C X D f .xk /
k 1 N
27 N 2
1
k 1
D
9
k 1
N
9k N
27 .N / N
j
k 1
D
C
27 N 2
N 2 2
C
N 2
!
27 D 812 C 2N : The area under the graph is lim RN
D N lim !1
N
!1
81 2
C
27 2N
D 812 :
The region under the graph is a trapezoid with base width 3 and heights 9 and 18. The area of the region is then which agrees with the value obtained from the limit of the ri ght-endpoint approximations.
1 2 .3/.9
C 18/ D 812 ,
D 12 x C 2, Œ0; Œ0; 4 D N 4 . Also, let SOLUTION Let f .x/ D 12 x C 2 on Œ0 Let f .x/ on Œ0;; 4. Let N Let N > 0 be an integer, and set a D 0; b D 4, and x D .4 0/=N D D 0 ; 1 ; : : : ; N 1 be the left endpoints of the N xk D 0 C kx D 4k the N subintervals. subintervals. Then N ; k
47.
lim LN , f.x/
N
!1
N 1
LN
D x D 8C
X
f .xk /
8 N 2
k 0
D
D
4 N
.N 1/2 2
N 1
X
k 0
D
C
1 2
N 1 2
4k N
!
C2 D
8 N
N 1
X
k 0
D
1
C
8 N 2
N 1
X
k
k 0
D
D 12 N 4 :
The area under the graph is lim LN
N
!1
D 12:
The region under the curve over Œ0;4 is a trapezoid with base width 4 and heights 2 and 4. 4 . From this, we get that the area is 1 4/ 12, 12, which agrees with the answer obtained from the limit of the left-endpoint approximations. 2 .4/.2
C D
SECTION
Approximat Approximating ing and Computing Computing Area
5.1
D 4x 2, Œ1;3 D SOLUTION Let f Let f .x/ D 4x 2 on Œ1 on Œ1;; 3. Let N Let N > 0 be an i nteger, and set a set a D 1; b D 3, and x D .3 1/=N D 2k xk D a C kx D 1 C N ; k D 0 ; 1 ; : : : ; N 1 be the left endpoints of the N the N subintervals. subintervals. Then
48.
557
lim LN , f.x/
N
!1
N 1
D x
LN
D N 162
X D
f .xk /
k 0
D
.N 1/2 2
2 N
N 1
X D !
k 0
8k N
C2 D
16 N 2
N 1
X
k
k 0
D
C
4 N
2 N .
Also, let
N 1
X
1
k 0
D
C N 2 1 C N 4 .N 1/
D 12 12 N The area under the graph is lim LN
D 12:
N
!1
The region under the curve over Œ1; Œ1; 3 is a trapezoid trapezoid with base width 2 width 2 and heights 2 heights 2 and 10. 10 . From this, we get that the area is 1 .2/.2 10/ 12, 12 , which agrees with the answer obtained from the limit of the left-endpoint approximations. 2
C
D
D x , Œ0;2 SOLUTION Let f Let f .x/ .x/ D x on Œ0; Œ0; 2. Let N > 0 be an integer and set a D 0, b D 2, and x D .b a/=N D 1 2k 1 xk D 0 C .k 2 /x D N ; k D 1 ; 2 ; : : : N , be the midpoints of the N the N subintervals subintervals of Œ0; Œ0; 2. Then
49.
lim M N , f.x/
N
!1
2 . N
Also, let
N
M N N
D x 2
D N 2 The area under the curve over Œ0 over Œ0;; 2 is
2 N
N
2k 1 N
X D X 0D X 1 D @ AD
f .xk /
k 1
k 1
N
2
4
k N
N 2
k 1
D
D N 22
N 2 2
C N 2
N
X !D
.2k 1/
k 1
2 N
D 2:
lim M N N
D 2:
N
!1
The region under the curve over Œ0 over Œ0;; 2 is a triangle with base and height 2, and thus area 2, which agrees with the answer obtained from the limit of the midpoint approximations.
D 12 4x, 4x , Œ2;6 SOLUTION Let f D N 4 . Also, let Let f .x/ D 12 4x on 4x on Œ Œ2; 2; 6. Let N Let N > 0 be an integer and set a D 2, b D 6, and x D .b a/=N D xk D a C .k 12 /x D 2 C 4kN 2 ; k D 1 ; 2 ; : : : N , be the midpoints of the N the N subintervals subintervals of Œ2; Œ2; 6. Then
50.
lim M N N , f.x/
N
!1
N
D x
M N N
D 16 N
X D X
f .xk /
k 1 N
D
k 1
D
64 D 16 .N / 2 N N
X
4
k 1
D
N
64 N 2
1
N
4 N
C N 322
16k 8 N
N
X
k
N 2 2
C N 2 C N 322 .N / D 16:
k 1
D !
X
1
k 1
D
The area under the curve over Œ over Œ2; 2; 6 is lim M N N
N
!1
D 16:
The region under t he curve over Œ2 over Œ2;; 6 consists of a tri angle of base 1 and height 4 above the axis and a tri angle of base 3 and height 12 below the axis. The area of thi s region is therefore 1 1 .1/.4/ .3/.12/ 2 2
D 16;
which agrees with the answer obtained from the limit of the midpoint approximations.
558
THE INTEGRAL
CHAPTER 5
51. Show, for f for f .x/
D 3x2 C 4x over 4x over Œ0 Œ0;; 2, that 2 N
RN
D
N
X
j 1
D
24j 24j 2 N 2
C
!
16j N
Then evaluat evaluatee lim RN . N
!1
SOLUTION
.2 0/=N
D D
Let f Let f .x/ .x/ 3x 2 2=N . 2=N . Also, let x let xj
D
C 4x on 4x on Œ0; Œ0; 2. Let N be N be a positive integer and set a D 0, b D 2, and x D .b a/=N D D a C jx D 2j=N , 2j=N , j D 1 ; 2 ; : : : ; N be be the right endpoints of the N the N subintervals subintervals of Œ0; Œ0; 3. Then N
D x
RN
D N 2
X D X
f .xj /
j 1 N
D
12j 2
j 1
D
N
2 N
X !D
2j N
3
j 1
N 2
C 8jN
2
16
2
C
2j 4 N
!
N 2
!
Continuing, we find N
24
RN
X D
D N 3
C N 2
j 1
N 3 3
24 N 3
D
j
N
C
N 2 2
X
j
N 6
!
j 1
D
C
C
16 N 2
N 2 2
C
D 16 C 20 C N 42 N Thus, lim RN
D N lim !1
N
!1
52. Show, for f for f .x/
D 3x3 x 2 over Œ1 over Œ1;; 5, that 4 N
RN
D
N
X
16
192j 3
j 1
D
C 20 C N 42 D 16: N
C
N 3
128j 2 N 2
C
!
28j N
C2
Then evaluat evaluatee lim RN . N
!1
SOLUTION
.5 1/=N Then
D D
Let f Let f .x/ .x/ 3x 3 x 2 on Œ1;5 on Œ1;5. Let N be N be a positive integer and set a 1, b 5, and x .b a/=N 4=N . 4=N . Also, let x 1 4j=N , 4j=N , j 1 ; 2 ; : : : ; N be Œ1; 5. let x j a jx be the right endpoints of the N the N subintervals subintervals of Œ1;
D
D C f .xj /
D C
3
D3 1C
4j N
12j N
D3 1C
D
D
C
1
C
48j 2 N 2
3
2
f .xj /x
4 N
4j N
C
D
2
64j 3 N 3
!
1
C
8j N
C
16j 2 N 2
!
D 192j C 128j C 28j C 2: N N 3 N 2 and N
X D
RN
j 1
D
D
N
X
j 1
D
192j 3 N 3
C
128j 2 N 2
C
28j N
!
C2
Continuing, we find
D 768 N 4
RN
D 768 N 4
N
X D
j3
j 1
N 4 4
C 512 N 3 3
N
X
j2
j 1
D
2
!
C 112 N 2
C N 2 C N 2 C 512 N 3
N
8 N
N
X C X D D ! j
j 1
N 3 3
2
1
j 1
C N 2 C N 6
:
D
D
SECTION
C
112 N 2
N 2 2
!
N 2
C
Approximat Approximating ing and Computing Computing Area
5.1
C N 8 .N /
696 832 D 1280 C C : 3 N 3N 2 Thus, lim RN
D N lim !1
N
!1
1280 3
C
696 N
832 3N 2
C
D 1280 : 3
In Exercises Exercises 53–60, 53–60, find a formula formula for RN and compute the area under the graph as a limit.
D x 2 , Œ0; Œ0; 1 10 1 D SOLUTION Let f Let f .x/ D x 2 on the interval Œ0 interval Œ0;; 1. Then x D and a and a D 0. Hence, N N
53. f.x/
N
D x
RN
X
f .0
j 1
D
N
1 N
C jx/ D
X
j 1
D
1 j N 2 2
1 N 3
D
N 3 3
C
N 2 2
C
N 6
!
1 D 13 C 2N C 6N 1 2
and
lim RN
D N lim C !1
N
54. f.x/
D x2,
SOLUTION
1 3
!1
1 2N
1
C 6N 2 D 13 :
Œ1; 5
Let f Let f .x/
5 .1/ 6 D x 2 on the interval Œ D interval Œ1; 5. Then x D and a and a D 1. Hence, N N N
RN
D x D N 6 D
X D X
f .1
j 1 N
72 N 2
1
j 1
D
6 72 .N / 2 N N
N
C jx/ D N 6 N
X C j
j 1
N 2 2
C
N 2
1
j 1
216 N 3
D !
X D X
C 6jN
2
N
j2
j 1
D
216
C N 3
N 3 3
C
N 2 2
C
N 6
!
36 D 42 C 72 C N N 2 and lim RN
D N lim !1
N
!1
42
C
72 N
36
C N 2 D 42:
D 6x2 4, Œ2;5 52 3 D SOLUTION Let f Let f .x/ D 6x 2 4 on the interval Œ interval Œ2; 2; 5. Then x D and a and a D 2. Hence, N N
55. f.x/
N
RN
D x
X
f .2
j 1
D
D 60 C 216 N 2 D 60 C
216 N 2
C jx/ D
N
162 N 3
3 N
N
X C 6 2
j 1
D
j 1
N 2 2
C
N 2
2
N
! X D 3 N
4
j 1
D
N
j2
X C X D ! D j
3j N
j 1
C
162 N 3
N 3 3
C
N 2 2
C
N 6
!
D 222 C 189 C N 272 N and lim RN
N
!1
D N lim !1
222
C
189 N
C
27 N 2
D 222:
20
C
72j N
C
54j 2 N 2
!
559
560
THE INTEGRAL
CHAPTER 5
D x2 C 7x, 7x , Œ6;11 11 6 5 D SOLUTION Let f Let f .x/ .x/ D x 2 C 7x on 7x on the interval Œ6;11 interval Œ6;11. Then x D and a and a D 6. Hence, N N
56. f.x/
N
RN
D x D N 5
X D X D X D
f .6
j 1 N
j 1
N
D 125 N 3
j 1
N 3 3
D
C
6
j 1
5j N
2
C7 6C
5j N
#
C 95j C 78 N N
C 475 N 2
j3
125 N 3
C jx/ D
25j 25j 2 N 2
N
X " C D !
5 N
N 2 2
N
390 N
X C X D ! D j
j 1
C
N 6
C
1
j 1
475 N 2
N 2 2
C
C
125 6N 2
N 2
!
C 390
300 125 D 4015 C C 6 N 6N 2 and lim RN
D N lim !1
N
!1
4015 6
C
300 N
D 4015 : 6
D x3 x , Œ0; Œ0; 2 20 2 D SOLUTION Let f .x/ D x 3 x on the interval Œ0 Let f .x/ interval Œ0;; 2. Then x D and a and a D 0. Hence, N N
57. f.x/
N
D x
RN
X D X D
C jx/ D
f .0
j 1
D
16 N 4
D
16 N 4
N
j
3
j 1
N 4 4
2 N
C
4 N 2
N 3 2
N
X
2j N
j 1
D
3
2j N
N
! X D 2 N
j 1
D
8j 3 N 3
2j N
!
N
X
j
j 1
D
C
N 2 2
!
4 N 2
N 2 2
C
N 2
!
D 2 C N 6 C N 82 and
58. f.x/
D 2x3 C x2,
SOLUTION
Let f Let f .x/ .x/
6 8 C D 2: lim RN D lim 2C N N 2 N !1 N !1 Œ2; 2 2 .2/ 4 D 2x3 C x 2 on the interval Œ D interval Œ2; 2. Then x D and a and a D 2. Hence, N N N
RN
D x D
4 N
X D X D
f .2
j 1 N
j 1
512
D N 4
C jx/ D
128j 3 N 3
N 4 4
C
N 3 2
176j 2 N 2
C
N 2 4
N
X"
4 N
!
C
2
2
j 1
D
80j N
704 N 3
C
4j N
3
C
2
C
4j N
!
12
N 3 3
C
N 2 2
C
!
C N 2
D 163 :
N 6
320
32 D 163 C 64 C N 3N 2 and lim RN
N
!1
D N lim !1
2
#
16 3
C
64 N
C
32 3N 2
N 2 2
C
N 2
!
48
SECTION
Approximat Approximating ing and Computing Computing Area
5.1
D 2x C 1, Œa; Œa; b (a; b constants with a with a < b ) b a SOLUTION Let f Let f .x/ D 2x C 1 on the interval Œa interval Œa;; b . Then x D . Hence, N
59. f.x/
N
RN
D x
X
f .a
j 1
D
.b a/ .2a N
D
C jx/ D N
C 1/
X
j 1
D
1
N
.b a/ N
C
X C D X D ! 2 a
2
a/
N
j 1
2.b 2 .b a/2 N 2
a/ D .b N C 2.bN 2a/ .2a C 1/N C
j
.b
C1
N
j
j 1
N 2 2
C N 2
D .b a/.2a C 1/ C .b a/2 C .b N a/
2
and lim RN
D N lim !1
N
!1
.b a/.2a
2
C 1/ C .b a/ C
.b
a/
2
N
!
D .b a/.2a C 1/ C .b a/2 D .b2 C b/ .a2 C a/:
D x 2 , Œa;b (a; b constants with a with a < b ) b a SOLUTION Let f Let f .x/ D x 2 on the interval Œa interval Œa;; b . Then x D . Hence, N
60. f.x/
N
RN
D x Da D
X
j 1
D
f .a
C jx/ D
2 .b a/ N
N
X
j 1
D
N
X C C D X C X D ! D
.b a/ N
a2
j 1 N
2a.b a/ 1C N 2
2
2a.b a/2 N 2
N 2 2
a2 .b a/ N N
C C
.b 2aj
.b
j
N 2
C
.b
a/
N
j
N 2
!
j2
j 1
a/
3
N 3
2
2 2 .b a/
N
a/3
N 3
j 1
C
N 3 3
3
C
N 2 2
C
N 6
3
a/ a/ a/ D a2 .b a/ C a.b a/2 C a.b N C .b 3 a/ C .b 2N C .b6N 2
!
3
and lim RN
D N lim !1
N
!1
a2 .b a/
C a.b a/2 C
a.b
a/2
N
C
.b
3
a/3
C
3
D a2 .b a/ C a.b a/2 C .b 3 a/ D 13 b3 13 a3 : In Exercises Exercises 61–64, 61–64, describe describe the area area repre represented sented by the limits. N
1 61. lim N !1 N
j 1
SOLUTION
The limit
X D
j N
4
lim RN
N
!1
represents the area between the graph of f of f .x/ 3 62. lim N !1 N
N
X C 2
j 1
D
3j N
4
D
1 lim N N !1
N
X
j 1
D
j N
4
D x 4 and the x the x-axis -axis over the interval Œ0 interval Œ0;; 1.
.b
a/
2N
3
C
.b
a/
6N 2
3
!
561
562
THE INTEGRAL
CHAPTER 5
The limit
SOLUTION
lim RN
D
N
!1
5 N !1 N lim
N 1
X
e
j 1
D
4
j 0
D
D
!1
represents the area between the graph of y of y
N
X
!1 2N j D1
SOLUTION
C
N
lim
j
2 5j=N
lim LN
N
2
3 N
The limit
SOLUTION
64.
N
X C
D x 4 and the x the x-axis -axis over the interval Œ interval Œ2; 2; 5.
represents the area between the graph of f of f .x/ .x/ 63.
3 lim N !1 N
sin
3
4N
C
j 2N
N 1
5 lim N !1 N
X
e
2 5j=N
C
j 0
D
D ex and the x the x-axis -axis over the interval Œ interval Œ2; 3.
The limit
lim
N
X
!1 2N j D1
N
represents the area between the graph of y of y
sin
3
C
4N
j 2N
D sin x and the x the x-axis -axis over the interval Πinterval Π3 ; 56 .
In Exercises Exercises 65–70, 65–70, express express the area under under the graph as as a limit using using the appr approximatio oximation n indicated indicated (in summation summation notation), notation), but but do not evaluate. 65. RN , f.x/ SOLUTION
D sin x over Œ0; over Œ0; D =N . Let f Let f .x/ .x/ D sin x over Œ0; over Œ0; and set a set a D 0, b D , and x D .b a/=N D =N . Then N
D x
X
RN
f .xk /
k 1
D
N
X
D N
sin
k 1
D
k N
:
Hence lim RN
D N lim !1 N
N
!1
is the area between the graph of f of f .x/ .x/ 66. RN , f.x/ SOLUTION
Dx
1
.x/ Let f Let f .x/
N
X
sin
k 1
D
k N
D sin x and the x the x-axis -axis over Œ0; over Œ0; .
over Œ1 over Œ1;; 7
Dx
1
over the interval Œ1 interval Œ1;; 7. Then x
1 D 7 N D N 6 and a and a D 1. Hence,
N
RN
D x
X
f .1
j 1
D
C jx/ D
6 N
N
X C 1
j 1
D
6 j N
and lim RN
N
!1
is the area between the graph of f of f .x/ .x/ 67. LN
Dx p , f.x/ D 2x C 1 over Œ7 over Œ7;; 11
1
D
6 lim N !1 N
N
X C 1
j 1
D
and the x the x-axis -axis over Œ1 over Œ1;; 7.
6 j N
1
1
SECTION
SOLUTION
Approximat Approximating ing and Computing Computing Area
5.1
563
11 7 4 D p 2x C 1 over the interval Œ7 D interval Œ7;; 11. Then x D and a and a D 7. Hence, N N
Let f Let f .x/
N 1
X
LN
D x
f .7
j 0
D
N 1
X r
4 N
C jx/ D
2.7
j 0
D
C j N 4 / C 1
and lim LN
D
N
!1
N 1
X r C
4 lim N !1 N
15
j 0
D
p 2x C 1 and the x D the x-axis -axis over Œ7 over Œ7;; 11.
8j N
is the area between the graph of f of f .x/ .x/
SOLUTION
D cos x over 8 ; 4 Let f Let f .x/ D cos x over the interval 8 ; 4 . Then x D
68. LN , f.x/
N 1
X
LN
D x
f
j 0
D
C jx 8
N N 1
D
D N 8 D 8N and a and a D 8 , Hence:
4 8
8N
X
cos
j 0
D
C j 8 8N
and D N lim 8N !1
N 1
X
lim LN
N
!1
is the area between the graph of f of f .x/ .x/
cos
j 0
D
C j 8N 8
D cos x and the x the x-axis -axis over Πover Π8 ; 4 .
D tan x over 12 ; 1 1 1 Let f Let f .x/ D tan x over the interval Πinterval Π12 ; 1. Then x D N D 2N and a and a D 12 . Hence
69. M N N , f.x/ SOLUTION
N
D x
M N N
X
f
j 1
D
1 2
j
C
1 2
x
D
1 2
1 2N
N
X
tan
j 1
D
1 2
1 j 2N
C
1 2
and so lim M N
D
N
!1
is the area between the graph of f of f .x/ .x/ 70. M N N , f.x/ SOLUTION
Dx
2
Let f Let f .x/
1 lim N !1 2N
N
X
tan
j 1
D
1 2
1 j 2N
C
1 2
D tan x and the x the x-axis -axis over Πover Π12 ; 1.
over Œ3 over Œ3;; 5
Dx
2
D 5N 3 D N 2 and a and a D 3. Hence
over the interval Œ3 interval Œ3;; 5. Then x N
M N N
D x
X
f 3
j 1
D
j
C
1 2
x
D
2 N
N
X C
2 j N
2 j N
3
j 1
D
1 2
2
and so lim M N N
D
N
!1
is the area between the graph of f of f .x/ .x/ 1 71. Eva Evalua luate te lim N !1 N SOLUTION
X s N
1
j 1
D
Dx
j N
2
2 lim N !1 N
N
X C 3
j 1
D
2
1 2
and the x the x-axis -axis over Œ3 over Œ3;; 5.
2
by interpreting it as the area of part of a familiar geometric figure.
The limit lim RN
N
!1
D
1 lim N !1 N
p
X s N
1
j 1
D
j N
2
represents the area between the graph of y of y f .x/ .x/ 1 x 2 and the x the x -axis over the interval Œ0; Œ0; 1. This is the portion of the 2 1 2 2 circular disk x disk x y 1 that lies in the first quadrant. Accordingly, its area is 4 .1/ 4 .
C
D
D
D
564
THE INTEGRAL
CHAPTER 5
In Exercises Exercises 72–74, 72–74, let f f .x/ .x/
D x2 and let R RN , L N , and M M N Œ0; 1. N be the approximations for the interval Œ0;
1 1 D 13 C 2N C 6N 1 2 . Interpret the quantity 2N C 6N 1 2 as the area of a region. 1 Let f Let f .x/ .x/ D x 2 on Œ0 on Œ0;; 1. Let N Let N > 0 be 0 be an integer and set a D 0, b D 1 and x D 1N 0 D N . Then
Show that R that RN
72. SOLUTION
N
RN
D x
X
f .0
j 1
D
N
1 N
C jx/ D
X
j 1
D
1 j N 2 2
D
in
RN
1 N 3
N 3 3
C
N 2 2
C
N 6
!
1 D 13 C 2N C 6N 1 2 :
The quantity 1 2N
C N 62
1 D 13 C 2N C 6N 1 2
represents the collective area of the parts of the rectangles that lie above the graph of f f .x/: .x/: It is the error between R between RN and the true 1 area A area A 3.
D
y
1 0.8 0.6 0.4 0.2 x
0.2
0.4
0.6
0.8
1
73. Show that
1 D 13 2N C 6N 1 2 ;
1 D 13 12N 2
LN
M N
Then rank the three approximations R approximations R N , LN , and M and M N in order of increasing accuracy (use Exercise Exercise 72). Let f Let f .x/ x 2 on Œ0 on Œ0;; 1. Let N Let N be be a positive integer and set a 1 kx k=N , k=N , k 0 ; 1 ; : : : ; N and and let x let x k a .k .k 2 / x
D
SOLUTION
xk
D a C
D
D
D C C
N 1
LN
D x D N 13
X D
f .xk /
k 0
D N 1
.N 1/3 3
D x D N 13 D
1 N 3
N 1
X
k 0
D
k N
2
D N 13
N 1
X
k2
k 1
D
!
2
1 C .N 2 1/ C N 6 1 D 13 2N C 6N 1 2
N 1
M N
D 0, b D 1, and x D .b a/=N D 1=N . 1=N . Let 1 D C 2 /=N , /=N , k D 0 ; 1 ; : : : ; N 1. Then
N 1
1 2
2
N 1
X D X C ! D X 0D0 X 1 0 XD 1 0 X 11D @@ A C @ A C @ AA D D D ! 1 N
f .xk /
k 0
N 1
k
N
k 0
N 1
k2
1 4
k
k 1
k 1
.N 1/3 3
C
1 N 3
.N 1/2 2
C
k2
k 0
C k C 14
N 1
1
k 0
N 1 6
C
.N 1/2 2
C
N 1 2
! !
1 C 14 N D 13 12N 2
1 1 1 1 1 , the error of L of L is given by and the error of M M is i s given by . N N N 2N 2N 6N 2 6N 2 12N 2 Of the three approximations, R approximations, R N is the least accurate, then L then LN and finally M finally M N is i s the most accurate. accurate. The error of R RN is given by
C
C
R N , LN , and M and M N , find the smallest integer N integer N for which the error is less than 0.001. 74. For each of R SOLUTION
For R For RN , the error is less than 0:001 than 0:001 when: when: 1 2N
C 6N 1 2 < 0:001:
We find an adequate solution in N in N :: 1 2N
C 6N 1 2 < 0:001
SECTION
Approximat Approximating ing and Computing Computing Area
5.1
565
C C 1 < 0:006.N 0:006.N 2 /
3N
0 < 0:006N 2
p 3C 9:024
3N 1;
N > 0:012 500:333. 500:333. Hence R A . in particular, if N Hence R 501 is within 0.001 of A. For L For LN , the error is l ess than 0:001 than 0:001 if if
D
ˇˇˇ
1 2N
We solve this equation for N for N ::
ˇˇˇ ˇˇ ˇ 1 2N
C
1 6N 2
ˇˇˇ ˇˇ ˇ
ˇˇˇ
1 < 0:001: 6N 2
< 0:001
3N 1 < 0:001 6N 2 3N 1 < 0:006N 2 0 < 0:006N 2
p 3C 9
0:024 0:012 1 For M For M N N , the error is given by 12N 2 ,
N > which is satisfied if N
3N
C C 1;
D 499:666. 499:666. Therefore, L A . Therefore, L500 is within 0.001 units of A.
so the error is less than 0.001 if 1 < 0:001 12N 2 1000 < 12N 2 9:13 < N
Therefore, M Therefore, M 10 10 is withi n 0.001 units of the correct answer. In Exercises Exercises 75–80, 75–80, use the the Graphical Graphical Insight Insight on page page 291 to to obtain obtain bounds bounds on the area area..
D p x over Œ0 over Œ0;; 1. Prove that 0:51 that 0:51 A 0:77 by 0:77 by computing R computing R4 and L and L4 . Explain your reasoning. 1 0 1 1 1 3 4 For n For n D 4, x D 4 D 4 and fxi gi D0 D f0 C i x g D f0; 4 ; 2 ; 4 ; 1g. Therefore, p p 4 1 1 2 3 C C C 1 0:768 R4 D x f .xi / D 4 2 2 2 i D1 p p 3 1 1 2 3 C L4 D x f .xi / D 0 C C 0:518: 4 2 2 2
75. Let A Let A be the area under f .x/ .x/ SOLUTION
X X
i
D0
! !
In the plot below, you can see the rectangles whose area is represented by L 4 under the graph and the top of those whose area is A under the curve is somewhere between L represented by R by R 4 above the graph. The area A under between L4 and R and R4 , so 0:518 A 0:768:
L4 , R4 and the graph of f of f .x/ .x/. y 76. Use R Use R5 and L and L5 to show that the area A area A under under y
2
0:0244. over Œ10 over Œ10;; 13 satisfies 0:0218 satisfies 0:0218 A 0:0244.
Let f Let f .x/ .x/ x 2 over the interval Œ10; Œ10; 13. Because f is f is a decreasing function over this interval, it follows that A LN for all N all N .. Taking N Taking N 5, we have x 3=5 and 3=5 and
SOLUTION
RN
D x
D
D D
D
3 R5 D 5
3 5
1 10:62
1 1 1 C 11:2 C 11:8 C 12:4 C 1312 D 0:021885: 2 2 2
Moreover, L5
D
1 102
C
1 10:62
C
1 11:22
C
1 11:82
C
1 12:42
D 0:0243344:
566
THE INTEGRAL
CHAPTER 5
Thus, 0:0218 < R5
A L5 < 0:0244:
77. Use R Use R4 and L and L4 to show that the area A area A under under the graph of y y
D sin x. f .x/ .x/ is increasing over the interval Œ0; interval Œ0; =2, so the Insight on page 291 applies, which indicates =2 0 D 8 and fxi gi4D0 D f0 C i x gi4D0 D f0; 8 ; 4 ; 38 ; 2 g. From this, that L that L4 A R4 . For n For n D 4, x D 4 SOLUTION
Let f Let f .x/ .x/
D sin x over 0; 2 satisfies 0:79 satisfies 0:79 A 1:19. 1:19.
L4
D
8
3
X
i
f .xi / 0:79;
R4
D0
D
8
4
X
i
f .xi / 1:18:
D1
Hence A Hence A is is between 0:79 between 0:79 and and 1:19 1:19..
Left and Right endpoint approximations to A to A..
D x 1 over Œ1 over Œ1;; 8 satisfies 1 1 1 1 1 1 1 1 1 1 1 1 1 2 C 3 C 4 C 5 C 6 C 7 C 8 A 1 C 2 C 3 C 4 C 5 C 6 C 7
78. Show that the area A area A under under f f .x/ .x/
Let f Let f .x/ .x/ x 1 , 1 x 8. 8. Since f Since f is is decreasing, the left endpoint approximation L approximation L 7 overestimates the true area between the graph of f and f and the x the x-axis, -axis, whereas the right endpoint approximation R approximation R 7 underestimates it. Accordingly, Accordingly,
D
SOLUTION
1 2
C 13 C 14 C 15 C 16 C 17 C 18 D R7 < A < L 7 D 1 C 12 C 13 C 14 C 15 C 16 C 17 1
Left endpoint approximation, n = 7
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
1
2
3
4
5
6
7
8
0
Right endpoint approximation, n = 7
1
2
3
4
5
6
7
8
79. Show that the area A area A under under y y x 1=4 over Œ0;1 over Œ0;1 satisfies L satisfies L N A R N for all N all N .. Use a computer algebra system to calculate L calculate L N and R N for f or N N 100 and 100 and 200 200,, and determine A determine A to to two decimal places. SOLUTION
D D D On Œ0 On Œ0;; 1, f .x/ .x/ D x 1=4 is an increasing function; therefore, L therefore, LN A RN for all N all N .. We find L100 D 0:793988 and R100 D 0:80399;
while L200 Thus, A Thus, A
D 0:797074
and
R200
D 0:802075:
D 0:80 to 0:80 to two decimal places.
80. Show that the area A area A under under y y 4=.x 2 1/ over 1/ over Œ0 Œ0;; 1 satisfies R satisfies R N A L N for all N all N .. Determine A Determine A to to at least three decimal places using a computer algebra system. Can you guess the exact value of A? A?
D
C
On Œ0 On Œ0;; 1, the function f function f .x/ .x/ 4=.x 2 1/ is 1/ is decreasing, so R so RN A L N for all N all N .. From the values in the table 3:142 to three decimal places. It appears that the exact value of A below, we find A find A 3:142 to of A is is . SOLUTION
D
D
C
N 10 100 100 1000 1000 10000 0 1000 100000 00
RN 3.03993 3.13158 3.1 3.1405 4059 3.14 3.1414 149 9 3.14 3.1415 158 8
LN 3.23992 3.15158 3.1 3.14259 4259 3.14 3.1416 169 9 3.14 3.1416 160 0
SECTION
81. In this exercise, we evaluate evaluate the area A under A under the graph of y of y sum (valid for r for r 1):
¤
1
D N 1
that LN (a) Show that L
(b) Apply Eq. (8) with r with r (c) Compute A Compute A
C r C r C C r
567
D ex over Œ0 over Œ0;; 1 [Figure 5(A)] using the formula for a geometric N 1
2
Approximat Approximating ing and Computing Computing Area
5.1
N 1
r j
X
D
j 0
D
N
D rr 11
8
N 1
X
ej=N .
j 0
D
e1 D e1=N to prove L prove L N D . N.e 1=N 1/
D N lim L using L’Hˆopital’s opital’s Rule. !1 N y
y y = e x y = ln x
3
e
1
2 1
A
B x
x
1
1
(A)
e
(B)
FIGURE 5 SOLUTION
(a) Let f Let f .x/ .x/
D ex on Œ0 D 1=N and on Œ0;; 1. With n With n D N , N , x D .1 0/=N D and D a C jx D N j
xj for j for j
D 0 ; 1 ; 2 ; : : : ; N . Therefore, N 1
LN
D x
(b) Applying Eq. (8) with r with r
X
f .xj /
j 0
D
N 1
D N 1
X
ej=N :
j 0
D
D e1=N , we have 1=N N
1 D N 1 .ee1=N / 1 1 D N.ee1=N : 1/
LN Therefore, A
1 D N lim LN D .e 1/ lim : 1=N N !1 N.e !1 1/
(c) Using L’Hˆopital’s opital’s Rule, 1
2
N N D .e 1/ N lim D D .e 1/ N lim .e 1/ lim e 1=N D e 1: 1=N !1 e !1 N !1 N 2 e 1=N 1 82. Use the result of Exercise 81 to show that the area B under B under the graph of f f .x/ .x/ D ln x over Œ1 over Œ1;; e is equal to 1. Hint: Relate B Relate B
A
in Figure 5(B) to the area A area A computed computed in Exercise 81.
Because y Because y ln x and y e x are inverse functions, we note that if the area B is reflected across the line y x and then combined with the area A area A,, we create a rectangle of width 1 and height e. e . The area of this rectangle is therefore e, e , and it follows that the area B area B is is equal to e to e minus minus the area A area A.. Using the result of Exercise 81, the area B area B is is equal to SOLUTION
D
D
D
e .e 1/
D 1:
568
CHAPTER 5
THE INTEGRAL
Further Insights and Challenges 83. Although the accuracy of R R N generally improves as N as N increases, this need not be true for small values of N of N .. Draw the graph of a positive continuous function f function f .x/ .x/ on an interval such that R 1 is closer than R than R2 to the exact area under the graph. Can such a function be monotonic? 1 4.
Let ı Let ı be be a small positive number less than f.x/ on f.x/ on Œ0 Œ0;; 1 by SOLUTION
f.x/
8ˆ ˆ< D ˆˆ:
(In t he figures below, ı below, ı
1
if 0 0
1 x 2ı ı x 1 ı 2ı
if 12
1
if 12
D 101 . But imagine ı imagine ı being being very tiny.) Define
1 2 ı 1 ı x < 2 1 x < ı 2
if 12
x<
C
C ı x 1
Then f Then f is is continuous on Œ0 on Œ0;; 1. (Again, just look at the figures.)
1 9 D 1 12 bh D 1 12 .2ı/.1/ D 1 ı. (For ı (For ı D 10 , we have A have A D 10 .) 1 1 With R With R1 D 1, the absolute error is jE1 j D jR1 Aj D j1 .1 ı/ j D ı . (For ı (For ı D 10 , this absolute error is jE1 j D 10 .) 1 1 1 1 1 With R With R2 D 2 , the absolute error is jE2 j D jR2 Aj D 2 .1 ı/ D ı 2 D 2 ı . (For ı (For ı D 10 , we have jE2 j D 25 .)
The exact area between between f and f and the x the x-axis -axis is A is A
ˇˇ
ˇˇ ˇˇ ˇˇ
Accordingly, R Accordingly, R 1 is closer to the exact area A area A than than is R is R2 . Indeed, the tinier ı is, ı is, the more dramatic the effect.
For a monotonic function, this phenomenon cannot occur. Successive Successive approximations from either side get progressively more accurate. x) Graph of f ( x
1
1
Right endpt approx, n = 1
1
Right endpt approx, n = 2
0.8 0.6
0.5
0.4
0.5
0.2 0
0.2 0.2 0.4 0.6 0.6 0.8 1
0
0.5
0
1
0.5
1
x
84. Draw the graph of a positive continuous function on an interval such that R that R2 and L and L2 are both smaller than the exact area under the graph. Can such a function be monotonic? SOLUTION In the plot below, .x/ is 3, whereas L 2 R2 2. Thus L below, the area under the saw-tooth saw-tooth function f function f .x/ Thus L 2 and R 2 are both smaller than the exact area. Such a function cannot be monotonic; if f if f .x/ .x/ is increasing, then L then L N underestimates and R N overestimates the area for all N , N , and, if f f .x/ .x/ is decreasing, then L then LN overestimates and R and RN underestimates the area for all N all N ..
D
D
Left/right-endpoint approximation, n = 2 2
1
1
85.
2
Explain graphically: The endpoint approximations are less accurate when f 0 .x/ is large .
When f When f 0 is large, the graph of f of f is steeper and hence there is more gap between f and L and L N or R or R N . Recall that the top line segments of the rectangles involved in an endpoint approximation constitute a piecewise constant function. If f If f 0 is large, then f then f is increasing more rapidly and hence is less like a constant function. SOLUTION
y
y
3
3
Smaller f'
2
2
1
1
0
x
0
1
2
3
4
0
Larger f'
x
0
1
2
3
4
SECTION
Approximat Approximating ing and Computing Computing Area
5.1
569
86. Prove that for any any function f function f .x/ .x/ on Œa on Œa;; b , a D b N .f.b/ f .a// .a//
RN LN
9
For any f any f (continuous (continuous or not) on I on I Œa; Œa; b , partition I partition I into N into N equal equal subintervals. Let x .b a/=N and and set xk a kx; k 0 ; 1 ; : : : N . Then we have the following approximations to the area between the graph of f of f and and the x the x-axis: -axis: the left endpoint approximation L approximation LN x kN D01 f .xk / and right endpoint approximation R approximation R N x kN D1 f .xk /. Accordingly,
D
SOLUTION
D C
D
D
D
P D P 0 X 10 X 1 A@ A D @ 0 D 0 X D1 0 X 11 A @ AA D @ C@ N
RN LN
x
N 1
f .xk /
x
k 1
f .xk /
k 0
N 1
x
f .xN /
N 1
f .xk /
f .x0 /
k 1
f .xk /
k 1
D
D
a D x .f.xN / f .x0// D b N .f.b/ f.a//:
a D b N .f.b/ f .a// .a//..
In other words, R words, R N LN
In this exercise exercise,, we prove prove that lim RN and and lim lim LN exist and are equal if f if f .x/ .x/ is increasing [the case of f of f .x/
87.
N
N
!1
!1
decreasing is similar]. We use the concept of a least upper bound discussed in Appendix B. 1 . (a) Explain with a graph why L why L N R M for all N; all N; M 1. (b) By (a), the sequence LN is bounded, so it has a least upper bound L. By definition, definition, L L is the smallest number such that LN L for all N all N .. Show that L that L RM for all M all M .. (c) According to (b), L (b), LN L RN for all N all N .. Use Use Eq. Eq. (9) to show show that that lim LN L and and lim lim RN L.
f g
N
!1
D
N
!1
D
SOLUTION
Let f .x/ .x/ be positive and increasing, and let N let N and and M be M be positive integers. From the figure below at the left, we see that L N (a) Let f underestimates the area under the graph of y y f.x/, f.x/, while from the figure below at the right, we see that R M overestimates the area under the graph. Thus, for all N; M 1, LN RM .
D
y
y
x
x
(b) Because the sequence sequence LN is bounded above by R by RM for any M any M ,, each R each RM is an upper bound for the sequence. Furthermore, the sequence LN must have a least upper bound, call it L it L.. By definition, the least upper bound must be no greater than any other upper bound; consequently, consequently, L RM for all M all M .. (c) Since L Since L N L RN , R N L RN LN , so RN L RN LN . From this,
f g
f g
j
lim
N
!1
j j
j
jRN Lj N lim jR L j : !1 N N
By Eq. (9), 1 j j.b a/.f.b/ f .a// RN LN j D lim .a//j D 0; N !1 N !1 N so lim lim jRN Lj jRN LN j D 0, hence hence lim RN D L. N !1 N !1 Similarly, jLN Lj D L LN RN LN , so lim
a/ jLN Lj jRN LN j D .b N .f.b/ f .a//: .a//:
This gives us that 1 j j.b a/.f.b/ f .a// LN Lj lim .a//j D 0; N !1 N !1 N lim
so lim lim LN N
!1
D L.
This This prove provess lim LN N
!1
D N lim R D L: !1 N
570
THE INTEGRAL
CHAPTER 5
Use Eq. (9) to show that if f f .x/ .x/ is positive and monotonic, then the area A under A under its graph over Œa over Œa;; b satisfies
88.
a jRN Aj b N jf.b/ f.a/j
10
SOLUTION Let f Let f .x/ .x/ be continuous, positive, and monotonic on Œa on Œa;; b . Let A Let A be be the area between the graph of f of f and and the x the x-axis -axis over Œa over Œa;; b . For specificity, say f is f is increasing. (The case for f f decreasing on Œa; Œa; b is similar.) As noted in the text, we have LN A RN . By Exercise 86 and the fact that A that A lies lies between L between L N and RN , we therefore have a D b N .f.b/ f.a//:
0 RN A R N LN Hence
ba a .f.b/ f.a// D jRN Aj b N jf.b/ f.a/j ; N where f where f .b/ .b/ f.a/ D jf.b/ f.a/j because f because f is is increasing on Œa on Œa;; b . In Exercises Exercises 89 and and 90, use use Eq. (10) (10) to find a value of of N such such that jRN Aj < 10 4 for the given given function function and and interval. interval. p 89. f.x/ D x , Œ1;4 p on Œ1;; 4. Then b SOLUTION Let f Let f .x/ .x/ D x on Œ1 Then b D 4, a D 1, and
We need
3 N
90. f.x/
D
SOLUTION
We need
9 N
< 10
1 3 3 jRN Aj 4 N .f .4/ .4/ f .1// .1// D .2 1/ D : N N p on Œ1;; 4. 4 , which gives N 4 for f gives N > 30;000. 30;000. Thus jR for f .x/ D x on Œ1 Aj < 10
p
30;001
9 x 2 , Œ0; Œ0; 3
Let f Let f .x/ .x/
< 10
D
p
9 x 2 on Œ0 on Œ0;; 3. Then b Then b
D 3; a D 0, and
a jRN Aj b N jf.b/ f.a/j D N 3 .3/ D N 9 : p 4 , which gives N gives N > 90;000. 90;000. Thus jR90;001 Aj < 10 4 for f for f .x/ D 9 x 2 on Œ0 on Œ0;; 3.
91. Prove that if f f .x/ .x/ is positive and monotonic, then M then M N between RN and L and LN and is closer to the actual area under N lies between R the graph than both R both R N and L and L N . Hint: In the case that f .x/ is x/ is increasing, Figure 6 shows that the part of the error in R N due to the i the i th rectangle is the sum of the areas A areas A B D, and for M for M N it is B E . On the other hand, A E .
C C
j
A
D E
B
F
C
x i
!
1
j
midpoint x i
x
FIGURE 6
SOLUTION
a D b N ,
Suppose f Suppose f .x/ .x/ is monotonic increasing on the interval Œa interval Œa;; b , x
fxk gkN D0 D fa; a C x; a C 2x ; : : : ; a C .N 1/x; bg and N 1 xk k 0
˚
D
D
a
C .a C x/ ; .a C x/ C .a C 2x/ ; : : : ; .a C .N 1/x/ C b : 2
2
2
Note that x that x i < x i < x i C1 implies f implies f .xi / < f.xi / < f.x i C1/ for all 0 all 0
0 @
LN
D
b a N
X 1A 0@
N 1
f .xk /
k 0
D
<
M N N
D
b a N
i < N because N because f f .x/ .x/ is monotone increasing. Then
X 1A 0@
N 1
f .xk /
k 0
D
<
RN
D
b a N
X 1A N
f .xk /
k 1
D
SECTION
5.2
The Defini Definite te Integr Integral al
571
Similarly, if f f .x/ is x/ is monotone decreasing,
0 @
ba N
LN
D
X 1A 0@
N 1
f .xk /
>
M N N
k 0
D
D
b a N
X 1A 0@
N 1
f .xk /
>
RN
D
k 0
D
ba N
X 1A N
f .xk /
k 1
D
Thus, if f f .x/ .x/ is monotonic, then M then M N always lies in between R between R N and LN . Now, as in Figure 6, consider the typical subinterval Œx subinterval Œx i 1 ; xi and its midpoint x midpoint x i . We let A;B;C;D; A;B;C;D; E , and F and F be be the areas as shown in Figure 6. Note that, by the fact that x that xi is the midpoint of the interval, A interval, A D E and F and F B C . C . Let E Let ER represent C F E ) the right endpoint approximation error ( A B D), let E let EL represent the left endpoint approximation error ( and let E let EM represent r epresent the midpoint approximation approximation error ( B E ).
D C C
D C
Dj
D D C
j
D C C C C
If B B > E , then E then EM
D B E . In this case, ER EM D A C B C D .B E/ D A C D C E > 0;
so E so ER > EM , while EL EM
D C C C F C C E .B E/ D C C C .B C C / C E .B E/ D 2C C C 2E > 0;
so EL > EM . Therefore Therefore,, the midpoint midpoint approxima approximation tion is more accurate accurate than than either either the left or the right endpoint endpoint approxim approximation. ation.
If B B < E , then E then EM
D E B . In this case, ER EM D A C B C D .E B/ D D C E C D .E B/ D 2D C B > 0;
so that E that ER > EM while EL EM
D C C C F C C E .E B/ D C C C F C C B > 0;
so EL > EM . Therefore Therefore,, the midpoint midpoint approxima approximation tion is more accurate accurate than than either either the right or the left endpoint endpoint approxim approximation. ation.
D E , the midpoint approximation is exactly equal to the area. Hence, for B < E , B > E , or B D E , the midpoint approximation is more accurate than either the left endpoint or the right
If B B
endpoint approximation.
5.2 The Definite Integral Preliminary Questions 5
Z Z Z D D
dx [the dx [the function is f is f .x/
1. What is
3
5
SOLUTION
3
7
2. Let I Let I
D 1]?
5
dx
Z D
1 dx
3
D 1.5 3/ D 2.
f.x/dx, f.x/dx , where f where f .x/ .x/ is continuous. State whether tr ue or false:
2
(a) I is is the t he area between the graph and the x the x-axis -axis over Œ2 over Œ2;; 7.
f .x/ .x/ 0, then I then I is i s the area between the graph and the x the x-axis -axis over Œ over Œ2; 2; 7. (b) If f (c) If f f .x/ .x/ 0, then I is is the area between the graph of f f .x/ and x/ and the x the x-axis -axis over Œ2 over Œ2;; 7. SOLUTION
b
R
(a) False. a f .x/ dx is dx is the signed area area between the graph and the x-axis. (b) True. (c) True.
3. Explain graphically:
Z
cos x dx
0
D 0.
x / cos x , the “negative” area between the graph of y Because cos. cos. x/ of y exactly cancels the “positive” area between the graph and the x-axis over Œ0; over Œ0; . 2
D
SOLUTION
5
negative, 4. Which is negative,
Z
1
8 dx or dx or 1
Z
8 dx ? 5
5
SOLUTION
Because 5 .1/
D 4,
Z
8 dx is dx is negative. 1
D cos x and the x-axis over Π2 ;
572
THE INTEGRAL
CHAPTER 5
Exercises In Exercises Exercises 1–10, 1–10, draw a graph of of the signed signed area area represen represented ted by the integral integral and and compute it using geometry. geometry. 3
1.
Z
2x dx 3
y 2x and the x The region bounded by the graph of y the x -axis over the i nterval Πnterval Π3; 3 consists of two right triangles. 1 One has area 9 below the axis, and the other has area 2 .3/.6/ 9 above the axis. Hence,
D
SOLUTION
1 2 .3/.6/
D
D
3
Z
2x dx
D 9 9 D 0:
3
y
6 4 2 x
3
2
!
1
!
!
1
2 !4 !6 !
2
3
3
2.
Z
.2x 2
C 4/dx
The region bounded by the graph of y y 1 triangle of area 2 .5/.10/ 25 above 25 above the axis. Hence,
D 2x C 4 and the x the x -axis over the interval Πinterval Π2; 3 consists of a single right
SOLUTION
D
3
Z
.2x
C 4/dx D 25:
2
y
10 8 6 4 2 x
2
!
1
1
!
2
3
1
3.
Z
.3x 2
C 4/dx
The region region bounded bounded by the graph graph of of y y 3x 4 and the x -axis over over the interval interval Œ Œ2; 1 consists of two right triangles. 2 1 7 49 One has area 3 below the axis, and the other has area 2 . 3 /.7/ 6 above the axis. Hence,
D C
SOLUTION
1 2 2 . 3 /.2/
D
1
Z
.3x 2
D
C 4/dx D 496 23 D 152 : y
8 6 4 2 x
2
!
1
!
2
1
!
1
4.
Z
4 dx 2
The region region bounded by the graph of y y above the axis. Hence, SOLUTION
D 4 and the x the x-axis -axis over the int erval Œ erval Œ2; 1 is a rectangle of area .3/.4/ area .3/.4/ D 12 1
Z
4 dx 2
D 12:
SECTION
5.2
The Defini Definite te Integr Integral al
573
y
4 3 2 1 x
2
1
!
1
!
8
5.
Z
.7 x/dx
6
The region region bounded by the graph of of y y 7 x and the x the x-axis -axis over the i nterval Œ6 nterval Œ6;; 8 consists of two right triangles. 1 1 1 1 One triangle has area 2 .1/.1/ 2 above the axis, and the other has area 2 .1/.1/ 2 below the axis. Hence,
D
SOLUTION
D
D
8
Z
.7 x/dx
6
D 12 12 D 0:
y
1 0.5 x !
2
0.5
4
6
8
1
!
3=2
6.
Z
sin x dx
=2
3 The region region bounded by the graph of y y sin x and the x the x-axis -axis over the interval Πinterval Π2 ; 2 consists of two parts of equal area, one above the axis and the other below the axis. Hence,
D
SOLUTION
3=2
Z
sin x dx
=2
D 0:
y
1 0.5 x
1
0.5
2
3
4
1
!
5
7.
Z p
25 x 2 dx
0
The region region bounded by the graph of y y radius 5. Hence, SOLUTION
D
p
25 x 2 and the x the x-axis -axis over the int erval Œ0 erval Œ0;; 5 is one-quarter of a circle of
5
Z p
25 x 2 dx
0
D 14 .5/2 D 254 :
y
5 4 3 2 1 x
1
2
3
4
5
3
8.
Z j j
x dx
2
y x and the x The region region bounded by the graph of y the x -axis over the interval Πinterval Π2; 3 consists of two right triangles, 1 1 9 both above the axis. One triangle has area 2 .2/.2/ 2, and the other has area 2 .3/.3/ 2 . Hence, SOLUTION
D j j D 3
Z j j
x dx
2
D
D 92 C 2 D 132 :
574
THE INTEGRAL
CHAPTER 5
y
3 2 1 x
2
1
!
1
!
2
3
2
9.
Z
.2 x / dx
jj
2
SOLUTION The region region bounded by the graph of y y with base 4 and height 2. Consequently, Consequently,
D 2 jx j and the x the x-axis -axis over the interval Πinterval Π2; 2 is a triangle above the axis
2
Z
.2 x / dx
jj
2
D 12 .2/.4/ D 4:
y 2 1
x
2
!
!
1
1
2
5
10.
Z
.3 2
C x 2jx j/ dx
The region region bounded by the graph of y y 3 x 2 x and the x the x-axis -axis over the i nterval Πnterval Π2; 5 consists of a triangle below the axis with base 1 and height 3, a triangle above the axis of base 4 and height 3 and a tri angle below the axis of base 2 and height 2. Consequently, Consequently,
D C
SOLUTION
5
Z
.3 2
jj
C x 2jx j/ dx D 12 .1/.3/ C 12 .4/.3/ 12 .2/.2/ D 52 : y
3 2 1
−2
4
x
2
−1 −2 −3
10
11. Calculate
Z
.8 x/dx in x/dx in two ways:
0
(a) As the limit limit lim RN N
!1
(b) By sketching the relevant relevant signed area and using geometry 10 10 10 10 D 8 x over Œ0;10 over Œ0;10. Consider the integral 0 f.x/dx D 0 .8 x/dx. x/dx . (a) Let N Let N be be a positive integer and set a D 0, b D 10, 10, x D .b a/=N D 10=N . 10=N . Also, let x k D a C kx D 10k=N , 10k=N , k D 1 ; 2 ; : : : ; N be the right endpoints of the N the N subintervals subintervals of Œ0;10 Œ0;10. Then SOLUTION
R
Let f Let f .x/ .x/
N
RN
D x D 10 N
X D
f .xk /
k 1
8N
10 N
N
10 N
X D D !!
8
k 1
N 2 2
C N 2
10k N
D
R
10 N
D 30 50 : N
0 0 X 1 0 X 11 @ @ A @ AA N
8
1
k 1
D
10 N
N
k
k 1
D
50 Henc Hencee lim RN lim 30 30. 30. N N !1 N !1 8 x and the x (b) The region bounded by the graph of y of y the x -axis over the interval Œ0;10 consists of two right triangles. One 1 triangle has area 2 .8/.8/ 32 above 32 above the axis, and the other has area 12 .2/.2/ 2 below the axis. Hence,
D
D
D
D
D
10
Z 0
.8 x/dx
D 32 2 D 30:
SECTION
5.2
The Defini Definite te Integr Integral al
575
y
8 6 4 2
10 2
4
6
x
8
4
12. Calculate
Z
1
.4x 8/dx in 8/dx in two two ways: ways: As the the limit limit lim RN and using geometry. N
!1
4
4
Z
Z
D 4x 8 over Œ over Œ1; 4. Consider the integral f.x/dx D .4x 8/dx. 8/dx . 1 1 Let N Let N be be a positive integer and set a D 1, b D 4, x D .b a/=N D 5=N . 5=N . Then x Then x k D a C kx D 1 C 5k=N , 5k=N , k D 1 ; 2 ; : : : ; N are are the right endpoints of the N the N subintervals subintervals of Œ Œ1; 4. Then
SOLUTION
Let f Let f .x/
N
RN
D x
X
f .xk /
D
k 1
D
100 D 60 .N / C 2 N N
5 N
N
X
4
k 1
D
N 2 2
C N 2
C
20k N
8
D
60 N
!
0X 1 0X 1 @ AC @ A N
1
k 1
D
100 N 2
N
k
k 1
D
50 D 60 C 50 C 50 D : 10 C N N 50 10 10. Henc Hencee lim RN lim 10. N N !1 N !1 The region bounded by the graph of y y 4x 8 and the x the x-axis -axis over the interval Œ interval Œ1; 4 consists of a triangle below the axis with base 3 and height 12 and a triangle above the axis wit h base 2 and height 8. Hence,
D
C
D
D 4
Z
.4x
8/dx
1
D 12 .3/.12/ C 12 .2/.8/ D 10: y
−1
5 x
1
−5
2
3
4
−10
In Exercises Exercises 13 and and 14, refer refer to Figur Figuree 1. y y = f ( x x)
x
2
4
6
FIGURE 1 The two parts of the graph are semicircles. 2
13. Evaluate: Evaluate: (a)
Z
6
f.x/dx
0
SOLUTION
(b)
Z
f.x/dx
0
Let f Let f .x/ be x/ be given by Figure 1. 2
R
(a) The definite integral 0 f.x/dx is f.x/dx is the signed area of a semicircle of radius 1 which lies below the x-axis. x -axis. Therefore, 2
Z
f.x/dx
0
D 12 .1/2 D 2 :
6
R
x/dx is the signed area of a semicircle of radius 1 which lies below the x-axis x -axis and a semicircle of (b) The definite integral 0 f .x/dx is radius 2 which lies above the x the x-axis. -axis. Therefore, 6
Z 0
f.x/dx
D 12 .2/2 12 .1/2 D 32 :
576
THE INTEGRAL
CHAPTER 5
4
14. Evaluate: Evaluate: (a)
Z
6
f.x/dx
Z j
f.x/ dx
(b)
1
SOLUTION
j
1
Let f Let f .x/ .x/ be given by Figure 1. 4
R
(a) The definite integral 1 f.x/dx is f.x/dx is the signed area of one-quarter of a circle of radius 1 which lies below the x -axis and one-quarter of a circle of radius 2 which lies above the x the x-axis. -axis. Therefore, 4
Z
f.x/dx
1
6
R j
D 14 .2/2 14 .1/2 D 34 :
(b) The definite integral 1 f.x/ dx is dx is the signed area of one-quarter of a circle of radius 1 and a semicircle of radius 2, both of which lie above the x the x-axis. -axis. Therefore,
j
6
Z j
f.x/ dx
j
1
D 12 .2/2 C 14 .1/2 D 94 :
In Exercises Exercises 15 and and 16, refer refer to Figur Figuree 2.
y
y = g (t )
2 1
t
1
2
3
4
5
1
!
2
!
FIGURE 2 3
15. Evaluate
Z
5
g.t/dt and g.t/dt and
0
Z
g.t/dt .
3
SOLUTION
The region bounded bounded by the curve y curve y g.t/ and g.t/ and the t the t -axis over the interval Œ0 interval Œ0;; 3 is comprised of two right triangles, one with 3 area 12 below the axis, and one with area 2 above the axis. The definite integral is therefore equal to 2 12 2.
D
D
The region bounded by the curve y curve y g.t/ and g.t/ and the t the t -axis over the interval Œ3 interval Œ3;; 5 is comprised of another two right triangles, one with area 1 above the axis and one with area 1 below the axis. The definite integral is therefore equal to 0.
D
a
16. Find a Find a,, b, b , and c and c such such that
Z
c
g.t/dt and g.t/dt and
0
SOLUTION
To make the value of
Z
g.t/dt are g.t/dt are as large as possible.
b
a
Z
g.t/dt as g.t/dt as large as possible, we want to include as much positive area as possible. This
0
happens when we take a take a
c
D 4. 4 . Now, to make the value of
Z
g.t/dt as g.t/dt as large as possible, we want to make sure to include all of
b
the positive area and only the positive area. This happens when we take b
D 1 and c and c D 4.
17. Describe the partition P partition P and and the set of sample points C for C for the Riemann sum shown in Figure 3. Compute the value of the Riemann sum. y
34.25
20 15 8 x
0.5 1
2 2.5 3 3.2
4.5 5
FIGURE 3 SOLUTION
The partition P partition P is is defined by x0
D0
<
x1
D 1 < x2 D 2:5 < x3 D 3:2 < x4 D 5 D fc1 D 0:5;c 0:5;c 2 D 2; c3 D 3; c4 D 4:5g. Finally, the value of the Riemann sum is The set of sample points is given by C by C D 34:25.1 0/ C 20.2:5 1/ C 8.3:2 2:5/ C 15.5 3:2/ D 96:85:
SECTION
18. Compute R. Compute R.f;P;C f;P;C / for f for f .x/ .x/
5.2
The Defini Definite te Integr Integral al
577
D x 2 C x for the partition P partition P and and the set of sample points C points C in in Figure 3.
SOLUTION
R.f;P;C/
D f .0:5/.1 .0:5/.1 0/ C f .2/.2:5 .2/.2:5 1/ C f .3/.3:2 .3/.3:2 2:5/ C f .4:5/.5 .4:5/.5 3:2/ D 0:75.1/ C 6.1:5/ C 12.0:7/ C 24:75.1:8/ D 62:7
R.f;P;C / for the given function, partition, and choice of sample points. Also, In Exercises 19–22, calculate the Riemann sum R.f;P;C R.f; P;C /. sketch the graph of f and the rectangles corresponding to R.f 19. f.x/
D x,
SOLUTION
P
D D f1;1:2; 1:5;2 D f1:1; 1:5;2 g, C D 1:1; 1:4; 1:9g Let f Let f .x/ D x . With D fx0 D 1; x1 D 1:2; P D 1:2; x2 D 1:5;x 1:5;x 3 D 2g
C
D D fc1 D 1:1; 1:1; c2 D 1:4; 1:4; c3 D 1:9g;
and
we get R.f;P;C/
D x1 f .c1/ C x2 f .c2 / C x3f .c3 / D .1:2 1/.1:1/ C .1:5 1:2/.1:4/ C .2 1:5/.1:9/ D 1:59:
Here is a sketch of the graph of f and f and the rectangles. y
2 1.5 1 0.5 x
0.5
20. f.x/
D 2x C 3,
1
1.5
D D f4; 1;1;4;8g, C D D f3;0;2;5g SOLUTION Let f Let f .x/ D 2x C 3. With D fx0 D 4; x1 D 1; x2 D 1; x3 D 4; x4 D 8g P D
2
2.5
P
C
D D fc1 D 3; c2 D 0; c3 D 2; c4 D 5g;
and
we get R.f;P;C/
D x1 f .c1/ C x2 f .c2 / C x3 f .c3 / C x4 f .c4/ D .1 .4//.3/ C .1 .1//.3/ C .4 1/.7/ C .8 4/.13/ D 70:
Here is a sketch of the graph of f and f and the rectangles. y
20 15 10 −4
5
−2
x
2
−5
21. f.x/
D x 2 C x,
SOLUTION
4
D D f2;3;4:5;5g, C D D f2;3:5;5g 2 Let f Let f .x/ D x C x . With D fx0 D 2; x1 D 3; x3 D 4:5;x P D 4:5;x 4 D 5g
6
8
P
C
D D fc1 D 2; c2 D 3:5;c 3:5;c 3 D 5g;
and
we get R.f;P;C/
D x1 f .c1/ C x2 f .c2 / C x3 f .c3 / D .3 2/.6/ C .4:5 3/.15:75/ C .5 4:5/.30/ D 44:625:
f and the rectangles. Here is a sketch of the graph of f and y
30 25 20 15 10 5 x
1
2
3
4
5
578
THE INTEGRAL
CHAPTER 5
22. f.x/
SOLUTION
D D 0; 6 ; 3 ; 2 , Let f Let f .x/ .x/ D sin x . With
D sin x ,
˚
P
P
D D
we get
n
x0
C
D D f0:4; 0:7; 0:7; 1:2g
D 0; x1 D 6 ; x3 D 3 ; x4 D 2
R.f;P;C/
o
C
D D fc1 D 0:4; 0:4; c2 D 0:7; 0:7; c3 D 1:2g;
and
D x1 f .c1 / C x2 f .c2 / C x3 f .c3/ D 6 0 .sin 0:4/ C 3 6 .sin 0:7/ C 2 3 .sin 1:2/ D 1:029225:
Here is a sketch of the graph of f and f and the rectangles. y
1 0.8 0.6 0.4 0.2 x
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
In Exercises Exercises 23–28, 23–28, sketch sketch the signed signed area area represen represented ted by the integral. integral. Indicate Indicate the regions regions of positive and negative negative area. area. 5
23.
Z
.4x
x
2
/ dx
0
SOLUTION
Here is a sketch sketch of the signed area represented represented by the integral y
5 2 0 .4x x / dx .
R
4 2
5 1
−2
2
3
x
4
−4 =4
24.
Z
tan x dx =4
SOLUTION
Here is a sketch sketch of the signed area represented represented by the integral y
=4 tan x =4
R
dx .
1.0 0.5 −0.6
+
−0.2
x
0.2 0.4 0.6 −0.5
−
−1.0
2
25.
Z
sin x dx
SOLUTION
Here is a sketch sketch of the signed area represented represented by the integral y
2
R
0.4 x
1 !
0.4
!
0.8
!
1.2
2
3
4
5 !
3
26.
Z 0
sin x dx
6
7
sin x dx .
SECTION
Here is a sketch sketch of the signed area represented represented by the integral
SOLUTION
y
3 0
R
5.2
The Defini Definite te Integr Integral al
579
sin x dx .
1 0.5
+
+
x
2 !
4
0.5
6
8
10 10
!
1
!
2
27.
Z
ln x dx
1=2
Here is a sketch sketch of the signed area represented represented by the integral
SOLUTION
2 1=2 ln x dx .
R
0.6 0.4 +
0.2 0.5
–0.2
–
1
1.5
2
–0.4 –0.6
1
28.
Z
tan
1
x dx
1
SOLUTION
Here is a sketch sketch of the signed area represented represented by the integral y
1 1 x dx . 1 tan
R
0.5 +
−1
−0.5 −
0 .5
1
x
−0.5
In Exercises Exercises 29–32, 29–32, determine determine the sign of the the integral integral without without calculat calculating ing it. Draw Draw a graph graph if necessary necessary.. 1
29.
Z
x 4 dx
2
SOLUTION
1
30.
Z
The integrand is always always positive. The integral must therefore therefore be positive, since the signed area area has only positive part.
x 3 dx
2
SOLUTION By symmetry, symmetry, the positive area from the interval Œ0 interval Œ0;; 1 is cancelled by t he negative area from Œ1; 0. With the i nterval Œ2; 1 contributing more negative area, the definite integral must be negative.
2
31.
Z
x sin x dx
0
As you can see from the graph below, below, the area below the axis is greater than the area above the axis. Thus, the definite integral is negative. negative. SOLUTION
y
0.2
+
x
1
2
32.
Z 0
sin x dx x
!
0.2
!
0.4
!
0.6
2
3
4
5 !
6
7
580
THE INTEGRAL
CHAPTER 5
From the plot below, below, you can see that the area above above the axis is bigger than the area below the axis, hence the integral
SOLUTION
is positive. y
1 0.8 0.6 0.4 0.2
+
1
2
3
4
x
!
5
6
In Exercises Exercises 33–42, 33–42, use properties properties of the integr integral al and the the formulas formulas in the summary summary to calculate calculate the the integrals. integrals. 4
33.
Z
.6t
3/dt
0
4
SOLUTION
Z
4
.6t
3/dt
0
2
34.
Z
.4x 3
D 6
4
Z
t dt
3
0
Z
1 dt
0
D 6 12 .4/2 3.4 0/ D 36. 36.
C 7/dx
SOLUTION
2
Z
2
.4x 3
D4
Z Z Z
D4
C 7/dx D 4
2
x dx
3
9
35.
dx
2
x dx 3
x dx
7.2 .3//
0
2
Z
7
3
0
D4
Z C ! Z C C Z ! C 3
x dx
0
x dx
35
0
1 2 1 2 .3/2 2 2
C 35 D 25:
x 2 dx
0
9
SOLUTION
By formula (5),
Z
x 2 dx
0
5
36.
Z
D 13 .9/3 D 243:
x 2 dx
2
5
SOLUTION
Z
5
Z D
2
x dx
2
1
37.
Z
2
2
x dx
0
Z
x 2 dx
0
D 13 .5/3 13 .2/3 D 39. 39.
.u2 2u/du
0
SOLUTION
1
Z
.u2 2u/du
0
1=2
38.
Z
.12y 2
0
1
D
Z
u2 du
0
1
2
Z
u du
0
D 13 .1/3 2
1 2
.1/2
D 13 1 D 23 :
C 6y/dy
SOLUTION
1=2
Z 0
.12y
2
1=2
C 6y/dy D 12 D
Z
1=2
2
y dy
0
1 12 3
Z C 6
y dy
0
1 2
3
D 12 C 34 D 54 :
C
1 6 2
1 2
2
SECTION
1
39.
Z
.7t 2
3
SOLUTION
The Defini Definite te Integr Integral al
5.2
C t C 1/dt First, write 1
Z
.7t
2
3
0
C t C 1/dt
Z D Z D
.7t
2
3
1
C t C 1/dt
3
.7t
Z C
2
0
.7t 2
0
1
Z C
C t C 1/dt
C t C 1/dt
.7t 2
0
C t C 1/dt
Then, 1
Z
.7t
2
3
C t C 1/dt D D
3
40.
Z
1 7 .3/3 3
C
1 .3/2 3 2
1 7 13 3
C
C
1 2 1 2
C1
C 92 3 C 73 C 12 C 1 D 196 : 3
63
.9x 4x 2 / dx
3
SOLUTION
First write 3
Z
0
Z D Z D
2
.9x 4x / dx
3
Z C
2
.9x 4x / dx
3
3
.9x 4x 2 / dx
0
3
2
.9x 4x / dx
0
3
Z C
.9x 4x 2 /dx:
0
Then, 3
Z
2
.9x 4x / dx
D
3
D 1
41.
Z
.x 2
a
SOLUTION
b 2 0 .x
First, 1
R
.x
2
a
42.
b 0
R Z D
C x/dx D
x 2 dx
0
C x/dx
.x
2
a
D a2
81 2
C 36 C
81 2
C
36
1 1 9 .3/2 4 .3/3 2 3
D 72:
C x/dx
Z Z
1 1 9 .3/2 4 .3/3 2 3
1 3 1 3
C
b 0
R
x dx
D 13 b3 C 12 b 2. Therefore
1
1
C x/dx
Z C
.x
1 2 1 2
1 .a/3 3
C
0
2
Z D
C x/dx C
.x
0
1 .a/2 2
2
a
C x/dx
Z
.x 2
0
D 13 a3 12 a2 C 56 :
x 2 dx
a
SOLUTION
a2
Z
x 2 dx
a
a2
D
Z
x 2 dx
a
0
Z
x 2 dx
0
D 13
3
a2
1 .a/ 3 3
In Exercises Exercises 43–47, 43–47, calculate calculate the the integral, integral, assuming assuming that that 5
Z
5
f.x/dx
0
D 5;
Z 0
g.x/dx
D 12
5
43.
Z
.f.x/
0
C g.x//dx 5
SOLUTION
Z
5
.f.x/
0
5
44.
Z 0
2f .x/
C g.x//dx D
1 g.x/ d x 3
Z 0
5
f.x/dx
C
Z 0
g.x/dx
D 5 C 12 D 17. 17.
D 13 a6 13 a3 :
C x/dx
581
582
THE INTEGRAL
CHAPTER 5
5
Z
SOLUTION
1 g.x/ d x 3
2f .x/
0
0
45.
Z
5
Z
D2
f.x/dx
0
1 3
5
Z
g.x/dx
0
D 2.5/ 13 .12/ D 6.
g.x/dx
5
0
Z
SOLUTION
5
g.x/dx
5
5
46.
Z
Z
D
g.x/dx
D 12. 12.
0
.f.x/ x/dx
0
5
Z
SOLUTION
5
.f.x/ x/dx
0
Z D
5
f.x/dx
0
47. Is it possible to calculate
x dx
0
5
Z
Z
D 5 12 .5/2 D 152 .
g.x/f.x/dx from g.x/f.x/dx from the i nformation given?
0
It is not possible to calculate calculate
SOLUTION
5 g.x/f.x/dx from 0 g.x/f.x/dx from
R
the information given.
48. Prove by computing computing the limit of right-endpoint approximations: b
Z
x 3 dx
0
Let f Let f .x/ .x/
SOLUTION
RN
D x b
Hence
Z
f .xk /
D
x dx
0
b N
D
k 1
3
9
D x 3 , a D 0 and x D .b a/=N D D b=N . b=N . Then
N
X
4
D b4
N
X
b3
k3
N 3
k 1
D
D N lim R D lim !1 N N !1
b4 4
!
0X 1 @ AD D ! N
b4
D N 4 4
b4
k3
N 4
k 1
4
N 4 4
C
N 3 2
C
N 2 4
!
4
4
4
b b D b4 C 2N C 4N : 2
4
b b C 2N C 4N D b4 . 2
In Exercises Exercises 49–54, 49–54, evaluate evaluate the integral integral using using the formulas formulas in the the summary summary and Eq. (9). 3
49.
Z
x 3 dx
0
3
SOLUTION
By Eq. (9),
Z
x 3 dx
0
3
50.
Z
4
D 34 D 814 .
x 3 dx
1
3
SOLUTION
Z
3
x dx
1
3
51.
Z
3
Z D
1
3
x dx
0
Z
x 3 dx
0
D 14 .3/4 14 .1/4 D 20. 20.
.x x 3/ dx
0
3
SOLUTION
Z
3
3
.x x / dx
0
1
52.
Z
.2x 3 x
0
Z D
3
x dx
0
Z
x 3 dx
0
D 12 32 14 34 D 634 .
C 4/dx
SOLUTION Applying the linearity of the definite integral, Eq. (9), the formula from Example 4 and and the formula for the definite integral of a constant:
1
Z
.2x
3
1
x
0
1
53.
Z
.12x 3
0
C 4/dx D 2
Z
3
x dx
1
0
Z
1
x dx
0
Z C 0
4 dx
D 2 14 .1/4 12 .1/2 C 4 D 4:
C 24x2 8x/dx
SOLUTION
1
Z 0
.12x
3
C 24x
2
1
8x/dx
D 12
Z 0
3
x dx
1
C 24
Z
x
2
1
8
0
D 12 14 14 C 24 13 13 8 12 12 D 3C84D 7
Z 0
x dx
SECTION
2
54.
Z
The Defini Definite te Integr Integral al
5.2
.2x 3 3x 2 / dx
2
SOLUTION
2
Z
.2x
3
0
2
3x / dx
2
Z D Z D Z D
.2x
3
2
2
3x / dx
2
.2x 3 3x 2 / dx
0
2
Z C Z
.2x 3 3x 2 / dx
0
2
.2x 3 3x 2 / dx
0
2
2
2
3
x dx
0
3
Z
2
2
x dx
0
2
Z
3
x dx
0
2
Z C 3
x 2 dx
0
D 2 14 .2/4 3 13 .2/3 2 14 .2/4 C 3 13 .2/3 D 8 8 8 8 D 16: In Exercises Exercises 55–58, 55–58, calculate calculate the the integral, integral, assuming assuming that that 1
Z
2
f.x/dx
0
Z
D 1;
4
f.x/dx
0
D 4;
Z
f.x/dx
1
D7
4
55.
Z
f.x/dx
0
4
SOLUTION
Z
1
f.x/dx
0
2
56.
Z
Z D
4
f.x/dx
0
Z C
f.x/dx
1
D 1 C 7 D 8.
f.x/dx
1
2
SOLUTION
Z
2
f.x/dx
1
1
57.
Z
Z D
1
f.x/dx
0
Z
f.x/dx
0
D 4 1 D 3.
f.x/dx
4
1
SOLUTION
Z
4
f.x/dx
4
4
58.
Z
Z
D
f.x/dx
1
D 7.
f.x/dx
2
SOLUTION
4 0 f.x/dx
R
From Exercise 55,
D 8. Accordingly, Accordingly,
4
Z
4
f.x/dx
D
2
Z
2
f.x/dx
0
Z
f.x/dx
0
D 8 4 D 4:
In Exercises Exercises 59–62, 59–62, express express each each integral integral as as a single integral. integral. 3
59.
Z
7
f.x/dx
0
Z C
3
SOLUTION
Z
9
60.
f.x/dx
7
f.x/dx
0
Z
f.x/dx
3
C
9
2
Z
SOLUTION
3
D
Z
f.x/dx. f.x/dx.
0
f.x/dx
Z
9
f.x/dx
2
Z
Z D
4
f.x/dx
4
9
61.
7
f.x/dx
4
9
Z
Z
9
f.x/dx
2
Z C
! Z
9
f.x/dx
4
4
f.x/dx
4
Z D
f.x/dx:
2
5
f.x/dx
2
Z
f.x/dx
2
9
SOLUTION
Z 2
5
f.x/dx
Z 2
Z D
5
f.x/dx
2
9
f.x/dx
Z C 5
! Z
5
f.x/dx
2
9
f.x/dx
Z D 5
f.x/dx:
583
584
3
62.
THE INTEGRAL
CHAPTER 5
Z
9
f.x/dx
7
Z C
f.x/dx
3
3
SOLUTION
9
Z
f.x/dx
7
Z C
Z
7
f.x/dx
3
Z
D
7
f.x/dx
C
3
9
Z C Z
f.x/dx
3
! Z D
9
f.x/dx
7
b
In Exercises Exercises 63–66, 63–66, calculate calculate the integ integral, ral, assuming assuming that that f is integrable and
f.x/dx:
7
f.x/dx
1
D 1b
1
for all b > 0 .
5
63.
Z
f.x/dx
1
5
SOLUTION
Z
f.x/dx
D 1 5 1 D 45 .
f.x/dx
Z D
1
5
64.
Z
f.x/dx
3
5
SOLUTION
5
Z 3
6
65.
Z
3
f.x/dx
1
Z
f.x/dx
1
D
1 5
1
1
1 3
D 152 .
.3f .3f .x/ 4/dx
1
6
SOLUTION
Z
6
.3f .3f .x/ 4/dx
D3
1
1
66.
Z
Z
6
f.x/dx
4
1
Z
1 dx
1
1
D 3.1 6
/ 4.6 1/
D 352 .
f.x/dx
1=2 1
SOLUTION
Z
1=2
f.x/dx
1=2
D
Z
f.x/dx
1
D
1
1 2
1
!
D 1. b
67.
Z
Explain the difference in graphical interpretation between
b
f.x/dx and f.x/dx and
a
SOLUTION
and the x the x -axis, whereas areas that were part of
b a
j
b f.x/dx represents the signed area between f between f .x/ .x/ a f.x/dx represents
R
f.x/ dx represents the total (unsigned) area between f .x/ .x/ and the x the x -axis. Any negatively signed
j
b x/dx are regarded as positive areas in a f .x/dx are
R
f.x/ dx .
a
When f When f .x/ .x/ takes on both positive and negative values values on Œa on Œa;; b ,
R j
Z j
x) Graph of f ( ( x
b a
R j
f.x/ dx . Here is a graphical example of this phenomenon.
j
x)| Graph of | f ( ( x
10
30 x
!
4
2
2
!
20
4
10
!
10
20
!
30
x
!
!
68.
4
2
2
!
4
Use the graphical interpretation of the definite integral to explain the inequality
ˇˇZ ˇ
b
ˇˇ Z ˇ j b
f.x/dx
a
f.x/ dx
j
a
where f where f .x/ .x/ is continuous. Explain also why equality holds if and only if either f either f .x/ .x/ 0 for all x all x or or f f .x/ 0 for all x all x.. Let A Let AC denote the unsigned area under the graph of y of y A denote the unsigned area when f when f .x/ < 0. 0. Then SOLUTION
D f.x/ over f.x/ over the interval Œa interval Œa;; b where f where f .x/ .x/ 0. 0. Similarly, let
b
Z
f.x/dx
a
D AC A
:
Moreover,
ˇˇˇZ ˇ
b
a
f.x/dx
ˇˇˇ ˇ
b
AC
CA
Z D j a
f.x/ dx: dx :
j
Equality holds if and only if one of the unsigned areas is equal to zero; in other words, if and only if either f .x/ .x/ f.x/ 0 for all x all x..
0 for 0 for
all x all x or or
SECTION
Let f Let f .x/ .x/
69.
The Defini Definite te Integr Integral al
585
D x. Find an interval Œa interval Œa;; b such that
ˇˇZ ˇˇ
b
f.x/dx
a
SOLUTION
5.2
If a a > 0, 0 , then f then f .x/ 0 for all x all x
ˇˇ ˇˇ D
b
1 2
j D 32
Z j
f.x/ dx
and
a
2 Œa;b, so
ˇˇˇZ ˇˇ Z j b
ˇˇˇ Z ˇˇ D j b
f.x/dx
a
f.x/ dx
j
a
by the previous exercise. We We find a si milar result if b if b < 0. 0 . Thus, we must have a have a < 0 and b and b > 0. 0 . Now, b
j D 12 a2 C 12 b2 :
f.x/ dx
a
Because b
Z
D 12 b2 12 a2 ;
f.x/dx
a
then
ˇˇZ ˇˇ
b
f.x/dx
a
If b b 2 > a 2 , then 1 2 a 2 yield a yield a
D 1 and b and b D
C 12 b2 D 32
ˇˇ ˇˇ D
1 2 b a2 : 2
j
j
and
1 2 .b a2 / 2
D 12
and
1 2 .a b 2 / 2
D 12
p
2. On the other hand, if b b 2 < a 2 , then 1 2 a 2
p D 2 and b and b D 1. 2 D 70. Evaluate I Evaluate I D 0 C J D D 2 . that I that I C
C 12 b2 D 32
yield a yield a
Z
2
2
sin
x dx and J D
Z
cos2 x dx as follows. First show with a graph that I
0
The graphs of f f .x/ .x/ sin 2 x and g.x/ and g.x/ the shaded areas in the two graphs are equal, thus
D cos2 x are shown below at the left and right, respectively. It is clear that
D
SOLUTION
2
Z D D
I
2
2
sin x dx
0
Z D
cos2 x dx
0
D J:
Now, Now, using the fundamental trigonometric identity, we find 2
I
C C J D D
Z
.sin2 x
0
C cos2 x/dx D
2
Z
1 dx
0
D 2:
Combining this last result with I with I
D D J yields I D J D D . yields I D y
y
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2 x
1
In Exercises Exercises 71–74, 71–74, calculate calculate the the integral. integral. 6
71.
Z j
3 x dx
0
j
D J . J . Then prove
2
3
4
5
6
x
1
2
3
4
5
6
586
THE INTEGRAL
CHAPTER 5
Over the interval, interval, the region between the curve curve and the interval Œ0 interval Œ0;; 6 consists of two triangles above the x the x axis, axis, each 9 9. 9 of which has height 3 and width 3, and so area 2 . The total area, hence the definite integral, is . SOLUTION
y
3 2 1 x
1
2
3
4
5
6
Alternately, 6
3
Z j
3 x dx
j
0
6
Z Z D C Z Z Z D C .3 x/dx
.x 3/dx
0
3
3
3
3
dx
6
x dx
0
0
Z ! Z 3
x dx
0
6
x dx
3
0
dx
3
D 9 12 32 C 12 62 12 32 9 D 9: 3
72.
Z j
2x 4 dx
j
1
The area between 2x 4 and the x the x axis axis consists of two triangles above the x the x-axis, -axis, each with width 1 width 1 and and height 2 height 2,, and hence with area 1 area 1.. The total area, and hence the definite integral, is 2. 2.
j
SOLUTION
j
y
2 1.5 1 0.5 x
0.5
1
1.5
2
2.5
3
Alternately, 3
2
Z j
2x 4 dx
1
3
Z Z D
.4 2x/dx
j D
1
2
dx
2
1
1
Z
Z ! C Z 1
x dx
0
D42 73.
.2x 4/dx
2
2
4
Z j
Z
C
3
x dx
2
0
1 2 1 2 2 1 2 2
C2
2
x dx
0
1 2 1 2 3 2 2 2
4
Z ! Z
3
x dx
0
4
dx
2
D 2:
x 3 dx
1
j
SOLUTION
jx 3 j D
(
x3
x
3 x
x < 0:
0
Therefore, 1
Z j 1
2
74.
Z j
x
3
0
j dx
Z D
x
3
1
1
dx
Z C
3
x dx
0
1
Z D
3
1
x dx
0
Z C
x 3 dx
0
x 2 1 dx
0
j
SOLUTION
jx
2
1
jD
(
x2 1
1x
2 .x 1/
0 x < 1:
Therefore, 2
Z j 0
x
2
1
1 dx
j
Z D 0
2
.1 x / dx
2
Z C 1
.x 2 1/dx
2
D 14 .1/4 C 14 .1/4 D 12 :
SECTION
1
1
Z Z D dx
x dx
0
0
1 .1/ 3
D1
C
Z
2
2
C
1
2
x dx
Z
0
2
x dx
0
1 1 .8/ .1/ 3 3
1
The Defini Definite te Integr Integral al
5.2
! Z
587
2
1 dx
1
D 2:
75. Use the Comparison Theorem to show that 1
Z
1
5
x dx
Z
0
SOLUTION
On the interval Œ0 interval Œ0;; 1, x 5
2
Z
4
x dx; dx ;
0
x
1
5
x dx
0
x 5 for x for x
SOLUTION
1 3
1 d x x
4
x 5 dx: dx :
1
6
1 dx d x 6
Z D 4
6
Z 4
1 d x: x
the interval Œ4 interval Œ4;; 6, so 6
6
1 d x x
Z 4
Therefore
Z
1 1 6 x , so, by Theorem 5,
1 3 1 1 x 4 on
x 4 dx: dx :
1 . 2
On the interval Œ interval Œ4; 4; 6,
On the other hand,
Z
2
x 4 dx
1
76. Prove that
1
2 Œ1;2, so, by the same Theorem, 2
6
x 5 dx
0
Z
Z
Z
4 , so, by Theorem 5, 1
x dx
1
Z On the other hand, x hand, x 4
2
4
1 dx d x 4
Z 4
D 14 .6 4/ D 12 :
6 1 1 1 3 4 x d x 2 , as desired.
R
0:3 0:3
R
77. Prove that 0:0198 that 0:0198 0:2 sin x dx
For 0 For 0 x 0:2 x 0:3, 0:3, we have SOLUTION
0:52, 6 0:52, m
0:0296. 0:0296. Hint: Show that 0:198 that 0:198 d .sin . sin x/ dx
we have
sin x
0:296 for 0:296 for x x in in Œ0:2;0:3 Œ0:2;0:3.
D cos x > 0. Hence sin x is increasing on Œ0: 2; 0:3. Accordingly, for on Œ0:2;
D 0:198 0:19867 sin 0:2 sin x sin 0:3 0:29552 0:296 D M
Therefore, by the Comparison Theorem, we have 0:3
0:0198
D m.0:3 0:2/ =4
78. Prove that 0:277 that 0:277
Z
cos x dx
Z D
0:3
m dx
0:2
Z
0:3
sin x dx
0:2
Z
M dx
0:2
D M.0:3 0:2/ D 0:0296:
0:363. 0:363.
=8
SOLUTION
cos x is decreasing on the interval Œ interval Œ=8; =4. Hence, for =8 x cos. cos.=4/ cos x
Since cos. cos.=4/
D
=4, =4,
cos. cos.=8/:
p
2=2, 2=2, 0:277
8
p
2 2
=4
Z D
=8
p
2 dx 2
=4
Z
cos x dx:
=8
0:924, Since cos. cos.=8/ 0:924, =4
Z
=8
Therefore 0:277 Therefore 0:277
=4 cos =8
R
x
0:363:
=4
cos x dx
Z
=8
0:924 0:924 dx
D 8 .0:924/ 0:363:
588
THE INTEGRAL
CHAPTER 5
=2
79. Prove that 0 that 0
Z
=4
sin x dx x
p
2 . 2
Let
SOLUTION
f.x/
D sinx x :
As we can seep in the sketch below, f .x/ is decreasing on the interval Œ interval Œ=4; =2. Therefore f Therefore f .x/ .x/ f .=4/ for =4/ for all x all x in in Œ=4; =2.
f .=4/
D2
2
, so:
=2
Z
=4
sin x dx x
=2
Z
=4
p
2 2 dx
D
p
2 2 4
D
p
2 : 2
y y = sin
x x
2 2/ 2/
x
/4
1
bounds for 80. Find upper and lower bounds
Z p
dx
5x 3
0
C4
Let
SOLUTION
/2
.
f.x/
1
D p
5x 3
C4
:
f.x/ is f.x/ is decreasing for x for x on on the interval Œ0 interval Œ0;; 1, so f.1/ so f.1/ f.x/ f.0/ for f.0/ for all x all x in in Œ0 Œ0;; 1. f.0/ 1
Z 0
1 dx d x 3 1 3
1
Z Z
1
f.x/dx
0
0
1
Z
f.x/dx
0
1 dx d x 2
1 : 2
.x/ g.x/ on g.x/ on Œa; Œa; b . By the Comparison Theorem, Suppose that f that f .x/ 0 g .x/ for .x/ for x x Œa;b ? If not, give a counterexample.
81. f 0 .x/
2
The assertion f assertion f 0 .x/ g 0 .x/ is .x/ is false. Consider a interval Œ0 interval Œ0;; 1, but f but f 0 .x/ 1 while g while g 0 .x/ 0 for all x all x.. SOLUTION
D
D
D 12 and f and f .1/ .1/ D 13 , so
b a f.x/dx
R
b g.x/dx . a g.x/dx.
R
Is it also true that
D 0, 0 , b D 1, 1 , f .x/ .x/ D x , g.x/ D 2. 2 . f .x/ .x/ g.x/ for g.x/ for all x all x in in the
State whether true or false. If false, sketch the graph of a counterexample. counterexample.
82.
b
(a) If f f .x/ > 0, 0, then
Z
f .x/dx .x/dx > 0. 0.
a
b
Z
(b) If
f .x/dx .x/dx > 0, 0, then f then f .x/ > 0. 0.
a
SOLUTION
(a) It is true that if f f .x/ > 0 for 0 for x x b
R
2 Œa;b, then
b .x/dx a f .x/dx
R
> 0. 0.
.x/dx > 0, 0, then f then f .x/ > 0 for 0 for x x (b) It is false that if a f .x/dx yet f yet f .2/ 2 < 0. 0 . Here is the graph from that exercise.
D
2 Œa;b. Indeed, in Exercise 3, we saw that y
6 4 2 x
2
!
1
!
2
!
1
1 2 .3x
R
C 4/dx D 7:5 > 0,0,
SECTION
The Defini Definite te Integr Integral al
5.2
589
Further Insights and Challenges 83. Explain graphically: If f f .x/ .x/ is an odd function, then a
Z
f.x/dx
D 0.
a
SOLUTION If f is f.x/ for f is an odd function, then f .x/ f.x/ for all x all x . Accordingly, for every positively signed area in the right half-plane where f where f is is above the x -axis, there is a corresponding negatively signed area in the left half-plane where f is f is below the x the x-axis. -axis. Similarly, for every negatively negatively signed area in the right half-plane where f where f is is below the x the x-axis, -axis, there is a corresponding positively signed area in the left half-plane where f where f is is above the x the x-axis. -axis. We conclude conclude that the net area between the graph of f of f and the x the x-axis -axis over Πover Πa; a is 0, since the positively signed areas and negatively negatively signed areas cancel each other out exactly. exactly.
D
y
4 2 1
! !
x
2
1
2
2
!
!
1
84. Compute
Z
4
sin. sin.sin.x//. sin.x//.sin sin2 .x/
C 1/dx. 1/dx . Let f Let f .x/ D sin. sin.sin.x//. sin.x//.sin sin2 .x/ C 1//. 1//. sin x is an odd function, while sin 2 x is an even function, so: f .x/ D sin. sin.sin. sin.x//.sin x//.sin2 .x/ C 1/ D sin. sin. sin.x//. sin.x//.sin sin2 .x/ C 1/ D sin. sin.sin.x//. sin.x//.sin sin2 .x/ C 1/ D f.x/: 1
SOLUTION
Therefore, f Therefore, f .x/ .x/ is an odd function. The function is odd and the interval is symmetric around the origin so, by the previous exercise, the integral must be zero. 85. Let k Let k and and b b be be positive. Show, by comparing the right-endpoint approximations, that b
Z
x k dx
0
SOLUTION Let k and 1 f.x/dx exist. Let N Let N 0 f.x/dx exist.
b be any positive numbers. Let f .x/ .x/
R
1
Z
D bkC1
be a positive integer and set x
x k dx
0
x k on Œ0; Œ0; b . Since f is f is continuous, both
D
b 0 f.x/dx
R
and
D .b 0/=N D D b=N . b=N . Let x Let x j D a C jx D bj=N , bj=N , j D 1 ; 2 ; : : : ; N b b Œ0; b . Then the right-endpoint approximation to 0 f.x/dx D 0 x k dx is dx is be the right endpoints of the N the N subintervals subintervals of Œ0; N
D x
X
RN
b In particular, if b
f .xj /
j 1
D
D
b N
N
X
j 1
D
k
bj N
D 1 above, then the right-endpoint approximation to N
S N N
D x
X
f .xj /
j 1
D
D
1 N
N
X
j 1
D
j N
k
D bk
0 C @
1 0 f.x/dx
R
D N k1C1
N kC1
D
N
jk
j 1
D
:
j 1
1 k 0 x
R
jk
X
R
N
1
1
R X 1A D dx is dx is
D b k1C1 RN
D bkC1 S N . Therefore,
In other words, R words, R N
b
Z
k
x dx
D N lim R D lim b !1 N N !1
0
86. Verify for 0 for 0
b
C
S N
D b
k 1
C
Z p
1 x 2 dx
0
The function f function f .x/ b 0
R
p D
p
D 12 b
p
1 b2
lim S N
D b
N
!1
1 by interpreting in terms of area: b
SOLUTION
k 1
C 12 sin
1
k 1
C
1
Z
x k dx :
0
b
1 x 2 is the quarter circle of radius 1 in the first quadrant. For 0
1 x 2 dx can dx can
represented by the integral be divided into two parts. The area of the triangular part is Pythagorean Theorem. The area of the sector with angle where where sin b , is given by 12 .1/2 . /. Thus
D D
b
Z p
1 x 2 dx
0
D 12 b
p
1 b2
C 12 D D 12 b
p
1 b2
C 12 sin
1
b:
p b
1 2 .b/
1, the area
1 b 2 using the
590
THE INTEGRAL
CHAPTER 5
y
1 θ
x
1
b
Suppose that f that f and g and g are are continuous functions such that, for all a, a,
87.
a
Z
a
f.x/dx a
Give an intuitive argument showing that that f f .0/ .0/ a f a
D
Z
g.x/dx a
D g.0/. g.0/. Explain your idea with a graph.
a g .x/dx. x/dx. a
R
D R
SOLUTION Let .x/dx Consider what happens as a decreases in size, becoming very close to zero. Intuitively, Intuitively, the areas of the functions become .a become .a .a//.f.0// 2a.f.0// and 2a.f.0// and .a .a .a//.g.0// 2a.g.0//. 2a.g.0//. Because we know these areas must be t he same, we have 2a.f .0// .0// 2a.g.0// and 2a.g.0// and therefore f.0/ therefore f.0/ g.0/. g.0/.
D
D
88. Theorem 4 remains remains true without the assumption a assumption a b SOLUTION
Suppose that we have b have b < a < c . By the additivity property, property, we have
R
c a b f.x/dx b f.x/dx
D R R
R R
b a f.x/dx
D R D R
c , we have
c f.x/dx . b f.x/dx.
C R R
c a f.x/dx
D
a c f.x/dx
b b a f.x/dx c f .x/dx
c a f.x/dx
R
c b f.x/dx
R
D
D
b a f.x/dx
R
a b f.x/dx
R
C
D R
R
C R
D
R
c f.x/dx . b f.x/dx.
C
R
C
R
c f.x/dx . Therefore, a f.x/dx.
R
b a c f.x/dx c f.x/dx b c f.x/dx . a f .x/dx b f.x/dx.
Now suppose that we have c < a < b . By the additivity property, we have Therefore,
c . Verify this for the cases b cases b < a < c and c and c < a < b .
The additivity property property of definite integrals integrals states for a for a b
c a f.x/dx
D
D
b f.x/dx . a f.x/dx.
b , and c Hence the additivity property holds for all real numbers a numbers a,, b, and c,, regardless of their relationship amongst each other.
5.3 The Fundamental Theorem of Calculus, Part I Preliminary Questions 1. Suppose that F that F 0 .x/ D f.x/ and f.x/ and F F .0/ .0/ D 3, F.2/ D 7. (a) What is the area under under y y
D f.x/ over f.x/ over Œ0 Œ0;; 2 if f f .x/ .x/ 0?
(b) What is the graphical graphical interpretation of F.2/ F.2/ F.0/ if F.0/ if f f .x/ takes x/ takes on both positive and negative values? SOLUTION
(a) If f f .x/ 0 over Œ0 over Œ0;; 2, then the area under y
D f.x/ is f.x/ is F.2/ F.2/ F.0/ D 7 3 D 4.
f .x/ takes x/ takes on both positive and negative values, then F.2/ F.0/ gives the signed area between y (b) If f then F.2/ F.0/ gives between y 2. Suppose that f that f .x/ .x/ is a negative function with antiderivative antiderivative F F such such that F that F .1/ .1/ number) between the x the x-axis -axis and the graph of f f .x/ .x/ over Œ1 over Œ1;; 3?
D f.x/ and f.x/ and the x the x-axis. -axis.
D 7 and F and F .3/ .3/ D 4. What is the area (a positive
3
SOLUTION
f.x/dx represents f.x/dx represents the signed area area bounded by t he curve and the interval Œ1 interval Œ1;; 3. Since f Since f .x/ .x/ is negative on Œ1 on Œ1;; 3,
1
3
Z
Z
f.x/dx is f.x/dx is the t he negative of the area. Therefore, if A is A is the area between the x-axis x -axis and the graph of f of f .x/ .x/, we have:
1
3
A
D
Z
f.x/dx
1
D .F.3/ F.1// D .4 7/ D .3/ D 3:
3. Are the following statements true or false? Explain. (a) FTC I is valid only for positive functions. (b) To use FTC I, you have to choose t he right antiderivative. (c) If you cannot find an antiderivative antiderivative of f f .x/ .x/, then the definite integral does not exist. SOLUTION
(a) False. The FTC I is valid for continuous functions. (b) False. The FTC I works for any antiderivative of the integrand. (c) False. If you cannot find an antiderivative of the integrand, you cannot use the FTC I to evaluate the definite integral, but the definite integral may still exist.
SECTION
9
4. Evaluate
Z 2
SOLUTION
f 0 .x/dx where .x/dx where f f .x/ .x/ is differentiable and f.2/ and f.2/ 9
Because f is is differentiable,
Z
f 0 .x/dx
2
The Fundamental Fundamental Theorem Theorem of Calculus, Calculus, Part I
5.3
D f.9/ D 4.
D f.9/ f.2/ D 4 4 D 0.
Exercises In Exercises Exercises 1–4, 1–4, sketch sketch the regio region n under under the graph graph of the function function and find find its area area using FTC I. 1. f.x/
D x 2 , Œ0; Œ0; 1
SOLUTION y
1 0.8 0.6 0.4 0.2 x
0.2
0.4
0.6
0.8
1
We have the area 1
A
Z D
2
x dx
0
2. f.x/
2 x ,
D 2x
D
1 3 x 3
Œ0;2
1
ˇˇ D ˇ
1 : 3
0
SOLUTION y
1 0.8 0.6 0.4 0.2 x
0.5
1
1.5
2
Let A Let A be be the area indicated. Then: 2
A
Z D
2
.2x x / dx
0
3. f.x/
Dx
2,
2
Z D
2
2x dx
0
Z
2
ˇˇˇ
2
Dx
2
1.6
1.8
x dx
0
Œ1;2
0
1 3 x 3
2
ˇˇˇ D 0
SOLUTION y
1.0 0.8 0.6 0.4 0.2
x
1.0
1.2
1.4
2.0
We have the area 2
A
Z D 1
4. f.x/
D cos x ,
SOLUTION
0;
2
x
2
dx
D
x
1
1
2
ˇˇˇ ˇD 1
1 2
C 1 D 12 :
.4 0/
8 3
0
D 43 :
591
592
THE INTEGRAL
CHAPTER 5
y
1 0.8 0.6 0.4 0.2 x
0.2 0.4 0.6 0.6 0.8 0.8 1 1.2 1.2 1.4 1.6
Let A Let A be be the shaded area. Then =2
=2
AD
Z
cos x dx D sin x
0
In Exercises Exercises 5–42, 5–42, evaluate evaluate the integr integral al using FTC I.
ˇˇ ˇ
D 1 0 D 1:
0
6
5.
Z
x dx
3
6
Z
SOLUTION
x dx
D
3
9
6.
Z
6
ˇˇ D ˇ
1 2 x 2
3
2 dx
1 2 1 2 .6/ .3/ 2 2
D 272 .
0
9
9
Z
SOLUTION
2 dx
0
1
7.
Z
.4x
9x
2
D 2x
/ dx
ˇˇ D ˇ
2.9/ 2.0/
0
D 18. 18.
0
1
SOLUTION
Z
2
.4x 9x / dx
0
2
8.
Z
D .2x
2
1
ˇˇ D ˇ
3
3x /
.2 3/ .0 0/
0
u2 du
D 1.
3
2
SOLUTION
Z
2
u du
D
3
2
9.
Z
.12x 5
0
C 3x2 4x/dx 2
SOLUTION
Z
.12x
5
0
2
10.
Z
.10x 9
2
C 3x
Z
.10x
9
2
0
11.
Z
2
2
ˇˇ ˇ
3
D 13 .2/3 13 .3/3 D 353 .
6
4x/dx
D .2x C x
3
2
C 3x
5
/ dx
D
x
10
C
1 6 x 2
.2t 3 6t 2 / dt
2
ˇˇ D ˇ
2x /
.128
0
C 3x5 / dx 2
SOLUTION
1 3 u 3
2
ˇˇ
ˇ
2
D
10
2
C
C 8 8/ .0 C 0 0/ D 128. 128.
1 6 2 2
10
2
C
1 6 2 2
D 0:
3
0
SOLUTION
Z
.2t
3
2
6t / dt
3
1
12.
Z
.5u4
1
Z
4
13.
4
.5u 1
Z p
1 4 t 2t 3 2
C u2 u/du 1
SOLUTION
D
2
Cu
u/du
D
5
u
C
ˇˇ
0
ˇD
.0 0/
3
1 3 1 2 u u 3 2
y dy
ˇˇ
ˇ
1 1
81 2
D 272 .
1 3
1 2
54
D 1C
0
SOLUTION
4
4
Z p D Z y dy
0
0
ˇˇ
4
D 23 y3=2 D 23 .4/3=2 23 .0/3=2 D 163 .
y 1=2 dy
0
1
1 3
1 2
D 83 .
SECTION
8
14.
Z
The Fundamental Fundamental Theorem Theorem of Calculus, Calculus, Part I
5.3
x 4=3 dx
1
8
SOLUTION
Z
x
4=3
dx
1
1
15.
Z
D
D
4 5=4 t 5
t 1=4 dt
1=16 1
SOLUTION
Z
t
1=4
dt
1=16
1
16.
Z
8
ˇˇ D ˇ ˇˇ D ˇ
3 7=3 x 7
t 5=2 dt
3 .128 1/ 7
1
1
1=16
4 5
1 40
D 381 . 7
D 31 . 40
4
1
SOLUTION
Z
t
5=2
dt
4
3
17.
D
dt t2
Z 1
3
SOLUTION
Z
x
4
3
dt t2
Z 1
4
18.
2 7=2 t 7
Z D
t
dx
1 x 3
2
1
ˇˇ D ˇ
dt
1
2 .1 128/ 7
4
D t
1
3
ˇˇ D ˇ
1
dx
D 254 . 7
1 3
C 1 D 23 .
1
4
SOLUTION
Z
x
4
D
1
1
19.
1=2
1
SOLUTION
8 d x x3
Z
1=2
1
20.
2
1
SOLUTION
Z
21.
.x 2 x
1 x3
2
2
8x
3
dx
1=2
1 .4/ 3
3
D 4x
C 13 D 21 . 64
2
1 d x x3
Z
1
Z D
1
8 d x x3
Z Z
4
ˇˇˇ D
3
2
d x
D
1 x 2
1
2
ˇˇˇ
1 1=2
D 4 C 16 D 12. 12.
D 12 .1/ 2 C 12 .2/ 2 D 38 .
2
/ dx
ˇˇ ˇ
1
2
SOLUTION
Z
.x
2
x
2
/ dx
1
9
22.
Z
t
1=2
D
1 3 x 3
Cx
1
2
ˇˇˇ
ˇ
dt
1
9
SOLUTION
Z
t
1=2
dt
1
27
23.
Z
t
1
D 2t
1=2
9
ˇˇ D ˇ
2.9/1=2
1
D 83 C 12
2.1/1=2
1
C 1 dt p t
1 3
C 1 D 116 :
D 4.
SOLUTION
27
Z 1
1
24.
Z
10t 4=3
8=27
8t
t
C 1 dt D p t
27
Z
.t
1=2
Ct
1=2
/ dt
2 3=2 t 3
1=2
27
ˇˇˇ
C 2t 1 2 p p 2 p 81 D 3 .81 3/ C 6 3 3 C 2 D 60 3 3 : D
ˇ
1=3
dt
t2
SOLUTION
1
Z
8=27
10t 4=3 8t 1=3 dt t2
1
Z D
.10t
2=3
8t
5=3
/ dt
8=27
D .30t
1=3
C 12t
2=3
ˇˇ ˇ
1
/
8=27
D .30 C 12/ .20 C 27/ D 5:
593
594
THE INTEGRAL
CHAPTER 5
3=4
25.
Z
sin d d
=4
3 =4
Z
SOLUTION
ˇˇ ˇ
sin d d
D D cos
=4
4
26.
Z
sin x dx
3=4
D
=4
p
2 2
p
2 2
C
D
p
2:
2 4
4
ˇ Z ˇˇ D D D Z ! Z ! D D ˇˇˇ D ˇ Z p ˇ Z ˇ D D ˇ D Z Z D ˇˇ D p C p D p Z p ˇ Z D D ˇˇ D D Z ˇ Z ˇˇ D p D p D Z ˇ Z ˇˇˇ D C D D Z Z D ˇˇˇ D Z Z D ˇˇˇ D C Z Z D ˇˇ D C D Z sin x dx
SOLUTION
cos x
2
=2
27.
1 .1/
0:
2
1 d 3
cos
0
=2
1 d 3
cos
SOLUTION
0
=2
1 3
3 sin
3 . 2
0
5=8
28.
cos 2x dx
=4
5 =8
1 sin 2x 2
cos 2x dx
SOLUTION
=4
=6
29.
sec2 3t
=6
sec2 3t
SOLUTION
1 5 sin 2 4
=4
1 sin 2 2
2 4
1 . 2
d t
6
0
5=8
0
6
1 tan 3t 3
d t
=6
6
1 3
0
1
3
4
3
3 3
=6
30.
sec tan tan d d
0
=6
=6
sec tan tan d d
SOLUTION
sec
0
sec
0
6
2 3 3
sec 0
1.
=10
31.
csc 5x cot 5x cot 5x dx
=20
=10
5x cot 5x dx csc 5x cot
SOLUTION
=20
=14
32.
=10
1 csc 5x 5
=20
1 1 5
1 . 2 1/. 1/. 5
2
csc2 7y dy
=28
=14
SOLUTION
csc2 7y dy
=28
1
33.
1 cot 7y 7
=14
=28
1 cot 7 2
1 cot 7 4
1 . 7
e x dx
0
1
SOLUTION
ex dx
ex
0
5
34.
e
4x
1
e 1.
0
dx
3
5
e
SOLUTION
4x
dx
3
3
35.
e1
6t
1 e 4
4x
5
3
1 e 4
20
1 e 4
12
.
dt
0
3
0
3
36.
2
e 4t
e1
6t
SOLUTION
3
dt
dt
1 1 e 6
6t
3
0
1 e 6
17
1 e 6
1 .e e 6
17
/.
.
SECTION
3
Z
SOLUTION
e
4t 3
dt
2
10
37.
2
10
Z
SOLUTION
2
4
38.
12
Z
SOLUTION
4
12
1
39.
C1
1
Z
SOLUTION
4
Z
C4
4
Z
SOLUTION
41.
1 9 1 5 e e . 4 4
D ln 10 ln 2 D ln 5.
2
4
D ln j4j ln j12j D ln 13 D ln 3.
12
1
ˇ j C jˇˇ D
C 1 D ln t dt
D 5t C 4
1
0
Z
2
D ln x
3
ˇˇˇ D ˇ
10
ˇj jˇˇ ˇ
dt 5t
1
dx x
dt t
0
40.
D ln x
dt t
0
ˇj jˇ ˇ
dx x
dx x
Z
D
3
dx x
Z Z
1 4t e 4
The Fundamental Fundamental Theorem Theorem of Calculus, Calculus, Part I
5.3
1
ln 2 ln 1
D ln 2.
0
4
ˇ j C jˇˇ D
1 ln 5t 5
.3x 9e 3x / dx
4
1
1 1 ln 24 ln 9 5 5
D 15 ln 249 .
2
0
Z
SOLUTION
.3x 9e
3x
/ dx
2
6
42.
Z C Z C
D
ˇˇˇ
3 2 x 3e 3x 2
ˇ
1 d x x
x
2
6
x
SOLUTION
2
1 d x x
1 2 x 2
D
0 2
D .0 3/ .6 3e
6
/
D 3e
6
9.
6
ˇˇ
C ln jx j
ˇD
.18
2
C ln 6/ .2 C ln 2/ D 16 C ln 3.
In Exercises Exercises 43–48, 43–48, write the integral integral as as a sum of of integrals integrals without without absolute absolute values and and evaluate. evaluate. 1
43.
Z j j
x dx
2
SOLUTION
1
0
Z j j D Z
1
x dx
2
5
44.
Z j
.x/dx 2
Z C
x dx
D
0
1 2 x 2
3 x dx
j
0
0
ˇˇ
2
C
1 2 x 2
1
ˇˇ D
0
0
1 .4/ 2
C 12 D 52 :
SOLUTION
5
Z j
3
3 x dx
0
j
Z D
3
45.
.3 x/dx
0
D
Z j
5
Z C
.x 3/dx
3
9
9 2
0
C
25 2
D
15
9 2
3x
9
1 2 x 2
3
ˇˇˇ 0
C
C
1 4 x 4
ˇ
D 132 :
1 2 x 3x 2
x 3 dx
2
j
SOLUTION
3
Z j 2
x
3
0
j dx
Z D
3
.x / dx 2
3
Z C 0
3
x dx
D
1 4 x 4
D 0 C 14 .2/4 C 14 34 0 D 974 :
ˇˇˇ
0 2
ˇˇˇ
3 0
5
ˇˇˇ
ˇ
3
595
596
3
46.
THE INTEGRAL
CHAPTER 5
Z j
x 2 1 dx
j
0
SOLUTION
3
Z j
x
2
1
1
j dx
0
Z D
3
Z C
2
.1 x / dx
0
D
.x
2
1/dx
D
1
1
1 3
0
C .9 3/
1 3
1
x
1
ˇˇ
1 3 x 3
ˇ
D 223 :
0
C
1 3 x x 3
ˇˇ
ˇ
3 1
47.
Z j
cos x dx
j
0
SOLUTION =2
Z j
cos x dx
j
0
5
48.
Z j
x 2 4x
0
Z D
Z C
cos x dx
0
. cos x/dx
D sin x
=2
C 3j dx
ˇˇ ˇ
=2
sin x
ˇˇ ˇ
=2
0
D 1 0 .1 0/ D 2:
SOLUTION
5
Z j
x
2
5
4x
0
C 3j dx
Z D j Z D
.x 3/.x 1/ dx
j
0
1
.x
2
3
4x
0
D D
1 3 x 2x 2 3 1 3
2
C 3/dx C 3x
.x
2
5
4x
1
1
C3
Z C
ˇˇ
ˇ
0
1 3 x 2x 2 3
0 .9 18
C 9/ C
C 3/dx 3
C 3x
1 3
2
Z C
.x 2 4x
C 3/dx
3
ˇˇ ˇ C
1 3 x 2x 2 3
1
C3 C
125 3
50
5
ˇˇ ˇ
C 3x C 15
D 283 : In Exercises Exercises 49–54, 49–54, evaluate evaluate the integral integral in terms terms of the constants. constants. b
49.
Z
x 3 dx
1
b
Z
SOLUTION
3
x dx
1
a
50.
Z
D
1 4 x 4
x 4 dx
b
a
Z
SOLUTION
x 4 dx
b
b
51.
Z
D 15 x 5
x 5 dx
b
ˇˇˇ D ˇ ˇˇˇ D 1
a b
1 4 1 4 b .1/ 4 4
D 14
b 4 1 for any number b number b..
1 5 1 5 a b for any numbers a b . numbers a,, b. 5 5
1
b
Z
SOLUTION
5
x dx
1
x
52.
Z
.t 3
x
D
1 6 x 6
C t/dt
b
ˇˇˇ D 1
1 6 1 6 b .1/ 6 6
D 16 .b6 1/ for 1/ for any number b number b..
SOLUTION
x
Z
.t
3
x
5a
53.
Z a
C t / d t D
dx x
SOLUTION
5a
Z a
dx x
ˇj jˇˇ ˇ
D ln x
1 4 t 4
C
1 2 t 2
ˇˇˇ
ˇ
x
5a a
x
D ln j5aj ln jaj D ln 5.
D
1 4 x 4
C
1 2 x 2
1 4 x 4
C
1 2 x 2
D 0:
3
.9 18
C 9/
SECTION
b2
54.
Z b
The Fundamental Fundamental Theorem Theorem of Calculus, Calculus, Part I
5.3
597
dx x b2
SOLUTION
Z Z
dx x
b
ˇj jˇ ˇ
D ln x
3
55. Calculate
b2 b
D ln jb2 j ln jbj D ln jbj.
f.x/dx, f.x/dx, where 2
f.x/
D
(
12 x 2
for x for x
x3
for x for x > 2
3
Z
2
SOLUTION
3
Z
2
2
Z
f.x/dx D
2
f.x/dx C 1 3 x 3
D
12x
f.x/dx D
2
2
ˇˇ ˇ C
1 4 x 4
2
1 3 2 3
12.2/
D
2
Z
3
ˇˇ ˇ
.12 x 2
3
/ dx C
Z
x 3 dx
2
2
1 .2/3 3
cos x
for x for x
cos x sin 2x
for x for x >
12.2/
2
C 14 34 14 24
65 707 D 128 C D : 3 4 12 2
56. Calculate
Z
f.x/dx, f.x/dx , where
0
f.x/
D
(
SOLUTION
2
Z
2
f.x/dx
0
Z D
f.x/dx
0
ˇˇ C ˇ
sin x
0
D .0 0/ C 1
57. Use FTC I to show show that
1
SOLUTION
x n dx
0
C
2
Z D ˇˇ ˇ
f.x/dx
cos x dx
0
D sin x
Z
Z C
C 12 cos 2x 1 2
0
C
Z C
.cos x sin 2x/dx
2
1 2
D 0:
D 0 if n is n is an odd whole number. Explain graphically.
We have 1
Z
n
x dx 1
Because n Because n is is odd, n odd, n
D
x nC1 n 1
1
ˇˇ C ˇ
n 1
C
1
D .1/ nC1
.1/nC1 : n 1
C
C 1 is even, which means that . that .1/nC1 D .1/nC1 D 1. Hence .1/nC1 n 1
C
.1/nC1 n 1
D n C1 1 n C1 1 D 0:
C
Graphically speaking, for an odd function such as as x x 3 shown here, the positively signed area from x negatively negatively signed area from x from x to x 0. 1 to x
D
D
y
1 0.5 !
0.5
x
1
0.5
!
!
0.5 1
!
1
D 0 to x D 1 cancels the
598
THE INTEGRAL
CHAPTER 5
58. Plot the function f function f .x/ .x/ graph of f f .x/ .x/ in the first quadrant.
D sin 3x x . Find the positive root of f f .x/ .x/ to three places and use it to find the area under the
SOLUTION The graph of f f .x/ .x/ sin 3x x is shown below at the l eft. In the figure below at the right, we zoom in on the positive f .x/ .x/ and find that, to three decimal places, this root is approximately x 0:760. 0:760. The area under the graph of f .x/ in the root of f of f .x/ first quadrant is then
D
D
0:760
Z
.sin 3x x/dx
D
0
1 1 2 cos 3x x 3 2
0:760
ˇˇˇ
ˇ
0
1 1 D 13 cos.2:28/ cos.2:28/ .0:760/2 C 0:262 2 3 y
0.5 x !
0.2 !
0 .4 .4
0 .6 .6
0 .8 .8
x
1
0.75 0.756 6
0.75 0.758 8
0.76 0.76
0.76 0.762 2
0.76 0.764 4
0.5
59. Calculate F.4/ Calculate F.4/ given given that F.1/ that F.1/ SOLUTION
0 .2 .2
D 3 and F .4/ F.1/ as F.1/ as a definite integral. and F 0 .x/ D x 2 . Hint: Express F Express F .4/
By FTC I, 4
F.4/ F.1/
D
Z
3 13
D4
x 2 dx
1
3
D 21
.4/ Therefore F Therefore F .4/
D F.1/ C 21 D 3 C 21 D 24. 24. 60. Calculate G.16/ Calculate G.16/,, where dG=dt where dG=dt D t 1=2 and G.9/ and G.9/ D 5.
SOLUTION
By FTC I, 16
G.16/ G.9/
Z D
t
1=2
dt
9
Therefore G.16/ Therefore G.16/
D 5 C 2 D 3. 1
61.
D 2.161=2 / 2.91=2 / D 2
Does
Z
x n dx get dx get larger or smaller as as n n increases? increases? Explain graphically. graphically.
0
Let n possibility.) Now SOLUTION
0 and consider 1
Z 0
n
x dx
D
1 0
x n dx . (Note: for n < 0 the integrand x n
R
1 n
C1
x
1
n 1
C
1
ˇˇ
ˇD 0
1 n
n 1
C1
.1/
C
1 n
C1
! 1 as x ! 0C, so we exclude this n 1
.0/
C
D n C1 1 ;
which decreases as n as n increases. increases. Recall that 0 x n dx represents dx represents the area between the positive curve f curve f .x/ .x/ x n and the x the x-axis -axis over the interval Œ0 interval Œ0;; 1. Accordingly, this area gets smaller as n gets n gets larger. This is readily evident in the following graph, which shows curves for several values of n of n..
R
D
y
1 1/4 1/2 1 2 4
8 x
0
1
62. Show that the area of the shaded parabolic arch in Figure 1 is equal to four-thirds the area of the triangle shown. y
x a
a + b
b
2
y FIGURE 1 Graph of y
D .x a/.b x/. x/ .
SECTION
SOLUTION
The Fundamental Fundamental Theorem Theorem of Calculus, Calculus, Part I
5.3
599
We first calculate the area area of the parabolic arch: arch: b
Z
b
.x a/.b x/dx
a
D
Z
D
b
.x a/.x b/dx
D
a
D 16 D 16 D 16 D 16
1 3 a 2 b 2 x x x 3 2 2
bx
C ab/dx
b
.2b 3 3ab 2 3b 3
a3
.x 2 ax
a
C abx
2x 3 3ax 2 3bx 2
.b 3
Z
ˇˇ
ˇ
a
C 6abx
ˇˇ
b a
C 6ab2 / .2a3 3a3 3ba 2 C 6a2 b/
C 3ab2 / .a3 C 3a2 b/
C 3ab2 3a2 b b3 D 16 .b a/3 :
The indicated triangle has a base of length b length b a and a height of
a
C b a b a C b D b a 2 : 2
2
2
Thus, the area of the triangle is 1 .b a/ 2
b a 2
2
D 18 .b a/3 :
Finally, we note that 1 .b a/3 6
D 43 18 .b a/3 ;
as required.
Further Insights and Challenges 63. Prove a famous result of Archimedes (generalizing (generalizing Exercise 62): For r For r < s , the area of the shaded region in Figure 2 is equal to four-thirds the area of triangle ACE, ACE , where C where C is is the point on the parabola at which the tangent line is parallel to secant line AE. AE .
4
(a) Show that C that C has x has x-coordinate -coordinate .r .r s/=2. s/=2. (b) Show that ABDE that ABDE has has area .s area .s r/ 3 =4 by =4 by viewing it as a parallelogram of height s height s r and base of length CF . CF . 3 ACE has area .s =8 by observing t hat it has the same base and height as the parallelogram. (c) Show that ACE has area .s r / =8 by (d) Compute the shaded area as the area under the graph minus the area of a tr apezoid, and prove Archimedes’ result.
C
4
y B
C
D
A
F
E
r
r + s
x
s
2
FIGURE 2 Graph of f f .x/ .x/
D .x a/.b x/. x/ .
SOLUTION
(a) The slope of the secant secant line AE line AE is is
f.s/ f.r/ sr
D .s a/.b s/s .rr a/.b r/ D a C b .r C s/
and the slope of the tangent line along the parabola is f 0 .x/
D a C b 2x:
If C is C is the point on the parabola at which the tangent line is parallel to the secant line AE, AE , then its x its x-coordinate -coordinate must satisfy a
C b 2x D a C b .r C s/
or
x
D r C2 s :
600
THE INTEGRAL
CHAPTER 5
(b) Parallelogram ABDE Parallelogram ABDE has has height s height s
r and
base of length CF . CF . Since the equation of the secant line AE line AE is is
y
D Œa C b .r C s/ .x r/ C .r a/.b r/;
the length of the segment CF segment CF is is
C s a b r C s Œa C b .r C s/ r C s r .r a/.b r/ D .s r /2 :
r
2
2
2
4
.s r/ 3 . 4
Thus, the area of ABDE is ABDE is
2 r (c) Triangle ACE ACF and Triangle ACE is is comprised of and CEF . CEF . Each of these smaller triangles has height s base of length .s 4r/ . 2 and base ACE is ACE is Thus, the area of
1sr 2 2
.s r/ 2 4
2
C 12 s 2 r .s 4 r / D .s 8r / between x D r and x and x D s is (d) The area under the graph of the parabola between x s
Z
.x a/.b x/dx
r
D
abx
C
1 .a 2
C b/x
2
1 3 x 3
3
:
s
ˇˇ
ˇ
r
D abs C 12 .a C b/s 2 13 s3 C abr 12 .a C b/r 2 C 13 r 3 D ab.r s/ C 12 .a C b/.s r/.s C r / C 13 .r s/.r 2 C rs C s2 /; while the area of the tr apezoid under the shaded region is 1 .s r/Œ.s a/.b s/ 2
D 12 .s r /
C .r a/.b r/
h
2ab
C .a C b/.r C s/ r 2 s2
i
D ab.r s/ C 12 .a C b/.s r/.r C s/ C 12 .r s/.r 2 C s2/: Thus, the area of the shaded region is .r
s/
1 2 r 3
C
1 rs 3
C
1 2 1 2 1 2 s r s 3 2 2
D .s r /
1 2 1 r rs 6 3
C
1 2 s 6
D 16 .s r/ 3 ;
which is four-thirds the area of the triangle ACE triangle ACE . 64. (a) Apply the Comparison Theorem (Theorem 5 in Section 5.2) to the inequality sin x
1
x2 2
cos x
x (valid for x for x
0) to prove that
1
(b) Apply it again to prove that that
x (c) Verify these inequalities for x for x
x3 6
sin x
x
.for x for x
0/
D 0:3. 0:3.
SOLUTION
x
x
(a) We have
Z
sin t d t
0
Solving, this gives cos x
D cos t
1
x2 2 .
gives us 1 us 1 (b) The previous part gives
ˇˇˇ D
x
cos x
0
x
ˇˇˇ D 0
1 2 x . Hence 2
C 1 x2 :
1 foll ows automatically.
cos t
1, for t for t > 0. 0 . Theorem 5 gives us, after integrating over the interval Œ0; Œ0; x , x3 6
sin x
x:
D 0:3 into 0:3 into the inequalities obtained in (a) and (b) yields 0:955 0:955336489 1
respectively.
0
D
1 2 t 2 2
x (c) Substituting x Substituting x
tdt
cos x
cos x
t2 2
C 1 and
Z
and 0:2955
0:2955202069 0:3;
SECTION
The Fundamental Fundamental Theorem Theorem of Calculus, Calculus, Part I
5.3
601
65. Use the method of Exercise 64 to prove that
Verify these i nequalities for x for x cos x , yields:
These inequalities apply for x for x Having established that
x2 2
cos x
1
x2 2
C x24
x
x3 6
sin x
x
x3 6
x C 120 (for x (for x 0) 0 )
1 3
6t
0 . 0.
sin t
5
t for t for t > 0. 0 . Integrating this inequality over the interval Œ0 interval Œ0;; x , and then solving for
1 2 1 4 x x 2 24
1 cos x
1
1 2 x 2
cos x
Since cos x , 1
x2 2 ,
and 1 and 1
1 for all t all t
4
D 0:1. 0:1. Why have we specified x 0 for sin x but not for cos x ?
By Exercise 64, t 64, t
SOLUTION
1
0, 0, we integrate over the
t2 2
cos t
1 2 x 2
1
1 2 x 2
x2 2
C x24 are all even functions, they also apply apply for x for x 0.
C 241 x 4 :
4
1
t2 2
t C 24 ;
4
x
x3 6
x C 120 :
interval Œ0 interval Œ0;; x , to obtain: x
x3 6
sin x
5
1 5 The functions sin x , x 16 x 3 and x and x 16 x 3 for x < 0. 0. 120 x are all odd functions, so the inequalities are reversed for x Evaluating these inequalities at x at x 0:1 yields 0:1 yields
C
D
0:995000000 0:995004165 0:995004167 0:0998333333 0:0998334166 0:0998334167; both of which are true. 66. Calculate the next pair of inequalities for sin x and cos x by integrating the results of Exercise 65. Can you guess the general pattern? SOLUTION
Integrating t
t3 6
sin t
t
t3 6
5
t C 120
(for t (for t
x4 24
6
0)
over the interval Œ0 interval Œ0;; x yields x2 2
x4 24
1 cos x
4
6
x2 2
x C 720 :
Solving for cos x yields 1
x2 2
2
4
x x x cos x 1 C x24 720 C : 2 24
Replacing each x each x by by t t and and integrating over the interval Œ0 interval Œ0;; x produces x
x3 6
5
7
3
5
x x x x sin x x C 120 C : 5040 6 120
To see the pattern, it i s best to compare consecutive inequalities for sin x and those for cos x : 0 sin x
x
x
x3 6
sin x
x
x
x3 6
sin x
x
x3 6
5
x C 120 :
602
THE INTEGRAL
CHAPTER 5
Each iteration adds an additional term. Looking at the highest order terms, we get the following pattern: 0 x
x3 6
3
D x3Š
x5 5Š
We guess that the leading t erm of the polynomials are of t he form x 2n C1 : .2n 1/Š
.1/n
C
Similarly, for cos x , the leading terms of the polynomials in the inequality are of the form x 2n : .2n/Š Œa;b , then f.x/ f.a/ .1/n
67. Use FTC I to prove prove that if f 0 .x/
j
j K for x for x 2
j K jx aj for x for x 2 Œa; Œa; b .
j
Let a Let a > b be real numbers, and let f let f .x/ .x/ be such that jf 0 .x/ j K for x for x 2 Œa;b . By FTC,
SOLUTION
x
Z a
Since f Since f 0 .x/
K for for
all x all x
f 0 .t/dt
D f.x/ f.a/:
2 Œa; Œa; b , we get:
x
f.x/ f.a/
Z D a
Since f Since f 0 .x/
K for for all x all x
2 Œa;b, we get:
f 0 .t/dt
K.x
a/:
x
f.x/ f.a/
Z D a
f 0 .t/dt
K.x
a/:
Combining these two inequalities yields K.x
a/
f.x/ f.a/ K.x
a/;
so that, by definition,
jf.x/ f.a/j K jx aj: 68. (a) Use Exercise 67 to prove that j sin a sin b j ja b j for all a; all a; b . (b) Let f Let f .x/ .x/ D sin.x sin.x C a/ sin x . Use part (a) to show that t he graph of f f .x/ .x/ lies between the horizontal lines y D ˙a. (c) Plot f Plot f .x/ .x/ and the lines y lines y D ˙a to verify (b) for a for a D 0:5 and 0:5 and a a D 0:2. 0:2. SOLUTION
(a) Let f Let f .x/ .x/
D sin x, so that f that f 0 .x/ D cos x , and jf 0 .x/j 1
for all x all x.. From Exercise 67, we get:
jsin a sin bj ja bj: (b) Let f Let f .x/ .x/
D sin.x sin.x C a/ sin.x/ sin.x/.. Applying (a), we get the inequality: jf.x/j D jsin.x sin.x C a/ sin.x/ sin.x/j j.x C a x/ j D jaj:
This is equivalent, by definition, to the t wo inequalities: a
sin.x sin.x
C a/ sin.x/ sin.x/ a: (c) The plots of y y D sin.x sin.x C 0:5/ sin.x/ sin.x/ and and of y y D sin.x sin.x C 0:2/ sin.x/ sin.x/ are are shown below. below. The inequality i s satisfied in both
plots.
y
y
0.5
0.2
0.25
0.1 x
!
4
2 ! 0.25
!
!
0.5
2
4
x !
4
!
2
2
!
0.1
!
0.2
4
SECTION
The Fundamental Fundamental Theorem Theorem of Calculus, Calculus, Part II
5.4
603
5.4 The Fundamental Theorem of Calculus, Part II Preliminary Questions x
Z p C D
1. Let G.x/ Let G.x/
t3
1 dt .
4
(a) Is the FTC needed needed to calculate G.4/ calculate G.4/??
needed to calculate G calculate G 0 .4/? .4/? (b) Is the FTC needed SOLUTION
(a) No. G.4/ No. G.4/
R p C 4 4
t3
Dp 0. p (b) Yes. By the FTC II, G II, G 0 .x/ D x 3 C 1, so G so G 0 .4/ D 65. 65. 2. Which of the following is an antiderivative antiderivative F.x/ of F.x/ of f f .x/ .x/ D x 2 satisfying F.2/ satisfying F.2/ D 0? D
1 dt
x
(a)
Z
2
2t dt
(b)
2
Z
2
t dt
(c)
0
x
The correct answer is (c) :
SOLUTION
Z
x
Z
t 2 dt
2
t 2 dt .
2
3. Does every continuous function have an antiderivative? antiderivative? Explain. x
Yes. All continuous functions have an antiderivative, antiderivative, namely
SOLUTION
4. Let G.x/ Let G.x/
f.t/dt .
a
x3
Z D
Z
sin t dt . Which of the following st atements are correct?
4
(a) G.x/ is G.x/ is the composite function sin.x sin.x 3/. (b) G.x/ is G.x/ is the composite function A.x function A.x 3 /, where x
A.x/
Z D
sin.t/dt sin.t/dt
4
(c) G.x/ is G.x/ is too complicated to differentiate. (d) The Product Rule is used to differentiate differentiate G.x/ G.x/.. (e) The Chain Rule is used to differentiate differentiate G.x/ G.x/.. (f) G 0 .x/
D 3x2 sin.x sin.x 3 /.
SOLUTION
Statements (b) , (e), and (f) are correct.
Exercises 1. Write the area function of f f .x/ .x/ SOLUTION
D 2x C 4 with lower limit a limit a D 2 as an integral and find a formula for it. Let f Let f .x/ D 2x C 4. The area function with lower limit a limit a D 2 is x
Z
A.x/ D
x
f.t/dt D
a
Z
.2t
C 4/dt:
2
Carrying out the integration, we find x
Z
.2t 2
or .x or .x
ˇˇ ˇ
2
C 4/dt D .t C 4t /
C 2/2 . Therefore, A.x/ Therefore, A.x/ D .x C 2/2 .
x
D .x2 C 4x/ ..2/2 C 4.2// D x 2 C 4x C 4
2
2. Find a formula for the area function of f f .x/ .x/ SOLUTION
D 2x C 4 with lower limit a limit a D 0. The area function for f for f .x/ .x/ D 2x C 4 with lower limit a limit a D 0 is given by x
A.x/
Z D
.2t
0
3. Let G.x/ Let G.x/
x 2 1 .t 2/dt .
R
D
x
G.x/
.t
1
2
C 4/dt D .t C 4t /
0
x2
C 4x:
.1/ and G G 0 .2/. .2/. Then find a formula for G.x/. G.x/. Calculate G.1/ Calculate G.1/,, G 0 .1/ and
x 2 1 .t 2/dt .
R Z D
Let G.x/ Let G.x/ 2. Finally,
SOLUTION
G 0 .2/ D
D
x
ˇˇˇ D
2
2/dt
D
Then G.1/ Then G.1/
1 3 t 2t 3
D
1 2 1 .t 2/dt
R x
ˇˇ
ˇD 1
D 0. Moreover, G Moreover, G 0 .x/ D x 2 2, so that G that G 0 .1/ D 1 and
1 3 x 2x 3
1 3 .1/ 2.1/ 3
D 13 x3 2x C 53 :
604
THE INTEGRAL
CHAPTER 5
x
Z p C
4. Find F Find F .0/ .0/, F 0 .0/, .0/, and F and F 0 .3/, .3/, where F.x/ where F.x/ D
By definition, F.0/ definition, F.0/
p D
SOLUTION
p p F 0 .3/ D 32 C 3 D 12 D 2
3.
R p C 0 0
t2
By definition, G.1/ definition, G.1/
D
1 1 tan
R
t dt
p 2 x
C x, so that F 0 .0/ D
By definition, H. definition, H. 2/
SOLUTION
x
7.
Z
du
C 1. 2 du 1 1 D 0: By 0: By FTC, H FTC, H 0 .x/ D 2 , so H so H 0 .2/ D . 2 u C1 x C1 5 u2
2
u4 du
2
x
SOLUTION
F.x/
Z D
4
u
2
x
8.
Z
1 5 u 5
du D
.12t 2 8t/dt
x
ˇˇˇ D ˇ
1 5 32 x . 5 5
2
2
x
SOLUTION
F.x/
Z D
.12t
2
8t/dt
2
x
9.
Z
3
D .4t
2
x
ˇˇ D ˇ
4x 3 4x 2 16. 16.
4t /
2
sin u du
0
x
x
SOLUTION
F.x/
Z D
sin u du
0
x
10.
Z
ˇˇˇ D ˇ
D . cos u/
sec2 d d
1 cos x .
0
=4
x
SOLUTION
F.x/
Z D
2
x
11.
Z
ˇˇˇ ˇ
sec d d
D D tan
=4
e 3u du
x =4
D tan x tan. tan.=4/ D tan x C 1.
4
x
SOLUTION
F.x/
Z D
e
3u
du
4
0
12.
Z
e
t
D
1 3u e 3
dt
x
ˇˇ D ˇ
1 3x e 3
4
1 12 e . 3
x
0
SOLUTION
F.x/
Z D
e
t
dt
x
D e
t
13.
0
ˇˇ D ˇ
1
x
x2
Z
t dt
Ce
x
.
1
x2
SOLUTION
F.x/
Z D
t dt
1
x=4
14.
Z
D
1 2 t 2
sec2 u du
ˇˇ
x2 1
D 12 x 4 12 .
x=2 x=4
SOLUTION
F.x/
Z D
2
sec u du
x=2
9x 2
15.
Z C
e
u
ˇˇˇ ˇ
D tan u
du
x=4 x=2
D tan x4 tan x2 .
3x
SOLUTION
9x 2
F.x/
Z C D 3x
e
u
du
D e
u
D 0. By FTC, G FTC, G 0 .x/ D tan x , so that G that G 0 .0/ D tan 0 D 0 and G and G 0 . 4 / D tan 4 D 1.
In Exercises Exercises 7–16, 7–16, find formulas formulas for the functions functions repr represented esented by the integrals. integrals.
Z
C 0 D 0 and
tan t d t .
2
0
1
Z D
p 2
x
Z
x
6. Find H. Find H. 2/ and 2/ and H H 0 .2/, 2/, where H.x/ where H.x/ D
t d t.
D 0: By 0: By FTC, F FTC, F 0 .x/ D
tdt
5. Find G.1/ Find G.1/,, G 0 .0/, .0/, and G and G 0 . =4/, =4/, where G.x/ where G.x/ D SOLUTION
t2
0
9x 2
ˇˇ ˇ
3x
C
D e
9x 2
Ce
3x
.
SECTION
Z p
x
16.
The Fundamental Fundamental Theorem Theorem of Calculus, Calculus, Part II
5.4
dt t
2
Z p
x
SOLUTION
2
dt t
ˇj jˇˇp D ˇ x
D ln t
ln
2
p x ln 2 D 1 ln x ln 2. 2
f .x/ .x/ satisfying the given initial condition as an integral. In Exercises Exercises 17–20, 17–20, express express the the antiderivativ antiderivativee F.x/ of f 17. f.x/
p 3 x
D
D sec x, F.0/ D 0 x2 ,
F .4/
D sec x satisfying F.0/ satisfying F.0/ D 0 is F.x/ is F.x/
D0
F.x/ of f .x/ .x/ The antiderivative antider ivative F.x/ of f
SOLUTION
De
x2
satisfying F satisfying F .4/ x
Z D
F.x/
e
t2
D 0 is
dt :
4
In Exercises Exercises 21–24, 21–24, calculate calculate the the derivative. derivative. 21.
d dx
x
Z
.t 5 9t 3 / dt
0
d By FTC II, dx
SOLUTION
d 22. d
Z
d 23. dt
.t 5 9t 3 / dt
0
D x5 9x3:
cot u du By FTC II,
d d
t
Z
Z
cot u du
1
D cot :
sec.5x sec.5x 9/dx
100
By FTC II,
SOLUTION
d 24. ds
x
Z
1
SOLUTION
s
Z
tan 2
SOLUTION
1
C u2
1
D
Z
t
Z
sec.5x sec.5x 9/dx
100
D sec.5t sec.5t 9/:
d u
d By FTC II, ds x
Let A.x/ 25. Let A.x/
d dt
t3
1 dt .
5
x
Z 7
t t2
C 1 d t . C9
x
F.x/ of f .x/ .x/ The antiderivative antider ivative F.x/ of f
De
Z p C
D xx2CC19 satisfying F satisfying F .7/ .7/ D 0 is F.x/ is F.x/ D
The antiderivative antider ivative F.x/ F.x/ of of f f .x/ .x/
SOLUTION
20. f.x/
C 1 satisfying F satisfying F .5/ .5/ D 0 is F.x/ is F.x/ D
D xx2CC19 , F.7/ D 0
SOLUTION
19. f.x/
x
p
The antiderivative antider ivative F.x/ F.x/ of of x 3
SOLUTION
18. f.x/
C 1, F.5/ D 0
s
Z
tan 2
1 1
C u2
du
D tan 1 C1 s2
:
f . t / d t for t for f f .x/ in x/ in Figure 1.
0
(a) Calculate A.2/ Calculate A.2/,, A.3/, A.3/, A0 .2/, .2/, and A and A0 .3/. .3/.
for A.x/ on on Œ0 Œ0;; 2 and Œ2 and Œ2;; 4 and sketch the graph of A.x/ of A.x/.. (b) Find formulas for A.x/ y
4
y = f ( x x)
3 2 1
x
1
2
3
FIGURE 1
4
Z D 0
sec t d t .
605
606
THE INTEGRAL
CHAPTER 5
SOLUTION
D C 12 D 6:5, 6:5, the area under f .x/ .x/ from x from x D 0 D D D D D D Œ0; 2, the region under the graph of y y D f .x/ is x/ is a rectangle of length x and height 2; for each x 2 Œ2;4, the (b) For each x 2 Œ0;
(a) A.2/ 2 2 4, the area under f .x/ .x/ from x from x 0 to x to x 2, while A.3/ while A.3/ 2 3 to x to x 3. By the FTC, A FTC, A0 .x/ f.x/ so f.x/ so A A0 .2/ f.2/ 2 and A and A 0 .3/ f.3/ 3.
D
D
D
D
region is comprised of a square of side length 2 and a trapezoid of height x height x A.x/
D
2x ; 1 2 2x
2 and bases 2 and x. x . Hence,
0 x <2 2x4
C 2;
A graph of the area function A.x/ is A.x/ is shown below. y
10
Area Function A( x x)
8 6 4 2
x
1
2
3
4
x
26. Make a rough sketch of the graph of A.x/ A.x/
Z D
g.t/dt for g.t/dt for g.x/ g.x/ in in Figure 2.
0
y y = g( x x)
x
1
2
3
4
FIGURE 2 SOLUTION The graph of y y g.x/ lies g.x/ lies above the x the x-axis -axis over the interval Œ0 interval Œ0;; 1, below the x the x-axis -axis over Œ1 over Œ1;; 3, and above the x -axis over Œ3 over Œ3;; 4. The corresponding area function should therefore be increasing on .0;1/, .0;1/, decreasing on .1 on .1;; 3/ and 3/ and increasing on .3;4/ on .3;4/.. Further, it appears from Figure 2 that the local minimum of the area function at x 3 should 3 should be negative. One possible graph of the area function is the following.
D
D
y
4 3 2 1 x
1
!
1
2
3
4
2
!
3
!
x
27. Verify:
Z j j
t dt
0
SOLUTION
D 12 x jx j. Hint: Consider x Consider x 0 and x and x 0 separately.
Let f.t/ Let f.t/
D jt j D
(
t
for t for t
t
for t for t < 0
0
. Then x
x
F.x/
Z D
f.t/dt
0
8ˆZ ˆ < D ˆZ ˆ:
t dt
for x for x
0
0 x t d t
for x for x < 0
0
1 2 1 For x For x 0, 0 , we have F.x/ have F.x/ 2x 2 x x since x 1 Therefore, for all real x real x we we have F.x/ have F.x/ 2x x .
D
D
x2
Find G 0 .1/, .1/, where G.x/ where G.x/ D 28. Find G SOLUTION
D
jj
Z p C t3
jj
1 2 t 2
8ˆ ˆ < D ˆ ˆ:
x
ˇˇ D ˇ ˇ ˇˇ D 0
1 2 t 2
1 2 x 2
for x for x
x 0
1 2 x 2
0
for x for x < 0
j j D x , while for x for x < 0, we have F.x/ have F.x/ D 12 x 2 D 12 x jx j since jx j D x .
3 dt .
0
By combining the Chain Rule and FTC, G FTC, G 0 .x/
D
p 6 x
p C 3 2x, 2x , so G so G 0 .1/ D 1 C 3 2 D 4.
SECTION
The Fundamental Fundamental Theorem Theorem of Calculus, Calculus, Part II
5.4
In Exercises Exercises 29–34, 29–34, calculate calculate the the derivative. derivative. 29.
d dx
x2
Z
C
0
SOLUTION
30.
d dx
d ds
1=x
2
tdt t 1
3
x 2x D 2x D . 2 2 C x C1 x C1
Z 0
cos3 t d t
1
1=x
d By the Chain Chain Rule and and the FTC, dx
cos s
Z
Z
3
cos t d t
1
D cos
3
1 x
1 x2
D
1 cos 3 x2
u4 du
6
SOLUTION
d 32. dx
x2
d By the Chain Chain Rule and and the FTC, dx
Z
SOLUTION
31.
tdt t 1
x4
Z
x2
d By the Chain Chain Rule and and the FTC, ds
u4 du
6
D cos4 s. sin s/ D cos4 s sin s.
p
tdt
Hint for Exercise Exercise 32: F.x/ SOLUTION
cos s
Z
D A.x 4/ A.x2 /.
Let x4
F.x/
Z D
x4
p
tdt
x2
Z D
x2
p
tdt
0
Z
p
t dt:
0
Applying the Chain Rule combined with FTC, we have F 0 .x/ d 33. dx
p
x 4 4x 3
D
x2
Z
p x
SOLUTION
p
x 2 2x
D 4x5 2x jxj :
tan t d t Let x2
G.x/
Z D
x2
tan t d t
p x
Z p
x
Z D
tan t d t
0
tan t dt:
0
Applying the Chain Rule combined with FTC twice, we have
p 1 G 0 .x/ D tan.x tan.x 2 / 2x tan. tan. x/ x
1=2
2
34.
d du
D 2x tan 2x tan.x .x
2
/
p p
tan. tan. x/ : 2 x
3u
Z p C x2
1 dx
u
SOLUTION
Let
3u
G.x/
D
Z p C x2
3u
1 dx
u
D
Z p C x2
u
1 dx
0
p
D3
9u2
p C u2
1:
B.x/
Z D
C1C
x2
0
Applying the Chain Rule combined with FTC twice, we have G 0 .x/
Z p C
In Exercises Exercises 35–38, 35–38, with f .x/ as in Figure 3 let x
A.x/
Z D
x
f.t/dt
and
0
2
y
2
y = f ( x x)
1 0 1
!
1
2
3
4
2
!
FIGURE 3
5
6
x
f.t/dt .
1dx:
1 . x
607
608
THE INTEGRAL
CHAPTER 5
35. Find the min and max of A.x/ on A.x/ on Œ0 Œ0;; 6.
The minimum values values of of A.x/ on A.x/ on Œ0 Œ0;; 6 occur where A where A0 .x/ f.x/ goes f.x/ goes from negative to positive. This occurs at one place, where x where x 1:5. 1:5. The minimum value of A.x/ is A.x/ is therefore A.1:5/ therefore A.1:5/ 1:25. The maximum values of A.x/ of A.x/ on on Œ0 Œ0;; 6 occur 1:25. 0 where A where A .x/ f.x/ goes f.x/ goes from positive to negative. This occurs at one place, where x 4:5. 4:5. The maximum value of A.x/ of A.x/ is is therefore A.4:5/ therefore A.4:5/ 1:25. 1:25.
D D
SOLUTION
D D D
D
36. Find the min and max of B.x/ on B.x/ on Œ0 Œ0;; 6.
The minimum values values of of B.x/ on B.x/ on Œ0 Œ0;; 6 occur where B where B 0 .x/ f.x/ goes f.x/ goes from negative to positive. This occurs at one 1:5. 1:5. The minimum value of A.x/ is A.x/ is therefore B.1:5/ 0:25. The maximum values of B.x/ Œ0;; 6 occur 0:25. place, where x therefore B.1:5/ of B.x/ on on Œ0 0 where B where B .x/ f.x/ goes f.x/ goes from positive to negative. This occurs at one place, where x 4:5. 4:5. The maximum value of B.x/ of B.x/ is is therefore B.4:5/ therefore B.4:5/ 2:25. 2:25.
D D
SOLUTION
D D D
D
for A.x/ and and B.x/ B.x/ valid valid on Œ2 on Œ2;; 4. 37. Find formulas for A.x/ 2
SOLUTION
A.x/
On the interval Œ2 interval Œ2;; 4, A 0 .x/ D B 0 .x/ D f.x/ D 1. A.2/ D
D .x 2/ 1 and B.x/ 2/. and B.x/ D .x 2/.
Z
2
f.t/dt
0
D 1 and B.2/ and B.2/
Z D
f.t/dt
2
D 0. Hence
B.x/ valid 38. Find formulas for A.x/ for A.x/ and and B.x/ valid on Œ4 on Œ4;; 5. 4
SOLUTION
4
Z 2
f.t/dt
On the interval Œ interval Œ4; 4; 5, A 0 .x/
D B 0 .x/ D f .x/ D 2.x 4:5/ D 9 2x. 2x . A.4/ D
Z 0
f.t/dt
D 1 and B.4/ and B.4/ D
D 2. Hence A.x/ Hence A.x/ D 9x x 2 19 and 19 and B.x/ B.x/ D 9x x 2 18. 18.
39. Let A.x/ Let A.x/
x
D
Z
f . t / d t, t , with f .x/ as in Figure 4. with f .x/
0
(a) Does A.x/ Does A.x/ have have a local maximum at P ? P ? does A.x/ have have a local minimum? (b) Where does A.x/ (c) Where does A.x/ does A.x/ have have a local maximum? false? A.x/ < 0 for 0 for all x all x in in the interval shown. (d) True or false? A.x/ y R
S
x
y = f ( x x)
P
Q
FIGURE 4 Graph of f f .x/ .x/. SOLUTION
(a) In order for A.x/ for A.x/ to to have a local maximum, A maximum, A 0 .x/ f.x/ must f.x/ must transition from positive to negative. negative. As this does not happen at P at P ,, A.x/ does A.x/ does not have a local maximum at P . P . (b) A.x/ will A.x/ will have a local minimum when A when A 0 .x/ f.x/ transitions f.x/ transitions from negative to positive. This happens at R at R,, so A.x/ so A.x/ has has a local minimum at R at R.. A.x/ will have a local maximum when A 0 .x/ f.x/ transitions f.x/ transitions from positive to negative. (c) A.x/ will negative. This happens at S at S ,, so A.x/ so A.x/ has has a local maximum at S at S .. that A.x/ < 0 on 0 on I I since since the signed area from 0 to 0 to x x is is clearly always negative from the figure. (d) It is true that A.x/
D
D
D
x
40. Determine f Determine f .x/, x/, assuming that
Z 0
x
SOLUTION
Let F.x/ Let F.x/
Z D
f.t/dt
0
f.t/dt
D x 2 C x .
D x2 C x . Then F Then F 0 .x/ D f.x/ D 2x C 1.
41. Determine the function g.x/ function g.x/ and and all values of c such c such that x
Z
g.t/dt
c
SOLUTION
D x 2 C x 6
By the FTC II we have g.x/
d 2 D dx .x C x 6/ D 2x C 1
and therefore, x
Z c
We must choose c choose c so so that c that c 2
D x 2 C x .c2 C c/
g.t/dt
C c D 6. We can take c take c D 2 or c or c D 3.
SECTION
b
42. Find a Find a
b such that
Z
The Fundamental Fundamental Theorem Theorem of Calculus, Calculus, Part II
5.4
609
.x 2 9/dx has 9/dx has minimal value.
a
x 2 Then F a .x/ a .t 9/dt . Then F
D x 2 9, and the critical points are x D ˙3. Because 3 F a00 .3/ D 6 and F a00 .3/ D 6, we see that F a .x/ has .x/ has a minimum at x D 3. Now, we find a minimizing a .x 2 9/dx. 9/dx. Let 3 2 0 2 G.x/ D x .x 9/dx. 9/dx. Then G 9/, yielding critical points x Then G .x/ D .x 9/, points x D 3 or x or x D 3. With x With x D 3, Let a Let a be be given, and let F a .x/
SOLUTION
R
D
0
R
R
3
G.3/
Z D
.x
2
9/dx
3
With x With x
D 3,
3
Z D
G.3/
D
1 3 x 9x 3
.x 2 9/dx
3
b
Hence a Hence a
D 3 and b and b D 3 are the values minimizing x
In Exercises Exercises 43 and and 44, let let A.x/
Z D
Z
3
ˇˇ
ˇ
3
D 36:
D 0:
.x 2 9/dx. 9/dx .
a
f.t/dt .
a
Area Area Func Functi tio ons and and Conc Conca avity vity Explain why the following statements are true. Assume Assume f f .x/ is x/ is differentiable.
43. 43.
(a) If c is c is an inflection point of A.x/, A.x/, then f then f 0 .c/ (b) A.x/ is A.x/ is concave up if f f .x/ is x/ is increasing. (c) A.x/ is A.x/ is concave down if f f .x/ .x/ is decreasing.
D 0.
SOLUTION
x c is an inflection point of A.x/, A.x/, then A f 0 .c/ 0. (a) If x then A00 .c/ 00 (b) If A.x/ is A.x/ is concave up, then A .x/ > 0. 0. Since A.x/ Since A.x/ is is the area function associated with f with f .x/, x/, A 0 .x/ 00 0 0 A .x/ f .x/. .x/ . Therefore f Therefore f .x/ > 0, 0, so f so f .x/ .x/ is increasing. 00 A.x/ is concave down, then A then A .x/ < 0. 0. Since A.x/ Since A.x/ is is the area function associated with f with f .x/ .x/, A0 .x/ (c) If A.x/ is 00 0 0 A .x/ f .x/. .x/ . Therefore, f Therefore, f .x/ < 0 and 0 and so f so f .x/ is x/ is decreasing.
D D D
D
D
D f.x/ by f.x/ by FTC II, so D f.x/ by f.x/ by FTC II, so
44. Match the property of A.x/ with A.x/ with the corresponding property of the graph of f f .x/ .x/. Assume f Assume f .x/ .x/ is differentiable. Area function A.x/ (a) A.x/ is A.x/ is decreasing. A.x/ has a local maximum. (b) A.x/ has (c) A.x/ is A.x/ is concave up. (d) A.x/ goes A.x/ goes from concave up to concave down. Graph of f f .x/ .x/ (i) Lies below the x the x-axis. -axis. (ii) Crosses the x the x-axis -axis from positive to negative. negative. (iii) Has a local maximum. maximum. (iv) f.x/ is f.x/ is increasing. x
Let A.x/ Let A.x/ f.t/dt be an area function of f of f .x/. x/. Then A Then A 0 .x/ f.x/ and f.x/ and A A 00 .x/ f 0 .x/. .x/ . a f.t/dt be 0 (a) A.x/ is A.x/ is decreasing when A when A .x/ f .x/ < 0, 0, i.e., when f when f .x/ .x/ lies below the x the x-axis. -axis. This is choice (i) . (b) A.x/ has A.x/ has a local maximum (at x 0 ) when A when A 0 .x/ f .x/ .x/ changes sign from to 0 to as x as x increases increases through x through x 0 , i.e., when f.x/ crosses f.x/ crosses the x the x-axis -axis from positive to negative. This is choice (ii). A.x/ is concave up when A 0, i.e., when f .x/ is increasing. This corresponds to choice (iv). (c) A.x/ is when A00 .x/ f 0 .x/ > 0, when f .x/ (d) A.x/ goes A.x/ goes from concave up to concave down (at x 0 ) when A when A 00 .x/ f 0 .x/ changes .x/ changes sign from to 0 to as x increases through x through x 0 , i.e., when f .x/ has x/ has a local maximum at x 0 . This is choice (iii).
D
SOLUTION
R
D
D
D
C
D
D
C
x
45. Let A.x/ Let A.x/
D
Z
f . t / d t, t , with f .x/ as in Figure 5. Determine: with f .x/
0
(a) The intervals on which which A.x/ is A.x/ is increasing and decreasing (b) The values x values x where where A.x/ A.x/ has has a local min or max A.x/ (c) The inflection points of A.x/ (d) The intervals where A.x/ where A.x/ is is concave up or concave down y y = f ( x x)
x
2
D
4
6
8
FIGURE 5
10
12 12
610
CHAPTER 5
THE INTEGRAL
SOLUTION
A.x/ is increasing when A .x/ > 0, 0, which corresponds to the intervals .0;4/ and .0;4/ and .8;12/ .8;12/.. A.x/ is A.x/ is decreasing when (a) A.x/ is when A 0 .x/ f .x/ 0 A .x/ f .x/ < 0, 0, which corresponds to the intervals .4;8/ intervals .4;8/ and and .12; .12; /. A.x/ has a local minimum when A 0 .x/ f .x/ .x/ changes from to , corresponding to x 8. A.x/ has A.x/ has a local maximum (b) A.x/ has to x 0 when A when A .x/ f.x/ changes f.x/ changes from to , corresponding to x to x 4 and x and x 12. 12. A.x/ occur where A f 0 .x/ changes .x/ changes sign, or where f changes from increasing to decreasing (c) Inflection points of A.x/ occur where A00 .x/ where f decreasing or vice versa. Consequently, A.x/ Consequently, A.x/ has has inflection points at x 2, x 6, and x and x 10. 10. f 0 .x/ is positive or f.x/ is increasing, which corresponds .0;2/ and .6;10/. .6;10/. (d) A.x/ is concave up when A00 .x/ corresponds to the intervals intervals .0;2/ and Similarly, A.x/ Similarly, A.x/ is is concave down when when f f .x/ .x/ is decreasing, which corresponds to the intervals .2;6/ and .2;6/ and .10; .10; /.
D
D
D
1
D
C
D
D
D
D
.x/ 46. Let f Let f .x/
D x 2 5x 6 and F.x/ and F.x/
C D
D
D
D
1
x
Z D
f.t/dt .
0
(a) Find the critical points of F.x/ and F.x/ and determine whether they are local minima or local maxima. F.x/ and determine whether the concavity changes from up to down or from down to up. (b) Find the points of inflection of F.x/ and (c) Plot f Plot f .x/ .x/ and F.x/ and F.x/ on on the same set of axes and confirm your answers to (a) and (b). SOLUTION
x
2 (a) If F.x/ F.x/ then F 0 .x/ x 2 5x 6 and F and F 00 .x/ 2x 5. Solving F Solving F 0 .x/ x2 0 .t 5t 6/dt , then F 1 and x 7 < 0, there is a local maximum value of F critical points x points x and x 6. 6 . Since F Since F 00 .1/ of F at x at x 00 F .6/ 7 > 0, 0 , there is a local mini mum value of F at F at x 6. (b) As noted in part (a),
D
D
R
D
D D
D
D
D
D
D
5x 6 0 yields 1. Moreover, since
D
D x2 5x 6 and F 00 .x/ D 2x 5: .x/ changes from negative to positive at this A candidate point of inflection occurs where F where F 00 .x/ D 2x 5 D 0. Thus x Thus x D 52 . F 00 .x/ changes 5 point, so there is a point of inflection at x at x D 2 and concavity changes from down to up. (c) From the graph below, we clearly note that F.x/ has F.x/ has a local maximum at x D 1, a local minimum at x at x D 6 and a point of 5 inflection at x at x D 2 . F 0 .x/
y f ( x x) x !
2
2
4
6
F ( x x)
47. Sketch the graph of an increasing increasing function f function f .x/ such x/ such that both f both f 0 .x/ and .x/ and A.x/ A.x/ SOLUTION
D
If f If f 0 .x/ is decreasing, then f 00 .x/ must be negative. Furthermore, if A.x/ if A.x/
x f.t/dt are decreasing. 0 f.t/dt are x
R D
Z 0
f.t/dt is decreasin decreasing, g, then A0 .x/
D
f.x/ must f.x/ must also be negative. Thus, we need a function which is negative but increasing and concave down. The graph of one such function is shown below. below. y x
x
48.
Figure 6 shows the graph of f f .x/
D x sin x. Let F.x/ Let F.x/
Z D
t sin t d t .
0
(a) Locate the local max and absolute max of F.x/ on F.x/ on Œ0 Œ0;; 3 . (b) Justify graphically: F.x/ graphically: F.x/ has has precisely one zero in Œ; 2 . (c) How many zeros does F.x/ have F.x/ have in Œ0 in Œ0;; 3 ? F.x/ on Œ0 Œ0;; 3 . For each one, state whether the concavity changes from up to down or from down (d) Find the inflection points of F.x/ on to up. y
8 4 x
0 !
4
2
3 2
2
5 2
FIGURE 6 Graph of f f .x/ .x/
3
D x sin x .
SECTION
SOLUTION
Let F.x/ Let F.x/
f 0 .x/. .x/ .
D
x 0
R
t sin t d t . A graph of f f .x/ .x/
The Fundamental Fundamental Theorem Theorem of Calculus, Calculus, Part II
5.4
611
D x sin x is depicted in Figure 6. Note t hat F hat F 0 .x/ D f.x/ and f.x/ and F F 00 .x/ D
(a) For F For F to have a local maximum maximum at x at x 0 .0;3 / we must have F have F 0 .x0 / f .x0 / 0 and F and F 0 f must change sign from to . The absolute maximum of F.x/ on 0 to as x as x increases increases through x through x 0 . This occurs at x at x F.x/ on Œ0 Œ0;; 3 occurs at x at x 3 since (from 0 and x c is greatest for x c 3 . the figure) the signed area between x and x for x
2
D
D D D
D D
D
D
C
D
(b) At x , the value of F of F is is positi positive ve since since f .x/ > 0 on .0; /. As x increases along along the interval Œ; 2 , we see that that F decreases decreases as the negatively signed area accumulates. Eventually the additional negatively signed area “outweighs” the prior positively signed area and F attains F attains the value 0, say at b .; 2 /. Thereafter, on .b;2 /, we see that f is f is negative and thus F becomes F becomes and continues to be negative as the negatively signed area accumulates. Therefore, F.x/ takes F.x/ takes the value 0 exactly once in the interval Œ; 2 .
D
2
(c) F.x/ has F.x/ has two zeroes in Œ0 in Œ0;; 3 . One is described in part (b) and the other must occur in the interval Œ2 interval Œ2; 3 because F.x/ because F.x/ < 0 at x at x 2 but clearly the positively signed area over Œ2 over Œ2; 3 is greater than the previous negatively negatively signed area. (d) Since f f is differentiable, we have that F F is twice differentiable on I . I . Thus F.x/ has F.x/ has an inflection point at x0 provided 00 0 00 0 00 0 F .x0 / f .x0 / 0 and F F .x/ f .x/ changes .x/ changes sign at x at x 0 . If F F f changes sign from to 0 to at x at x 0 , then f then f has has a local maximum at x at x 0 . There is clearly such a value x value x0 in the figure in the interval Œ interval Œ =2; and another around 5 around 5 =2. =2. Accordingly, F has two inflection points where F.x/ f 0 changes sign from to 0 to at where F.x/ changes changes from concave up to concave down. If F If F 00 x0 , then f then f has a local minimum at at x 0 . From the figure, there is such an x an x 0 around 3 around 3=2; =2; so F so F has one inflection point where F.x/ where F.x/ changes from concave down to concave up.
D
D
D
D
D
C
D
C
Find the smallest positive critical point of
49.
x
F.x/
Z D
cos.t cos.t 3=2 / dt
0
and determine whether it is a local min or max. Then find the smallest positive inflection point of F.x/ of F.x/ and and use a graph of 3=2 y cos.x cos.x / to determine whether the concavity changes from up to down or from down to up.
D
D cos.x cos .x 3=2 / D 0. The smallest positive critical points occurs where D =2, x =2, so that x that x D . =2/2=3 . F 0 .x/ goes .x/ goes from positive to negative at this point, so x D . =2/2=3 corresponds to a local maximum.. Candidate inflection points of F.x/ F.x/ occur where F 00 .x/ D 0. By FTC, F 0 .x/ D cos.x cos.x 3=2 /, so F 00 .x/ D .3=2/x1=2 sin.x sin.x 3=2 /. 00 Finding the smallest positive solution of F F .x/ D 0, we get: 1=2 sin.x sin.x 3=2 / D 0 .3=2/x sin.x sin.x 3=2 / D 0 .since x since x > 0/ x 3=2 D x D 2=3 2:14503: From the plot below, below, we see that F 0 .x/ D cos.x cos.x 3=2 / changes from decreasing to increasing at 2=3 , so F.x/ changes from concave SOLUTION 3=2
A critical critical point point of F.x/ occurs F.x/ occurs where F 0 .x/
down to concave up at that point.
y
1 0.5 x !
1
0.5
2
3
1
!
Further Insights and Challenges 50. Proof Proof of FTC II The proof in the text assumes that f that f .x/ .x/ is increasing. To prove it for all continuous functions, let m.h/ and M.h/ and M.h/ denote denote the minimum and maximum of f f .t/ on t/ on Œx Œx;; x h (Figure 7). The continuity of f f .x/ .x/ implies that lim m.h/
C
lim M.h/
h
!0
D f.x/. f.x/. Show that for h 0, for h > 0, hm.h/ A.x
For h For h < 0, 0 , the inequalities are reversed. Prove that that A A0 .x/
C h/ A.x/ hM.h/
D f.x/. f.x/.
h
!0
D
612
THE INTEGRAL
CHAPTER 5
y y = f ( x x) M (h)
m (h) x
a
x
x + h
A.x FIGURE 7 Graphical interpretation of A.x
C h/ A.x/. A.x/.
.x/ be continuous on Œa M.h/ denote f on Let f Let f .x/ on Œa;; b . For h For h > 0, let m.h/ let m.h/ and and M.h/ denote the minimum and maximum values of f h. Since f Since f is continu continuous, ous, we have have lim m.h/ lim M.h/ f.x/. f.x/. If h If h > 0, then since m.h/ since m.h/ f .x/ .x/ M.h/ on M.h/ on
SOLUTION
Œx;x
C Œx;x C h, we have
!0C
x h
hm.h/
Z C D
In other words, hm.h/ words, hm.h/
D
x h
m.h/d m.h/d t
x
h
D h!0C
h
Z C
x
f.t/dt
x
A.x
! 0C yields
D A.x C h/ A.x/
h
Z C D
x h
f.t/dt
x
Z C
M.h/dt
D hM.h/:
x
A.x C h/ A.x/ C h/ A.x/ hM.h/. hM.h/. Since h Since h > 0, it follows that m.h/ that m.h/ M.h/. Letting M.h/. h A.x
f.x/ lim h
C h/ A.x/ f.x/; h
!0C
whence h
A.x
lim
C h/ A.x/ D f.x/ h
!0C
by the Squeeze Theorem. If h h < 0, 0 , then x
hm.h/
Z D
x
m.h/dt
x h h >
f.t/dt
x h
C
Since h Since h < 0, 0 , we have
x
Z
C
D A.x/ A.x C h/
Z D
x
f.t/dt
x h
Z
M.h/dt
x h
C
C
D hM.h/:
0 and thus m.h/
A.x/ A.x h
C h/ M.h/
or m.h/ Letting h Letting h
A.x
C h/ A.x/ M.h/: h
! 0 gives A.x
f.x/ lim h
!0
C h/ A.x/ f.x/; h
so that h
lim
A.x
C h/ A.x/ D f.x/ h
!0
by the Squeeze Theorem. Since the one-sided limits agree, we therefore have A.x C h/ A.x/ D hlim D f.x/: h !0 b F.a/ if F F 0 .x/ D f.x/. f.x/. Use FTC II to give a new proof of FTC I as a f.t/dt D F.b/ F.a/ if A0 .x/
R
FTC I asserts that x follows. Set A.x/ Set A.x/ a f.t/dt . (a) Show that F.x/ that F.x/ A.x/ C for for some constant. 51. Proof of FTC I
D
R
D
C b (b) Show that F.b/ that F.b/ F.a/ D A.b/ A.a/ D f.t/dt . a x SOLUTION Let F Let F 0 .x/ D f.x/ and f.x/ and A.x/ A.x/ D a f.t/dt . (a) Then by the FTC, Part II, A II, A0 .x/ D f.x/ and f.x/ and thus A.x/ thus A.x/ and and F.x/ F.x/ are are both antiderivatives antiderivatives of f f .x/ .x/. Hence F.x/ Hence F.x/ D A.x/ C C
Z R
for some constant C constant C .. (b)
F.b/ F.a/
D .A.b/ C C / .A.a/ C C / D A.b/ A.a/ b
D which proves the FTC, Part I.
Z a
a
f.t/dt
Z a
b
f.t/dt
D
Z a
b
f.t/dt
0
D
Z a
f.t/dt
SECTION
Net Change Change as the Integr Integral al of a Rate Rate
5.5
613
x
R
52. Can Every Every Antiderivati Antiderivative ve Be Expressed Expressed as as an Integral? Integral? The area area functi function on a f.t/dt is f.t/dt is an antiderivative antiderivative of f.x/ of f.x/ for every 1 2 a. a f f .x/ .x / x F.x/ C . C . value of . However, However, not all antiderivatives are obtained in this way. The general antiderivative of is F.x/ is 2x Show that F.x/ that F.x/ is is an area function if C C 0 but not if C C > 0 .
D
D
C
x x 1 2 SOLUTION Let f .x/ x . The general antiderivative of f .x/ is F.x/ C . C . Let A.x/ Let f .x/ of f .x/ a f.t/dt a t dt 2x x 1 2 1 2 1 2 of f .x/ .x/ x . To express F.x/ as F.x/ as an area function, we must find a value for a such 2t a 2 x 2 a be an area function of f 1 2 2C . 2C or that 12 x 2 12 a2 C , C , whence a 2C . If I f C C 0, then 2C 0 and we may choose either a 2x a 2C . 2C . However, if C C > 0, then there is no real solution for a for a and and F.x/ F.x/ cannot cannot be expressed as an area function.
D
ˇˇ
D D p
D
D
C
D
D p D ˙
C
R
D
D
R
D
p
53. Prove the formula
d dx SOLUTION
v.x/
Z
f.t/dt
u.x/
D f .v.x//v .v.x//v 0.x/ f .u.x//u .u.x//u0.x/
Write v.x/
Z
0
f.x/dx
u.x/
Z D
v.x/
v.x/
Z C
f.x/dx
u.x/
f.x/dx
0
u.x/
Z D
f.x/dx
0
Z
f.x/dx:
0
Then, by the Chain Rule and the FTC, v.x/
d dx
Z
f.x/dx
u.x/
D
d dx
v.x/
Z
f.x/dx
0
d dx
u.x/
Z
f.x/dx
0
D f .v.x//v .v.x//v 0 .x/ f .u.x//u .u.x//u0.x/: 54. Use the result of Exercise 53 to calculate
d dx SOLUTION
ex
Z
sin t d t
ln x
By Exercise 53, d dx
ex
Z
sin t d t
ln x
D ex sin ex x1 sin sin lnx: ln x:
5.5 Net Change as the Integral of a Rate Preliminary Questions 1. A hot metal object is submerged in cold water. The rate at which the object cools (in degrees per minute) is a function f.t/ of f.t/ of
time. Which quantity is represented by the integral T T 0
R
The definite integral being submerged in the cold water. SOLUTION
T T 0
R
f.t/dt ?
f.t/dt represents f.t/dt represents the total drop in temperature of the metal object in the first T minutes minutes after
v.t/ km/h, 2. A plane travels 560 travels 560 km km from Los Angeles to San Francisco in 1 hour. If the plane’s velocity at time t time t is is v.t/ km/h, what is the 1 value of 0 v.t/ v.t/ dt ?
R
1 v.t/dt represents the total distance traveled 0 v.t/dt represents 1 Los Angeles to San Francisco. Therefore the value of 0 v.t/dt is v.t/dt is 560 km. SOLUTION
The definite integral
R
by the airplane during the one hour flight from
R
3. Which of the following quantities would be naturally represented as derivatives and which as integrals? (a) Velocity of a train (b) Rainfall during a 6-month period
automobile (c) Mileage per gallon of an automobile (d) Increase in the U.S. population from 1990 to 2010 SOLUTION Quantities (a) and (c) involve rates of change, so t hese would naturally be represented as derivatives. derivatives. Quantities (b) and (d) involve an accumulation, accumulation, so these would naturally be represented as integrals.
614
THE INTEGRAL
CHAPTER 5
Exercises Water flows into an empty reservoir at a r ate of 3000 3000 1. Water 5 hours? SOLUTION
The quantity of water in the reservoir reservoir after five five hours is 5
Z
.3000
0
C 20t/dt
D
3000t
C 10t
2
ˇˇ
5
ˇD
15;250 gallons 15;250 gallons::
1 3 t 12
ˇˇ
0
C 10t C 0:25t 2 insects per day. Find the insect population after 3 days,
2. A population of insects increases at a rate of 200 of 200 assuming that there are 35 insects at t 0.
D
SOLUTION
C 20t liters 20t liters per hour. What is the quantity of water in the reservoir after
The increase in the insect population population over three days is 3
Z
200
C 10t C
0
1 2 t d t 4
D
2
200t
C 5t C
Accordingly, Accordingly, the population after 3 days is 35 is 35
3
ˇD 0
2589 4
D 647:25:
C 647:25 D 682:25 or 682:25 or 682 insects. 3. A survey shows that a mayoral candidate is gaining votes at a rate of 2000t of 2000t C 1000 votes 1000 votes per day, where t is the number of days since she announced her candidacy. How many supporters will the candidate have after 60 days, assuming that she had no supporters at t at t D 0? SOLUTION
The number of of supporters the candidate candidate has after 60 days is 60
Z
ˇˇ ˇ
2
.2000t
60
C 1000 D 3;660;000: 1000// dt D .1000t C 1000t/ 0 95 C 3t 2 t bicycles per week. How many bicycles were produced from the beginning 4. A factory produces bicycles bicycles at a rate of 95 0
of week 2 to the end of week 3?
The rate of production is r.t/ is r.t/ 95 3t 2 t bicycles per week and the period from the beginning of week 2 to the end of week 3 corresponds to the second and third weeks of production. Accordingly, the number of bikes produced from the beginning of week 2 to the end of week 3 is
D
SOLUTION
3
Z
C
3
Z D
r.t/dt
1
95
1
C 3t
2
t d t
D
95t
C t
3
1 2 t 2
bicycles.
The displacement displacement is given by 5
Z
.4t
3/dt
2
D .2t
2
ˇD
212
1
D 4t 3 m/s over the time interval Œ interval Œ2; 2; 5.
5. Find the displacement of a particle moving in a straight line with velocity v.t/ velocity v.t/ SOLUTION
3
ˇˇ
5
ˇˇˇ D
3t /
.50 15/ .8 6/
2
D 33m 33m:
6. Find the displacement over the time interval Œ1 interval Œ1;; 6 of a helicopter whose (vertical) velocity at time t time t is is v.t/ v.t/ SOLUTION
Given v.t/ Given v.t/
D 501 t 2 C t m= m =s, the change in height over Œ1 over Œ1;; 6 is 6
6
v.t/dt
Z D
1
Z
Ct
d t
1 3 t 150
1
6
9:8t dt
1=2
D 4:9t
2
ˇˇ ˇ
ˇ
1
D 4:9 1:225 D 3:675 m 3:675 m:: 8. A projectile is released with an initial (vertical) velocity of 100 m/s. 100 m/s. Use the formula v.t/ D 100 9:8t for velocity to 1=2
jv.t/j d t
Z D
1 2 t 2
ˇˇˇ
D C D 284 18:93 m 18:93 m:: 15 1 1 1 0:5 and t t D 1 s? 1 s? Use 7. A cat falls from a tree (with zero initial velocity) at time t D 0. How far does the cat fall between t D 0:5 and Galileo’s formula v.t/ formula v.t/ D 9:8t m/s. 9:8t m/s. SOLUTION Given v.t/ Given v.t/ D 9:8t m 9:8t m= =s, the total distance the cat falls during the interval Œ interval Œ 12 ; 1 is
Z
1 2 t 50
D 0:02t 2 C t m/s.
determine the distance traveled during the first 15 seconds. SOLUTION
1=2
The distance traveled is given by 15
Z 0
100=9:8
j100 9:8t j d t
Z D
15
.100 9:8t/ 9:8t/ dt
0
D
100t
Z C
.9:8t
100/d 100/d t
100=9:8
4:9t
2
ˇˇ
ˇ
100=9:8
C
4:9t
0
2
100t
ˇˇ
ˇ
15
100=9:8
622:9 m 622:9 m::
SECTION
Net Change Change as the Integr Integral al of a Rate Rate
5.5
615
In Exercises Exercises 9–12, 9–12, a particle moves moves in a straight straight line with the the given velocity velocity (in m/s). Find Find the displaceme displacement nt and distance distance traveled traveled over the time interval, and draw a motion diagram like Figure 3 (with distance and time labels). 9. v.t/
D 12 4t , Œ0; Œ0; 5
5
Displacement is given given by
SOLUTION
Z
.12 4 t / d t
D .12t 2t
0
5
3
Z j
12 4t d t
j
0
Z D
2
5
ˇˇ D ˇ
/
5
Z C
.12 4 t / d t
0
.4t
10 ft, 10 ft, while total distance is given by
0
12/ 12/ dt
3
2
D .12t 2t
3
ˇˇ C ˇ
/
.2t
5
ˇˇ D ˇ
2
12t/
0
The displacement diagram is given here.
26 ft 26 ft::
3
t = 5 t = 3 t = 0
0
10
Distance
18
D 36 24t C 3t 2 , Œ0;10 SOLUTION Let v.t/ Let v.t/ D 36 24t C 3t 2 D 3.t 2/.t 6/. 6/. Displacement is given by
10. v.t/
10
Z 0
10
ˇˇ ˇ
D 160 .36 24t C 3t 2 / dt D .36t 12t 2 C t 3 / 0
meters. Total distance traveled is given by 10
Z
2
j36 24t C 3t j dt
0
2
Z D
.36 24t
0
C 3t
2
D .36t 12t C t D 224 meters 224 meters::
3
2
6
Z C
/ dt
.24t
10
2
36 3t / dt
2
2
ˇˇ C ˇ
/
.12t
2
Z C
.36 24t
6
36t
3
ˇˇ ˇ
10
t /
0
2
C .36t 12t C t
0
3
C 3t 2/ dt
10
ˇˇ ˇ
/
6
The displacement diagram is given here. y t = 10
t = 6
t = 2
0
11. v.t/
Dt
SOLUTION
2 1,
20 40 60 80 100 1 20 20 1 40 40 1 60 60
x
Œ0:5; Œ0:5; 2 2
Displacement is given given by
Z
.t
2
1/dt
0:5
2
Z ˇˇˇ
t
2
1
ˇˇˇ D Z
1
0:5
dt
.t
D .t
1
2
2
1/dt
0:5
Z C
.1 t
2
2
t/
ˇˇ ˇ
0:5
/ dt
1
D 0 m, while total distance is given by
D .t
1
1
ˇˇ ˇ
t/
0:5
The displacement diagram is given here.
C .t C t
1
2
ˇˇ D ˇ
/
1 m: m:
1
t = 2 t = 1 t = 0
0
12. v.t/
D cos t , Œ0;3
SOLUTION
3
Z Z D
cos t d t
Displacement is given given by
0
3
Z 0
0.5
D sin t
=2
j cos t j dt
3 0
D 0 meters, while the total distance traveled is given by
3=2
cos t d t
0
Z
5=2
cos t d t
=2
=2
D sin t
ˇˇ
ˇˇ ˇ
0
sin t
ˇˇ ˇ
3=2 =2
Distance
C sin t
ˇˇ ˇ
C
Z
3 =2
5=2
sin t
Z
5=2
3
3=2
3
cos t d t
ˇˇ ˇ
5 =2
cos t ; d t
616
THE INTEGRAL
CHAPTER 5
D 6 meters: meters: The displacement diagram is given here. y t = 3π
5π 2
t = t =
3π 2 π
t = −1.0 −0.5
0.5
2
D 8t t 2 m/s2 .
13. Find the net change in velocity over Œ1 over Œ1;; 4 of an object with a.t/ with a.t/ SOLUTION
1.0
x
The net net change in velocity is 4
Z
4
Z D
a.t/dt
1
.8t
t
2
/ dt
1
D
4t
2
1 3 t 3
ˇˇ
4
ˇD
39 m 39 m= =s:
1
14. Show that if acceleration is constant, then the change in velocity is proportional to the length of the time interval. SOLUTION Let a.t/ Let a.t/ a be the constant acceleration. Let v.t/ be v.t/ be the velocity. Let Œt 1 ; t2 be the time interval concerned. We know that v that v 0 .t/ .t / a, and, by FTC,
D
D
t2
v.t2 / v.t1 /
Z D
a dt
t1
D a.t2 t1 /;
a.. So the net change in velocity is proportional to the length of the time interval with constant of proportionality proportionality a 15. The traffic flow rate past a certain point on a highway is q.t/ is q.t/ many cars pass by in the time i nterval from 8 to 10 A M? SOLUTION
D 3000 C 2000t 300t 2 (t in hours), where t where t D 0 is 8 AM . How
The number number of cars is given given by 2
Z
2
Z D
q.t/dt
D
2
.3000
2
3
2
ˇˇ
C 2000t 300t / dt 3000t C 1000t 100t 0 0 0 D 3000.2/ C 1000.4/ 100.8/ D 9200 cars 9200 cars:: 16. The marginal cost of producing producing x x tablet tablet computers is C is C 0 .x/ D 120 0:06x C 0:00001x2 What is the cost of producing 3000
ˇ
units if the setup cost is $90,000? If production is set at 3000 units, what is the cost of producing 200 additional units? SOLUTION
The production coot for producing producing 3000 units is 3000
Z
.120 0:06x
0
2
C 0:00001x
/ dx
1 0:00001x3 3
2
3000
ˇˇ
D 120x 0:03x C 0 D 360;000 270;000 C 90;000 D 180;000
ˇ
dollars. Adding in the setup cost, we find the total cost of producing 3000 units is $270,000. If production is set at 3000 units, the cost of producing an additional 200 units is 3200
Z
.120 0:06x
3000
2
C 0:00001x
/ dx
1 0:00001x3 3
2
3200
ˇˇ
D 120x 0:03x C 3000 D 384;000 307;200 C 109;226:67 180;000
ˇ
or $6026.67. 17. A small boutique produces wool sweaters at a marginal cost cost of 40 40 5ŒŒx=5 for 0 for 0 x 20, 20, where ŒŒx where ŒŒx is the greatest integer function. Find the cost of producing 20 producing 20 sweaters. sweaters. Then compute the average cost of the first 10 sweaters and the last 10 sweaters. SOLUTION
The total cost of producing 20 sweaters sweaters is 20
Z 0
5
.40 5ŒŒx=5/ dx
Z D 0
10
40dx
Z C 5
15
35dx
Z C
10
20
30 dx
Z C
25dx
15
D 40.5/ C 35.5/ C 30.5/ C 25.5/ D 650 dollars 650 dollars:: From this calculation, we see that the cost of the first 10 sweaters is $375 and the cost of the last ten sweaters is $275; thus, the average cost of the first ten sweaters is $37.50 and the average cost of the last ten sweaters is $27.50.
SECTION
Net Change Change as the Integr Integral al of a Rate Rate
5.5
617
18. The rate (in liters per minute) at which water drains from a tank is recorded at half-minute intervals. Compute the average of the left- and right-endpoint approximations to estimate the total amount of water drained during the first 3 minutes.
t (min)
SOLUTION
Let t
0
0.5
1
1.5
2
2.5
3
r .l/min/ l/min/ 50 50
48
46
44
42
40
38
D 0:5. 0:5. Then RN
D 0:5.48 C 46 C 44 C 42 C 40 C 38/ D 129:0 liters 129:0 liters LN D 0:5.50 C 48 C 46 C 44 C 42 C 40/ D 135:0 liters 135:0 liters The average of R RN and L N is i s 12 .129 C 135/ D 132 liters. 132 liters. 19. The velocity of a car is recorded at half-second intervals (in feet per second). Use the average of the left- and right-endpoint approximations to estimate the total distance traveled during the first 4 seconds.
t
SOLUTION
Let t
0
0. 0 .5
1
1.5
2
2.5
3
3.5
4
v.t/ 0
12
20
29
38
44
32
35
30
D :5. :5. Then RN
D 0:5 .12 C 20 C 29 C 38 C 44 C 32 C 35 C 30/ D 120 ft 120 ft:: LN D 0:5 .0 C 12 C 20 C 29 C 38 C 44 C 32 C 35/ D 105 ft 105 ft::
The average of R RN and L and L N is i s 112:5 ft. 112:5 ft. 20. To model the effects of a carbon tax on CO2 emissions, policymakers study the marginal cost of abatement B.x/, B.x/, defined as the cost of increasing CO2 reduction from x from x to to x x 1 tons (in units of ten thousand tons—Figure 1). Which quantity i s represented by the area under the curve over Œ0 over Œ0;; 3 in Figure 1?
C
B( x x ) ($/ton)
100 75 50 25 1
2
3
x
Tons reduced (in ten thousands)
abatement B.x/.. FIGURE 1 Marginal cost of abatement B.x/ SOLUTION The area under the curve curve over Œ0 over Œ0;; 3 represents the total cost of reducing the amount of CO 2 released into the atmosphere by 3 tons.
21. A megawatt of power is 10 is 10 6 W, or 3:6 or 3:6 109 J/hour. Which quantity is represented by the area under the graph in Figure 2? Estimate the energy (in joules) consumed during t he period 4 P M to 8 P M. Megawatts Megawatts (in thousands) 28 27 26 25 24 23 22 21 20 19 18
00 02 04 06 08 10 12 14 16 18 20 22 24 Hour of the day
FIGURE 2 Power consumption consumption over 1-day period in California (February 2010).
618
CHAPTER 5
THE INTEGRAL
SOLUTION The area under the graph in Figure 2 represents represents the total power consumption over one day day in California. Assuming t 0 correspond correspondss to midnight, midnight, the period period 4 PM to 8 PM corresponds to t 16 to t 20. 20. The left and right endpoint endpoint approximations are
D
D
D
L
D 1.22:8 C 23:5 C 26:1 C 26:7/ D 99:1 megawatt 99:1 megawatt hours R D 1.23:5 C 26:1 C 26:7 C 26:1/ D 102:4 megawatt 102:4 megawatt hours The average of these values is
D 3:627 1011 joules: joules:
100:75megawatt 100:75megawatt hours
22. Figure 3 shows shows the migration migration rate M.t/ rate M.t/ of of Ireland in the period 1988–1998. This is the rate at which people (in thousands per year) move into or out of the country. (a) Is the following integral positive or negative? negative? What does this quantity represent? 1998
Z
M.t/dt
1988
(b) Did migration in the period 1988–1998 result in a net influx of people into Ireland or a net outflow of people from Ireland? (c) During which two years could the Irish prime minister announce, “We’ve “We’ve hit an inflection point. We are still losing population, but the trend is now improving.” 30 ) s d 20 n a s 10 u o 0 h t n !10 i ( n !20 o i t !30 a r g ! 40 i M!50
1994 1988
1990 1990
1992 1992
1996
1998
2000
FIGURE 3 Irish migration rate (in thousands per year). year). SOLUTION
(a) Because there appears appears to be more area below the t -axis than above in Figure 3, 1998
Z
M.t/dt
1988
is negative. This quantity represents the net migration from Ireland during the period 1988–1998. (b) As noted in part (a), (a), there there appears appears to be more more area area below below the the t -axis than above above in Figure 3, so migration migration in the period 1988–1998 resulted in a net outflow of people from Ireland. M is at a local minimum, which appears to be in the years 1989 (c) The prime minister can make this statement when the graph of M and 1993. 23. Let N.d/ Let N.d/ be be the number of asteroids of diameter d kilometers. d kilometers. Data suggest that the diameters are distributed according to a piecewise power law:
N 0 .d / D
1:9 109 d
(
2:3
for d for d < 70
4
for d for d 70
2:6 1012 d
(a) Compute the number of asteroids with diameter between 0:1 between 0:1 and and 100 100 km. km. (b) Using the approximation N approximation N.d .d 1/ N.d/ N 0 .d /, estimate the number of asteroids of diameter 50 diameter 50 km. km.
C C
SOLUTION
(a) The number of asteroids with diameter between 0:1 between 0:1 and and 100 100 km km is 100
Z
0:1
D N 0 .d/dd D D
70
Z
9
2:3
1:9 10 d
100
Z C C
0:1
2:6 1012 d
4
dd
d d
70
1:9 109 d 1:3
70
1:3
ˇˇ ˇ
0:1
2:6 1012 d 3
3
ˇˇ ˇ
100
70
D 2:916 1010 C 1:66 106 2:916 1010 : (b) Taking d Taking d
D D 49:5, 49:5, N.50:5/ N.49:5/ N 0 .49:5/
D 1:9 109 49:5
Thus, there are approximately 240,526 asteroids of diameter 50 km.
2:3
D 240;525:79:
SECTION
Net Change Change as the Integr Integral al of a Rate Rate
5.5
619
24. Heat Capacity Capacity The heat capacity C. capacity C.T T / of a substance is the amount of energy (in joules) required to raise the temperature of 1 g by 1 by 1 ı C at temperature T temperature T .. (a) Explain why the energy required to raise the temperature from T from T 1 to T to T 2 is the area under the graph of C. of C.T T / over ŒT over ŒT 1 ; T 2 . (b) How much energy is required to raise the temperature temperature from 50 from 50 to to 100 100ı C if C.T C.T /
p D 6 C 0:2 T ? T ?
SOLUTION
Since C.T T / is the energy required to raise the temperature of one gram of a substance by one degree when its temperature is (a) Since C. T 2
T , T , the total energy required to raise the temperature from T from T 1 to T to T 2 is given by the definite integral the definite integral also represents the area under the graph of C.T C.T /. (b) If C.T C.T / or
Z
C . T / d T . As C. T / > 0, 0, As C.T
T 1
D 6 C :2p T D D 6 C 15 T 1=2 , then the energy required to raise the temperature from 50 from 50ı C to 100 to 100ı C is 100
Z
50
6
1 1=2 T d T 5
C
D D
2 3=2 T 15
6T
C C
p
ˇˇˇ
ˇ
100 50
D
6.100/
C
2 .100/3=2 15
6.50/
C
100 100 50 C . T / d T
R
2 .50/3=2 15
1300 100 2 386:19 Joules 386:19 Joules 3 25. Figure 4 shows the rate R.t/ of R.t/ of natural gas consumption (in billions of cubic feet per day) in the mid-Atlantic states (New York, New Jersey, Jersey, Pennsylvania). Express the total quantity of natural gas consumed in 2009 as an integral (with respect t o time t time t in days). Then estimate this quantity, given the following monthly values of R.t/: R.t/:
D
3.18 3.18,, 1.01 1.01,,
2.86 2.86,, 0.89 0.89,,
2.39 2.39,, 0.89 0.89,,
1.49 1.49,, 1.20 1.20,,
1.08 1.08,, 1.64 1.64,,
0.80 0.80,, 2.52 2.52
Keep in mind that the number of days in a month varies with the month. Natural gas consumption (109 cubic ft/day) 3 2 1
J F M A M J J A S O N D
FIGURE 4 Natural gas consumption consumption in 2009 in the mid-Atlantic states SOLUTION
The total quantity quantity of natural gas consumed consumed is given given by 365
Z
R.t/dt:
0
With the given data, we find 365
Z
R.t/dt
0
31.3:18/
C 28.2:86/ C 31.2:39/ C 30.1:49/ C 31.1:08/ C 30.0:80/ C31.1:01/ C 31.0:89/ C 30.0:89/ C 31.1:20/ C 30.1:64/ C 31.2:52/
D 605:05 billion 605:05 billion cubic feet: feet: 26. Cardiac Cardiac output is the rate R rate R of volume of blood pumped by the heart per unit time (in liters per minute). Doctors measure R measure R by injecting A injecting A mg mg of dye into a vein leading into the heart at t 0 and recording the concentration c.t/ of c.t/ of dye (in milligrams per liter) pumped out at short regular time intervals (Figure 5).
D
(a) Explain: The quantity of dye pumped out in a small time interval Œ interval Œt; t; t T T 0 c.t/dt ,
R
(b) Show that A that A R dye returns by recirculation.
D
(c) Assume A Assume A
C t is approximately Rc.t/ approximately Rc.t/t .
where T where T is large enough that all of the dye is pumped through the heart but not so large that the
D 5 mg. Estimate R Estimate R using using the fol lowing values of c.t/ recorded c.t/ recorded at 1-second intervals from t from t D 0 to t to t D 10: 10: 0, 0.4, 6.1, 4,
2.8, 2.3,
6.5, 1.1,
9.8, 0
8.9,
620
THE INTEGRAL
CHAPTER 5
Measure concentration here
Inject dye here
c (t ) (mg / l) l)
y = c (t )
Blood flow
t (s) (s)
FIGURE 5 SOLUTION
(a) Over a short time interval, interval, c.t/ c.t/ is is nearly constant. Rc.t/ constant. Rc.t/ is is the rate of volume of dye (amount of fluid concentration of dye in fluid) flowing out of the heart (in mg per minute). Over the short t ime interval Œ interval Œt; t; t t , the rate of flow of dye is approximately =minute. Therefore, the flow of dye over the interval is approximately Rc.t/t mg. constant at Rc.t/ at Rc.t/ mg mg= (b) The rate of flow flow of dye is Rc.t/ is Rc.t/.. Therefore the net flow between time t 0 and time t time t T is
C
D
T
Z
D
T
Rc.t/dt
0
D R
Z
c.t/dt:
0
If T T is great enough that all of the dye is pumped through the heart, the net flow is equal to all of the dye, so T
A
(c) In the table, t
DR
Z
c.t/dt:
0
D 601 minute, and N D 10. and N D 10. The right and left hand approximations of
T
Z
c.t/dt are: c.t/dt are:
0
R10
minute D 601 .0:4 C 2:8 C 6:5 C 9:8 C 8:9 C 6:1 C 4 C 2:3 C 1:1 C 0/ D 0:6983 mg liter
L10
minute D 601 .0 C 0:4 C 2:8 C 6:5 C 9:8 C 8:9 C 6:1 C 4 C 2:3 C 1:1/ D 0:6983 mg liter
Both L Both LN and R N are the same, so the average of L of L N and RN is 0.6983. Hence, T
A 5 mg R
DR D
Z
c.t/dt
0
mg minute R 0:6983 liter
5 liters liters D 0:6983 D 7:16 : minute minute
Exercises Exercises 27 and 28: A study suggests suggests that the extinction extinction rate r.t/ of marine animal families during the Phanerozoic Eon can be modeled by the function r.t/ 3130=.t 262/ for 0 t 544, where t is time elapsed (in millions of years) since the beginning 544 refers to the present time, t 540 is 4 million years ago, and so on. of the eon 544 million years ago. Thus, t
D
C D
27. Compute the average of R of R N and L and L N with N with N 100 t 150 and 150 and 350 350 t 400. 400.
D
D 5 to estimate the total number of families that became extinct in the periods
SOLUTION
(100 t
150) 150) For N For N
D D 5, t
The table of values r.t i /
f
D 150 5 100 D 10:
gi D0:::5 is given below: ti
10 100
r.t i / 8.6 8.646 4641 41
110
120
130
140
150
8.41 8.4139 398 8
8.19 8.1937 372 2
7.98 7.9846 469 9
7.78 7.7860 607 7
7.59 7.5970 709 9
The endpoint approximations are: RN
D 10.8:41398 C 8:19372 C 7:98469 C 7:78607 C 7:59709/ 399:756 families 399:756 families LN D 10.8:64641 C 8:41398 C 8:19372 C 7:98469 C 7:78607/ 410:249 families 410:249 families
The right endpoint approximation estimates 399.756 families became extinct in the period 100 t 150, 150, the left endpoint approximation estimates 410.249 families became extinct during this time. The average of the two i s 405.362 families.
SECTION
(350 t
Net Change Change as the Integr Integral al of a Rate Rate
5.5
621
400) 400) For N For N
D D 10, 10, t
D 400 5 350 D 19:
The table of values r .ti /
f
gi D0:::5 is given below: ti
35 350
r .ti / 5.1 5.114 1438 38
360
370
380
390
400
5.03 5.0321 215 5
4.95 4.9525 253 3
4.87 4.8753 539 9
4.80 4.8006 061 1
4.72 4.7281 810 0
The endpoint approximations are: RN
D 10.5:03215 C 4:95253 C 4:87539 C 4:80061 C 4:72810/ 243:888 families 243:888 families LN D 10.5:11438 C 5:03215 C 4:95253 C 4:87539 C 4:80061/ 247:751 families 247:751 families
The right endpoint approximation estimates 243.888 families became extinct in the period 350 t 400, 400 , the left endpoint approximation estimates 247.751 families became extinct during this time. The average of the two is 245.820 families. Estimate the total number of extinct families from t from t
28. SOLUTION
D 0 to the present, using M D 544. using M N with N D 544. N with N
We are estimat ing 544
Z
3130 d t .t 262/
C
0
544 0 D D 544. D 544, D 1 and fti gi D1;:::N D i t .t=2/ D i 12 . 544. If N N D 544, t D 544
using M using M N with N with N
N
D t
M N N
X
i
544
r .ti /
D1
D1
X
i
D1
3130 261:5 i
C D 3517:3021:
Thus, we estimate that 3517 families have become extinct over the past 544 million years.
Further Insights and Challenges 29. Show that a particle, located at the origin at t at t point x point x 2.
D
The particle’s particle’s velocity is v.t/ is v.t/ position at time t time t is is SOLUTION
D 1 and moving along the x the x-axis -axis with velocity v.t/ velocity v.t/ D t
D s 0 .t/ .t / D t
2,
an antiderivative for which is F.t/ t
t
s.t/
Z D 0
s .u/du
1
for all t all t
1. 1 .
ˇˇ D ˇ
D F.u/
Thus, the particle will never pass x pass x
F.t/ F.1/
1
D t
2 , will never pass the
1.
Hence, the particle’s
D 1 1t < 1
D 1, which implies it will never pass x pass x D 2 either. at t D 1 and moving along the x the x-axis -axis with velocity v.t/ velocity v.t/ D t 30. Show that a particle, located at the origin at t
1=2
moves arbitrarily
far from the origin after sufficient time has elapsed.
The particle’s particle’s velocity is v.t/ is v.t/ position at time t time t is is SOLUTION
D s 0.t / D t
1=2 ,
t
t
s.t/
Z D 0
s .u/du
1
for all t all t
1 . 1.
an antiderivative for which is F.t/
ˇˇˇ D ˇ
D F.u/
F.t/ F.1/
1
D 2 t 1=2 . Hence, the particle’s
D 2p t 1
Let S Let S > 0 denote 0 denote an arbitrarily l arge distance from the origin. We see that for t>
S
C 1 2 ; 2
the particle particle will be more than than S units from from the origin. In other words, words, the particle particle moves moves arbitrarily arbitrarily far from from the origin after after sufficien sufficientt time has elapsed.
622
THE INTEGRAL
CHAPTER 5
5.6 Substitution Method Preliminary Questions 1. Which of the following integrals is a candidate for the Substitution Method? (a)
Z
5x 4 sin.x sin.x 5 / dx
(b)
Z
sin5 x cos x dx
(c)
Z
x 5 sin x dx
The function in (c) : x 5 sin x is not of the form g.u.x//u form g.u.x//u0 .x/. .x/ . The function in (a) meets the prescribed pattern with 5 sin u and u.x/ and u.x/ x . Similarly, the function in (b) meets the prescribed pattern with g.u/ u5 and u.x/ and u.x/ sin x .
SOLUTION
g.u/
D
D
D
D
2. Find an appropriate choice of u for u for evaluating the following integrals by substit ution: (a)
Z
x.x 2
C 9/4 dx
(b)
Z
x 2 sin.x sin.x 3 / dx
(c)
Z
sin x cos 2 x dx
SOLUTION
C 9/4 D 12 .2x/.x2 C 9/4; hence, c hence, c D 12 , f .u/ .u/ D u4 , and u.x/ and u.x/ D x 2 C 9. .u/ D sin u, and u.x/ (b) x 2 sin.x sin.x 3 / D 13 .3x 2 / sin.x sin.x 3 /; hence, c hence, c D 13 , f .u/ and u.x/ D x 3 . (c) sin x cos2 x D . sin x/ cos2 x ; hence, c hence, c D 1, f.u/ D u2, and u.x/ and u.x/ D cos x . 2 3. Which of the following is equal equal to x 2 .x 3 C 1/dx for 1/dx for a suitable substitution? (a) x.x 2
Z 0
1 (a) 3
2
Z
9
u du
(b)
0
Z
1 (c) 3
u du
0
SOLUTION
With the substitution u substitution u
(c).
D x3 C 1, the definite integral
2 2 3 0 x .x
R
C
In Exercises Exercises 1–6, 1–6, calculate calculate du .
D x3 x2
SOLUTION
Let u Let u
D x 3 x2 . Then du 2x/dx . Then du D .3x 2 2x/dx.
D 2x4 C 8x 1 SOLUTION Let u Let u D 2x 4 C 8x 1 . Then du Then du D .8x 3 8x 2 / dx . 3. u D cos.x cos.x 2 / SOLUTION Let u Let u D cos.x cos.x 2 /. Then d Then du u D sin.x sin.x 2 / 2x dx D 2x sin 2x sin.x .x 2 / dx . 4. u D tan x SOLUTION Let u Let u D tan x . Then du Then du D sec2 x dx . 5. u D e 4x C1 SOLUTION Let u Let u D e 4x C1 . Then du Then du D 4e 4x C1 dx. dx . 6. u D ln.x ln.x 4 C 1/ 4x 3 SOLUTION Let u Let u D ln.x ln.x 4 C 1/. 1/. Then d Then du uD 4 dx d x . x C1
2. u
In Exercises Exercises 7–22, 7–22, write the the integral integral in terms terms of u and du d u. Then evaluate. 7.
Z
.x 7/3 dx, dx ,
SOLUTION
8.
Z
.x
Let u Let u
u
Dx7
D x 7. Then du Then du D dx. dx . Hence
Z
SOLUTION
2
C 25/
Let u Let u
dx ,
u
.x
3
7/
dx
D
Z
u3 du
D 14 u4 C C D D 14 .x 7/4 C C:
D x C 25
D x C 25. 25. Then d Then du u D dx and dx and
Z
.x
2
C 25/
dx
D
Z
u
2
du
D u 1 C C D D x C1 25 C C:
u du
1 9 1/dx becomes 1/dx becomes 13 1 u du. du . The correct answer is
Exercises 1. u
9
Z
R
SECTION
9.
Z p C t2
t
SOLUTION
Substituti Substitution on Method Method
5.6
D t2 C 1 Let u Let u D t 2 C 1. Then du Then du D 2t dt . Hence, 1 dt ,
u
Z p C t
Z
1 dt
D
1 2
Z
u1=2 du
D 13 u3=2 C C D D 13 .t 2 C 1/3=2 C C:
C 1/ cos.x cos.x 4 C 4x/dx, 4x/dx, u D x 4 C 4x SOLUTION Let u Let u D x 4 C 4x. 4x . Then d Then du u D .4x 3 C 4/dx D 4.x 3 C 1/dx and 1/dx and 10.
.x 3
t2
Z
.x
11.
Z
t3 .4 2t 4 /11 Let u Let u
SOLUTION
12.
Let u Let u
C 1/ cos.x cos.x C 4x/dx D
x.x
u
Z
cos u du
D 14 sin u C C D D 14 sin.x sin.x 4 C 4x/ C C:
t3 dt .4 2t 4 /11
D
1 8
Z
u
11
du
D 801 u
C C D D 801 .4 2t 4 /
10
10
C C:
D 4x 1
D 4x 1. Then du Then du D 4 dx or dx or 14 du D dx. dx . Hence p 1 1 2 3=2 1 1=2 C D 4x 1 dx D u du D u C D .4x 1/3=2 C C: 4 4 3 6
Z
C 1/9 dx , u D x C 1 Let u Let u D x C 1. Then x Then x D u 1 and du and du D dx . Hence
Z Z p
x 4x 1 dx ,
SOLUTION
1 4
D 4 2t 4
u
Z
SOLUTION
14.
4
D 4 2t 4 . Then du Then du D 8t 3 dt . Hence,
4x 1 dx ,
SOLUTION
13.
d t ,
Z
Z p Z
3
Let u Let u
u
x.x
9
C 1/
dx
Z D
9
.u 1/u du
Z D
.u10 u9 / du
D 111 u11 101 u10 C C D D 111 .x C 1/11 101 .x C 1/10 C C:
D 4x 1
D 4x 1. Then x Then x D 14 .u C 1/ and 1/ and d du u D 4 dx or dx or 14 du d u D dx . Hence, p 1 1 x 4x 1 dx D .u C 1/u1=2 du D .u3=2 C u1=2 / du 16 16
Z
Z
D
1 16
Z
2 5=2 u 5
C
1 16
2 3=2 u 3
C C
D 401 .4x 1/5=2 C 241 .4x 1/3=2 C C: 15.
Z
p
x2 x
SOLUTION
C 1 dx , u D x C 1 Let u Let u D x C 1. Then x Then x D u 1 and du and du D dx . Hence
Z
x
2
p
x
C 1 dx
Z D
2 1=2
.u 1/ u
du
Z D
.u5=2 2u3=2
C u1=2 / du
D 27 u7=2 45 u5=2 C 23 u3=2 C C D 27 .x C 1/7=2 45 .x C 1/5=2 C 23 .x C 1/3=2 C C:
16.
Z
, sin.4 sin.4 7/ d
SOLUTION
Let u Let u
u
D 4 7
D 4 7. Then du and Then du D 4 d and
Z
D D 14
sin.4 sin.4 7/ d
Z
sin u du
D 14 cos u C C D D 14 cos.4 cos.4 7/ C C:
623
624
17.
Z
, d sin2 cos cos d
Let u Let u
SOLUTION
18.
Z Z
D sin
D sin . Then d . Hence, Then du u D cos d d
sec2 x tan x dx , Let u Let u
u
2
sin cos cos d d
xe
x2
u
Let u Let u
2
sec x tan x dx
Z D
u
xe
x2
dx
D
1 2
Z
21.
22.
Z
Let u Let u
Z u
Z
C
u
D
Z
e u du
1
D tan
.ln x/ 2 dx x
Z D
D 1C1x
d x , and
u2 du
1x
x . Then du Then du
1 x/ 2
2
.tan x2
C1
dx
Z D
u2 du
D 13 u3 C C D D 13 .tan
24.
Z
Let u Let u
x/ 3
C C:
C C for for an appropriate choice of u.x/ and constant a .
D x 4 . Then du Then du D 4x 3 dx or dx or 14 d u D x 3 dx. dx . Hence
Z
3
4
x cos.x cos.x / dx
D
1 4
Z
cos u du
D 14 sin u C C D D 14 sin.x sin.x 4 / C C:
x 2 cos.x cos.x 3
SOLUTION
25.
1
x 3 cos.x cos.x 4 / dx
SOLUTION
Z
D eu C C D D etan t C C:
D 13 u3 C C D D 13 .ln x/ 3 C C:
sin.u.x// In Exercises Exercises 23–26, 23–26, evaluate evaluate the integral integral in the the form a sin.u.x// 23.
C C:
D tan
Z Z
x2
D ln x . Then du Then du D x1 d x , and
Let u Let u
D 12 eu C C D D 12 e
D ln x
.tan 1 x/ 2 dx , x2 1
SOLUTION
e u du
D tan t . Then du Then du D sec2 t d t and
.ln x/ 2 dx , x
SOLUTION
D 12 u2 C C D D 12 tan2 x C C:
D tan t .sec2 t /etan t dt
Z
u du
D x 2. Then du dx or 12 d u D x dx . Hence, Then du D 2x dx or
.sec2 t /e tan t dt ,
SOLUTION
D 13 u3 C C D D 13 sin3 C C:
u2 du
D x2
Z
Z
Z
D tan x . Then du Then du D sec2 x dx . Hence
dx ,
Let u Let u
D D
D tan x
Z
SOLUTION
20.
u
Z
SOLUTION
19.
THE INTEGRAL
CHAPTER 5
C 1/dx Let u Let u D x 3 C 1. Then du Then du D 3x 2 dx or dx or 13 d u D x 2 dx . Hence
Z
2
x cos.x cos.x
3
C 1/dx D
1 3
Z
cos u du
D 13 sin u C C:
x 1=2 cos.x cos.x 3=2 / dx
SOLUTION
Let u Let u
D x 3=2 . Then d Then du u D 32 x 1=2 dx or dx or 23 d u D x 1=2 dx . Hence
Z
x
1=2
cos.x cos.x
3=2
/ dx
D
2 3
Z
cos u du
D 23 sin D 23 sin sin u C C D sin.x .x 3=2 / C C:
SECTION
26.
Z
5.6
Substituti Substitution on Method Method
cos x cos. cos.sin x/dx Let u Let u
SOLUTION
D sin x . Then d Then du u D cos x dx . Hence
Z
Z D
cos x cos. cos.sin x/dx
cos u du
D sin u C C:
In Exercises Exercises 27–72, 27–72, evaluate evaluate the the indefinite indefinite integral. integral.
Z
C 5/9 dx SOLUTION Let u Let u D 4x C 5. Then du Then du D 4 dx and dx and 27.
.4x
Z
9
.4x
28.
Z
dx .x 9/5 Let u Let u
SOLUTION
29.
Z p
Z
C 12
Let u Let u
SOLUTION
dx
D
Z
u9 du
D 401 u10 C C D D 401 .4x C 5/10 C C:
D x 9. Then d u D dx and dx and Then du
dt
t
C 5/
1 4
dx .x 9/5
Z D
u
5
du
D 14 u 4 C C D D 4.x 1 9/4 C C:
D t C 12. 12. Then du Then du D dt and
Z p
dt
D t C 12
Z
Z
u
1=2
du
p D 2u1=2 C C D D 2 t C 12 C C:
C 2/2=3 dt SOLUTION Let u Let u D 9t C 2. Then d Then du u D 9 dt and 30.
.9t
Z
.9t
Z
u2=3 du
D 19 35 u5=3 C C D D 151 .9t C 2/5=3 C C:
C 1 d x C 2x/3 SOLUTION Let u 2x . Then du 2/dx or 12 du D .x C 1/dx. 1/dx . Hence Let u D x 2 C 2x. Then du D .2x C 2/dx or 1 2 xC1 1 1 1 1 2 1 2 C D C D C C: d x D d u D C D C D u .x C 2x/ 2 3 3 2 2 2 2 4 .x C 2x/ u 4.x C 2x/ 2 32. .x C 1/.x 2 C 2x/ 3=4 dx SOLUTION Let u 2x . Then du 1/dx and Let u D x 2 C 2x. Then du D .2x C 2/dx D 2.x C 1/dx and 31.
Z
x
C 2/2=3 dt D 19
.x 2
Z
Z
Z
Z
.x
33.
Z p
x
x2
SOLUTION
C9
34.
.4x 3
SOLUTION
3=4
dx
D
1 2
Z
u3=4 du
D 12 47 u7=4 C C
D 27 .x2 C 2x/7=4 C C:
dx
Let u Let u
2x 2
2
C 1/.x C 2x/
D x 2 C 9. Then du Then du D 2x dx or dx or 12 du D x dx . Hence p x 1 1 1 u p 2 dx D 2 p u du D 2 1 C C D D x C9 2
Z
Z
Z
p C x2
9
C C:
C x d x C 3x2/2 u D .12x 2 C 6x/dx or 6x/dx or 16 du D .2x 2 C x/dx. x/dx . Hence Let u Let u D 4x 3 C 3x 2 : Then d Then du
Z
.4x 3
C 3x2/
2
.2x 2
C x/dx D 16
Z
u
2
du
D 16 u 1 C C D D 16 .4x 3 C 3x2 / 1 C C:
625
626
35.
CHAPTER 5
Z
.3x 2
THE INTEGRAL
C 1/.x3 C x/ 2 dx Let u Let u D x 3 C x . Then d Then du u D .3x 2 C 1/dx. 1/dx . Hence
SOLUTION
Z
.3x
36.
Z
5x 4 2x d x .x 5 x 2 /3
C C
Let u Let u
SOLUTION
2
3
C 1/.x C x/
2
D 13 u3 C C D D 13 .x3 C x/3 C C:
u2 du
D x 5 C x2 . Then du Then du D .5x 4 C 2x/dx. 2x/dx . Hence 5x 4 C 2x 1 1 1 1 1 C D C C: d x D d u D C D 5 2 3 3 2 5 2u 2 .x C x 2 /2 .x C x / u
Z
Z
dx
Z D
Z
C 8/11 dx SOLUTION Let u Let u D 3x C 8. Then d Then du u D 3 dx and dx and 37.
.3x
Z
.3x
38.
Z
11
C 8/
dx
D
1 3
Z
u11 du
D 361 u12 C C D D 361 .3x C 8/12 C C:
C 8/11 dx
x.3x
Let u Let u
SOLUTION
u8 D 3x C 8. Then d u D 3 dx , x D Then du , and 3
Z
x.3x
11
C 8/
dx
D
1 9
Z
D
1 9
11
.u 8/u
du
1 13 2 12 u u 13 3
1 9
D
Z
.u12 8u11 / du
C C
1 2 D 117 .3x C 8/13 .3x C 8/12 C C: 27 39.
Z p C x2
x3
1 dx
Let u Let u
D x 3 C 1. Then du dx and Then du D 3x 2 dx and
SOLUTION
Z p C x
40.
Z p C x5
SOLUTION
2
x3
1 dx
D
1 3
Z
u1=2 du
D 29 u3=2 C C D D 29 .x3 C 1/3=2 C C:
x3
1 dx
Let u Let u
D x 3 C 1. Then du Then du D 3x 2 dx , x 3 D u 1 and p 1 1 x 5 x 3 C 1 dx D .u 1/ u du D
Z p
D
3
Z
1 3
3
2 5=2 u 5
2 3=2 u 3
Z
.u3=2 u1=2 / du
C C
D 152 .x 3 C 1/5=2 29 .x 3 C 1/3=2 C C: 41.
Z
dx .x 5/3
C
Let u Let u
SOLUTION
42.
Z
x 2 dx .x 5/3
SOLUTION
D x C 5. Then du Then du D dx and dx and
Z
dx
D .x C 5/3
Z
u
3
du
D 12 u 2 C C D D 12 .x C 5/ 2 C C:
C
Let u Let u
D x C 5. Then du Then du D dx , x D u 5 and
Z
x 2 dx .x 5/3
C D
Z
.u 5/2 du u3
D
Z
.u
1
2
10u
C 25u
3
/ du
SECTION
1
D ln juj C 10u 43.
Z
25 u 2
2
5.6
Substituti Substitution on Method Method
C C
25 D ln jx C 5j C x 10 C 5 2.x C 5/2 C C: z 2 .z 3
C 1/12 dz Let u Let u D z 3 C 1. Then du Then du D 3z 2 dz and dz and
SOLUTION
Z
Z
2
z .z
3
12
C 1/
dz
1 3
D
Z
D 391 u13 C C D D 391 .z3 C 1/13 C C:
u12 du
C 4z 2 /.z3 C 1/12 dz SOLUTION Let u Let u D z 3 C 1. Then du Then du D 3z 2 dz , z 3 D u 1 and 44.
.z 5
Z
.z 5
C 4z 2 /.z3 C 1/12 dz D 13 D
1 3
Z
.u
C 3/u12 du D 13
1 14 u 14
C
3 13 u 13
Z
.u13
C 3u12 / du
C C
D 421 .z 3 C 1/14 C 131 .z3 C 1/13 C C:
Z
C 2/.x C 1/1=4 dx SOLUTION Let u Let u D x C 1. Then x Then x D u 1, du D dx and dx and 45.
.x
Z
.x
1=4
C 2/.x C 1/
dx
Z D
.u
1=4
C 1/u
du
Z D
.u5=4
C u1=4 / du
D 49 u9=4 C 45 u5=4 C C
D 49 .x C 1/9=4 C 45 .x C 1/5=4 C C: 46.
Z
x 3 .x 2 1/3=2 dx Let u Let u
SOLUTION
Z
D x 2 1. Then u dx or 12 d u D x dx . Hence Then u C 1 D x 2 and du and du D 2x dx or 3
x .x
2
3=2
1/
dx
Z D Z D
x 2 x.x 2 1/3=2 dx
1 2
D 47.
Z
Let u Let u
C
1 2
D
2 5=2 u 5
.u5=2
C u3=2 / du
C C D D 17 .x2 1/7=2 C 15 .x 2 1/5=2 C C:
sin.8 sin.8 3 / d
D D
1 3
Z
sin u du
D 13 cos u C C D D 13 cos.8 cos.8 3 / C C:
sin. sin. 2 / d
Let u Let u
SOLUTION
49.
2 7=2 u 7
du
Z
D 8 3 . Then du and Then du D 3 d and
Z 48.
C 1/u
1 2
sin.8 sin.8 3 / d
SOLUTION
Z
1 2
.u
3=2
Z p p t
cos
SOLUTION
t
D 2 . Then du and Then du D 2 d d and
Z
2
sin. sin. / d
D D
1 2
Z
sin u du
D 12 cos u C C D D 12 cos. cos. 2 / C C:
dt
Let u Let u
D p t D t 1=2 . Then du Then du D 12 t 1=2 dt and p cos t p t dt D 2 cos u du D 2 sin u C C D D 2 sin p t C C:
Z
Z
627
628
50.
Z
x 2 sin.x sin.x 3
C 1/dx Let u Let u D x 3 C 1. Then du Then du D 3x 2 dx or dx or 13 du D x 2 dx. dx . Hence
SOLUTION
51.
Z
Z
Z
53.
Z
Z
tan.4 tan.4
Let u Let u
Let u Let u
1=5
57.
Z
D 14 ln j sec uj C C D D 14 ln j sec.4 C 9/j C C: sec.4 C
tan u du
Z D D
8
d sin cos cos d
D 19 u9 C C D D 19 sin9 C C:
u8 du
cot x dx
D
Z
cos x dx sin x
D
Z
du u
D ln juj C C D D ln j sin xj C C:
Let u Let u
4 D x 4=5 . Then du Then du D x 5 1=5
/ tan x
4=5
1=5
dx
D
dx and dx and 5 4
Z
tan u du
D 54 ln j sec uj C C D D 54 ln j sec x 4=5 j C C:
sec2 .4x
C 9/dx Let u Let u D 4x C 9. Then d Then du u D 4 dx or dx or 14 d u D dx. dx . Hence
Z
2
sec .4x
C 9/dx D
1 4
Z
sec2 u du
D 14 tan u C C D D 14 tan.4x tan.4x C 9/ C C:
sec2 x tan4 x dx Let u Let u
D tan x . Then du Then du D sec2 x dx . Hence
Z
p p x
sec2 . x/dx
SOLUTION
Z
C 9/ d D D
1 4
/ tan x 4=5 dx
Z
SOLUTION
Z
D 13 cos u C C D D 13 cos.x cos.x 3 C 1/ C C:
D sin x. Then du Then du D cos x dx , and
Z
SOLUTION
56.
sin u du
D sin . Then d and Then du u D cos d d and
.x
Z
Z
cot x dx
SOLUTION
55.
C 1/dx D
Z
.x
Z
1 3
sin8 cos cos d d
SOLUTION
54.
x sin.x sin.x
3
C 9/ d C 9. Then du and Let u Let u D 4 C Then du D 4 d and
SOLUTION
Z
2
tan.4 tan.4
SOLUTION
52.
THE INTEGRAL
CHAPTER 5
Let u Let u
2
4
sec x tan x dx
Z D
u4 du
D 15 u5 C C D D 15 tan5 x C C:
1 1 D p x. Then du Then du D p d x or 2 or 2 du D p d x . Hence, 2 x x p p sec2 . x/dx p x D 2 sec2 u dx D 2 tan u C C D D 2 tan. tan. x/ C C:
Z
cos 2x
Z
C sin 2x/2 d x SOLUTION Let u Let u D 1 C sin 2x. 2x . Then du Then du D 2 cos 2x or 2x or 12 du D cos 2x dx. dx . Hence 58.
.1
Z
.1
59.
Z
p
sin 4x cos 4x
C sin 2x/
C 1 dx
2
cos 2x dx
D
1 2
Z
2
u
du
D 12 u 1 C C D D 12 .1 C sin 2x/ 1 C C:
SECTION
Let u Let u
SOLUTION
60.
Z
Z
D cos 4x C 1. Then du Then du D 4 sin 4x or 4x or 14 du D sin 4x. 4x . Hence p 1 1 2 3=2 C C D D 16 .cos 4x C 1/3=2 C C: sin 4x cos 4x C 1 dx D u1=2 du D u 4 4 3
Z
Let u Let u
D 3 sin x 1. Then du Then du D 3 cos x dx or dx or 13 du D cos x dx . Hence
Z
.3 sin x 1/ cos x dx
Z Z
Let u Let u
Z
Z
u du
1 3
D
1 2 u 2
C C D D 16 .3 sin x 1/2 C C:
D sec 1. Then du and d Then du D sec tan tan d and
Z
Z
sec tan tan .sec 1/ d
D D
u du
D 12 u2 C C D D 12 .sec 1/2 C C:
cos t cos. cos.sin t / d t Let u Let u
SOLUTION
63.
D
1 3
sec tan tan .sec 1/ d
SOLUTION
62.
Substituti Substitution on Method Method
cos x.3 sin x 1/dx
SOLUTION
61.
5.6
D sin t . Then du Then du D cos t d t and
Z e 14x
7
cos u du
D sin u C C D D sin. sin.sin t / C C:
dx
Let u Let u
SOLUTION
cos t cos. cos.sin t / d t
Z D
1 D 14x 7. Then du Then du D 14dx or 14dx or 14 d u D dx. dx . Hence,
Z
Z
e
14x 7
dx
1 14
D
Z
e u du
D 141 eu C C D D 141 e14x 7 C C:
2
C 1/ex C2x dx SOLUTION Let u Let u D x 2 C 2x. 2x . Then du Then du D .2x C 2/dx or 2/dx or 12 d u D .x C 1/dx. 1/dx . Hence, 64.
.x
Z
.x
65.
Z
ex .e x
SOLUTION
66.
dx 1/4
C
Let u Let u
C2x dx D 1
2
ex .e x
C
1/4
d x
Z D
Let u Let u
D 12 eu C C D D 12 ex C2x C C: 2
4
u
du
D 3u13 C C D D 3.ex 1C 1/3 C C:
D tan . Then du , and Then du D sec2 d d
Z
2
.sec / e
Z
e2t
SOLUTION
68.
e u du
.sec2 / e tan d
SOLUTION
67.
Z
D ex C 1. Then d Then du u D e x dx, dx , and
Z
Z
C 1/e
x2
Z
e t dt 2e t
Z D D
d
e u du
D eu C C D D etan C C:
C1 Let u Let u D e t . Then d Then du u D e t dt , and
C
dx x.ln x. ln x/ 2
SOLUTION
tan
Let u Let u
Z
e 2t
e t dt 2e t
C
C1
Z D
u2
du 2u
C C1
Z D
du .u 1/2
1 1 D C D C C: C D t uC1 e C1 C
D ln x. Then du Then du D x1 d x , and
Z
dx x.ln x. ln x/ 2
Z D
u
2
du
D u1 C C D D ln1x C C:
629
630
69.
Z
.ln x/ 4 dx x
SOLUTION
70.
Z Z
D ln x . Then du Then du D x1 d x , and
Z
Let u Let u
Z
Let u Let u
.ln x/ 4 dx x
Z D
D 15 u5 C C D D 15 .ln x/ 5 C C:
u4 du
D ln x . Then du Then du D x1 d x , and
Z
tan. tan.ln x/ dx x
SOLUTION
72.
Let u Let u
dx x ln x
SOLUTION
71.
THE INTEGRAL
CHAPTER 5
dx x ln x
D
Z
du u
D ln juj C C D D ln j ln x j C C:
D cos. cos.ln x/. x/ . Then du Then du D x1 sin. sin .ln x/dx or x/dx or du D x1 sin. sin.ln x/dx. x/dx . Hence,
Z
tan. tan.ln x/ dx x
Z D
sin. sin.ln x/ dx d x x cos. cos.ln x/
D
Z
du u
D ln juj C C D D ln j cos. cos.ln x/ j C C:
.cot x/ ln. ln.sin x/dx
SOLUTION
Let u Let u
D ln. x/ . Then ln.sin x/. D sin1 x cos x D cot x;
du and
Z
.cot x/ ln. ln.sin x/dx
Z
u du
D 12 u2 C C D D 12 .ln . ln..sin x// 2 C C:
p x. Hint: Show that dx p D C using u using u 1 that dx D 2.u 1/du. 1/du. .1 C x/ 3 p Let u Let u D 1 C x . Then p d u D 2.u 1/du: 1 du D p d x or dx D 2 x du
73. Evaluate SOLUTION
D
Z
dx
2 x
Hence,
Z
dx
.1
Z
u1 du u3
Z
D 2 .u 2 u 3 / du C p x/3 D 2 D 2u 1 C u 2 C C D D 1 C2p x C .1 C 1p x/2 C C: Hannah uses the substitution u substitution u D tan x and Akiva uses u uses u D sec x to evaluate
74. Can They Both Both Be Right? Right? Show that they obtain different answers, and explain the apparent contradiction. SOLUTION
With the substitution u substitution u
Z
D tan x, Hannah finds d u D sec2 x dx and dx and finds du
tan x sec2 x dx
On the other hand, with the substitution u substitution u
Z
D
Z
u du
tan x sec2 x dx .
R
D 12 u2 C C 1 D 12 tan2 x C C 1 :
D sec x, Akiva finds du finds du D sec x tan x dx and dx and 2
tan x sec
x dx D
Z
sec x.tan x. tan x sec x/dx
D 12 sec2 x C C 2
Hannah and Akiva have each found a correct antiderivative. To resolve what appears to be a contradiction, recall that any two antiderivatives antiderivatives of a specified function differ by a constant. To To show that this is true in their case, note that
1 sec2 x 2
C C 2
1 tan2 x 2
C C 1 D 12 .sec2 x tan2 x/ C C 2 C 1
D 12 .1/ C C 2 C 1 D 12 C C 2 C 1 ; a constant Here we used the trigonometric identity tan2 x
C 1 D sec2 x .
SECTION
R
75. Evaluate sin x cos x dx using substitution in two different ways: first using u using u the two different answers. SOLUTION
First, let u let u
631
D sin x and then using u using u D cos x . Reconcile
D sin x: Then x: Then du du D cos x dx and dx and
Z Next, let u let u
Substituti Substitution on Method Method
5.6
sin x cos x dx
Z D
u du
D 12 u2 C C 1 D 12 sin2 x C C 1:
D cos x . Then d Then du u D sin x dx or dx or du D sin x dx . Hence,
Z
sin x cos x dx
D
Z
D 12 u2 C C 2 D 12 cos2 x C C 2:
u du
To reconcile reconcile these two seemingly different answers, recall that any two antiderivatives antiderivatives of a specified function differ by a constant. 1 C 1 / . 12 cos2 x C 2 / C 1 C 2 , a constant. Here we used the To show that this is true here, note that . 21 sin2 x 2 2 2 trigonometric identity sin x cos x 1.
C
C
D
C
D C
76. Some Choices Choices Are Better Than Others Others Evaluate
Z twice. First use u use u
sin x cos 2 x dx
D sin x to show that
Z
sin x cos 2 x dx
D
Z p
u 1 u2 du
and evaluate the integral on the right by a further substitution. Then show that u that u SOLUTION
Now let w let w
Consider the integral
sin x cos2 x dx .
R
Z
D cos x is a better choice. p If we let u let u D sin x , then cos x D 1 u2 and du and du D cos x dx . Hence 2
sin x cos x dx
Z p D
u 1 u2 du:
D 1 u2 . Then dw Then dw D 2udu or 2udu or 12 dw D u du. du . Therefore,
Z p u
1 u2 du
D
1 2
Z
w
1=2
dw
D
1 2
2 3=2 w 3
C C
D 13 w3=2 C C D D 13 .1 u2/3=2 C C
D 13 .1 sin2 x/ 3=2 C C D D 13 cos3 x C C: A better substitution choice is u is u D cos x . Then du Then du D sin x dx or dx or du D sin x dx . Hence 1 1 3 D cos x C C: sin x cos2 x dx D u2 du D u3 C C D 3 3 77. What are the new limits of integration if we apply the substitution u substitution u D 3x C to the integral 0 sin.3x sin.3x C / dx? dx ? SOLUTION The new new limits of integration are u.0/ are u.0/ D 3 0 C D and u. and u. / D 3 C D 4 . 8 78. Which of the following is the result result of applying the substitution u substitution u D 4x 9 to the integral 2 .4x 9/20 dx ?
Z
8
(a)
Z Z
20
u
Z
R
1 (b) 4 1 (d) 4
du
2
23
(c) 4
u20 du
1
SOLUTION
Hence
Let u Let u
8
Z Z
u20 du
2 23
u20 du
1
D 4x 9. Then d Then du u D 4 dx or dx or 14 d u D d x . Furthermore, when x D 2, 2 , u D 1, and when x when x D 8, 8 , u D 23. 23 . 8
Z 2
.4x
20
9/
dx
D
1 4
23
Z
u20 du:
1
The answer is therefore (d) . In Exercises Exercises 79–90, 79–90, use the the Change-of-V Change-of-Variable ariabless Formula Formula to evaluate evaluate the definite definite integr integral. al. 3
79.
Z 1
.x
R
C 2/3 dx
632
THE INTEGRAL
CHAPTER 5
Let u Let u
SOLUTION
D x C 2. Then du Then du D dx . Hence 3
Z
.x
1
6
80.
Z p C x
3
C 2/
5
dx
Z D
3
u du
D
3
1 4 u 4
3 dx
5
54 4
ˇˇ D 3
34 4
D 136:
1
Let u Let u
SOLUTION
D x C 3. Then du Then du D dx . Hence 6 p 9 p x C 3 dx D u du D
Z 1
1
81.
4
x
Z
.x 2
0
SOLUTION
Z
2
x .x 2
0
Z p
d x 1/3
C
2
1 2
D
Z 1
Z
4
1 2
D
1 u 2
2
ˇˇˇ
2
ˇD
1
ˇˇ ˇ
Z
1
1
Z p C x
2 .27 8/ 3
D 383 :
1 16
C 14 D 163 D 0:1875:
C 6 dx Let u Let u D 5x C 6. Then d Then du u D 5 dx or dx or 15 d u D dx. dx . Hence 2 2 p 1 16 p 1 2 3=2 16 42 D 5x C 6 dx D u du D u .64 1/ D : 5 5 3 15 5
SOLUTION
83.
1 d u u3
5x
1
4
9
ˇˇ D ˇ
C 1/3 d x dx or 12 d u D x dx . Hence Let u Let u D x 2 C 1. Then du Then du D 2x dx or 1
82.
2 3=2 u 3
Z
1
x2
9 dx
Let u Let u
D x 2 C 9. Then du Then du D 2x dx or dx or 12 d u D x dx . Hence 1 4 1 25 p 1 2 3=2 25 2 D .125 27/ D 98 : x x C 9 dx D u du D u
0
SOLUTION
Z p
2
0
2
84.
Z 1
Z
2
9
ˇˇ ˇ
3
4x .x 2
SOLUTION
3
9
3
C 12 dx C 6x C 1/2 6/dx and Let u Let u D x 2 C 6x C 1. Then du Then du D .2x C 6/dx and 2 17 4x C 12 2 17 2 D D d x 2 u du 2 2 u 8 1 .x C 6x C 1/ 8 D 172 C 14 D 689 :
Z
Z
ˇˇˇ ˇ
1
Z
C 1/.x2 C 2x/5 dx SOLUTION Let u Let u D x 2 C 2x. 2x . Then d Then du u D .2x C 2/dx D 2.x C 1/dx, 1/dx , and 85.
.x
0
1
Z
.x
0
17
86.
Z
2=3
.x 9/
2
C 1/.x C 2x/
5
dx
D
1 2
3
Z
5
u du
0
1 6 u 12
D
dx
3
ˇˇˇ D ˇ 0
729 12
D 243 : 4
10
SOLUTION
Let u Let u
D x 9. Then du Then du D dx. dx . Hence 17
Z
2=3
.x 9/
10
1
87.
Z
8
dx
Z D
2=3
u
dx
1
1=3
D 3u
tan. tan. 2 / d
8
ˇˇˇ D ˇ
3 .2 1/
1
D 3:
0
SOLUTION
1
Z 0
Let u Let u
D cos 2 . Then du or . Hence, Then du D 2 sin sin 2 d or 12 du D sin sin 2 d 2
1
Z D D
tan. tan. / d
0
sin. sin. 2 /
cos. cos. 2 /
d
D D
1 2
cos 1
Z 1
du u
D
ˇj jˇˇ ˇ
1 ln u 2
cos 1
1
1 D 12 Œln Œ ln..cos 1/ C ln 1 D ln. ln.sec 1/: 2
SECTION
=6
88.
Z
sec2 2x
0
SOLUTION
Let u Let u
6
d x
D 2x 6 . Then du Then du D 2 dx and dx and =6
Z
sec
2
0
2x
6
d x
1 2
D
D 12 =2
89.
Z
Substituti Substitution on Method Method
5.6
=6
Z p
sec u du
D =6 p p 3 3 3 C D : 3 3 3
ˇˇ
1 tan u 2
2
!
=6 =6
cos3 x sin x dx
0
SOLUTION
Let u Let u
D cos x . Then du Then du D sin x dx . Hence =2
Z
0
3
cos x sin x dx
D
0
=2
90.
Z
1
Z D
3
u du
1
Let u Let u
=2
cot2
=3
0
x x csc 2 d x 2 2
1
D 2 D
2
Z
D
1
ˇˇˇ D ˇ
1 4
0
0
D 14 :
r
5
Let u Let u
D5
u2 du
p
2 3 u 3
Z q p 0
SOLUTION
u du
1 4 u 4
1 x D cot x2 . Then du Then du D csc2 and 2 2
Z 91. Evaluate
3
x x csc2 d x 2 2
cot2
=3
SOLUTION
Z
4 r 2 dr .
3 1
ˇˇ D ˇp 3
p
2 .3 3 1/: 3
p
4 r 2 . Then du
D p r d r 2 D 5r d ru 4r
so that r dr
D .5 u/du:
Hence, the integral becomes: 2
Z q p r
5
4 r2 dr
0
5
5
Z D p
u.5 u/du
3
D
50 3
Z D
1=2
5u
3=2
u
3
p
p
5 10 5
p
10 3
18 5
p 3
d u
D
D 20 p
10 3=2 u 3
5
3
32 5
2 5=2 u 5
p
3:
92. Find numbers a numbers a and and b b such such that b
Z
.u
2
a
=4
C 1/du
Z D
sec4 d d
=4
and evaluate. Hint: Use the identity sec2 SOLUTION
D D tan2 C 1. . Moreover, because Let u Let u D tan . Then u Then u2 C 1 D tan2 C 1 D sec2 and d and du u D sec2 d d tan
it follows that a that a
D
4
1
an and
tan tan
4
D 1;
D 1 and b and b D 1. Thus, =4
Z
4
1
Z D D
d sec d =4
.u 1
2
C 1/du D
1 3 u 3
ˇˇˇ
Cu
ˇ
1 1
D 83 :
ˇˇ
ˇ
5 3
633
634
THE INTEGRAL
CHAPTER 5
93. Wind engineers have found that wind speed v (in v (in meters/second) at a given location follows a Rayleigh distribution of the type
D 321 ve
W.v/
v 2 =64
This means that at a given moment in ti me, the probability that v that v lies lies between a between a and and b b is is equal to the shaded area in Figure 1. (a) Show that the probability probability that v that v
2 Œ0; Œ0; b is 1 is 1 e (b) Calculate the probability that v that v 2 Œ2;5.
b 2 =64 .
y
0.1 y = W (v)
0.05
(m/s)
v
a
20
b
FIGURE 1 The shaded area is the probability that v that v lies lies betweena betweena and b and b.. SOLUTION
(a) The probability that v that v
2 Œ0;b is b
1 ve 32
Z 0
Let u Let u
v 2 =64
dv:
D v2=64. =64. Then du Then du D v=32dv and v=32dv and b
Z 0
(b) The probability that v that v that v that v Œ2;5 is
2
1 ve 32
=2
94. Evaluate
dv
D
Z
u
e du
0
D e
u
ˇˇ ˇ
b 2 =64
D e
0
b 2 =64
1e
sinn x cos x dx for dx for n n
25=64
1e
1=16
D
e
1=16
e
25=64
:
0.
0
SOLUTION
Let u Let u
D sin x. Then du Then du D cos x dx . Hence =2
Z
1
n
sin x cos x dx
0
Z D
n
u du
0
D
95.
0
1
n
C 1:
f.x/3 f 0 .x/dx
SOLUTION
Let u Let u
D f.x/. f.x/. Then du Then du D f 0 .x/dx. .x/dx . Hence
Z
3
f.x/
96.
1
ˇˇ C ˇˇ D
unC1 n 1
.x/. In Exercises Exercises 95–96, 95–96, use substitutio substitution n to evaluate evaluate the the integral integral in terms of f .x/
Z Z
f 0 .x/ d x f.x/2
SOLUTION
Let u Let u
f 0 .x/dx D
=6
97. Show that
0
Z
u3 du
D 14 u4 C C D D 14 f.x/4 C C:
D f.x/. f.x/. Then du Then du D f 0 .x/dx. .x/dx . Hence
Z 0 Z D D
f .x/ d x f.x/2
Z
C 1:
2 Œ2;5 is the probability that v that v 2 Œ0; Œ0; 5 minus the probability that v that v 2 Œ0;2. By part (a), the probability
Z
b 2 =64
v 2 =64
1=2
f .sin / d
0
f.u/
Z D p 1
u
2
1 u2
du
du. du .
1 D u 1 C C D D f.x/ C C:
SECTION
Let u Let u
SOLUTION
Substituti Substitution on Method Method
5.6
635
D sin . Then u. , so that Then u. =6/ D 1=2 and 1=2 and u.0/ u.0/ D 0, as required. Furthermore, du Furthermore, du D cos d d du D D cos :
d
D D u, then u D 1; so D then u2 C cos2 D 1; so that cos D
If sin
p
1 u2 . Therefore d Therefore d
D D du=
=6
Z
1=2
Z D D
f .sin / d
0
p
1 u2 . This gives
p 1
f.u/
du:
1 u2
0
Further Insights and Challenges
D 1 C x 1=n to show that 1 C x 1=n dx D n
98. Use the substitution u substitution u
Z p p
Evaluate for n for n
Z
u1=2 .u 1/n
D 2; 3. Let u D 1 C x 1=n . Then x D .u 1/n and dx D n.u 1/n
1 du. du .
SOLUTION
1
du
Accordingly,
1/n 1 du. du . For n For n 2, we have
Z p p C 1
x 1=n
dx
D n
Z
u1=2 .u
D
Z p p C 1
For n For n
x 1=2 dx
D2
Z
D2
u1=2 .u 1/1 du
2 5=2 2 3=2 u u 5 3
D2
Z
.u3=2
1=2
u
/ du
C C D D 45 .1 C x 1=2 /5=2 43 .1 C x 1=2/3=2 C C:
D 3, we have
Z p p C 1
x 1=3 dx
D3
Z
D3
u
1=2
2
.u 1/ du
2 7=2 .2/ u 7
2 5
D3
Z
.u5=2
5=2
u
C
3=2
2u
2 3=2 u 3
C u1=2 / du
C C
D 67 .1 C x 1=3/7=2 125 .1 C x 1=3/5=2 C 2.1 C x 1=3/3=2 C C: =2
Z Z C C D D
99. Evaluate I Evaluate I
check that I that I
D
0
1
=2
J
C
d . Hint: Use substitution to show that I that I is is equal to J to J tan6;000
=2
D
Z 0
d
1
C cot6;000 and then
. d
0
SOLUTION
To evaluate =2
Z D D
I
dx
C tan6000 x ; we substitute t substitute t D =2 x . Then d Then dtt D dx , x D =2 t , t.0/ D =2, =2, and t and t .=2/ D 0. Hence, 0
=2
I
D D
Let J Let J
D D
0
of the integral,
1
0
C
dx tan6000 x
0
D
Z
=2
=2
dt
D 1 C tan6000 .=2 t /
Z 0
1
C
dt : cot6000 t
dt
=2
R
Z
1
1
D J , C J D D 2I . . We know I know I D J , so I so I C 2I . On the other hand, by t he definition of I I and J and J and and the linearity C cot6000 .t/ .t / =2
Z C C D D Z D Z D
I
J
0
=2
0
=2
0
1
C 1
1
dx tan6000 x
C C
C 1 C
1 tan6000 x 1 tan6000 x
dx cot6000 x
C 1 C
=2
Z D 0
1 .1= tan6000 x/
1
C
dx
1 1/= tan6000 x
C .tan6000 x C
1 tan6000 x
dx
C 1 C
1 cot6000 x
dx
636
THE INTEGRAL
CHAPTER 5
Z D C C Z D =2
0
=2
1
1 tan6000 x
1 1
tan6000 x tan6000 x
C C J D D 2I D D =2, D =4. Hence, I Hence, I C =2, so I so I D =4. 0
tan6000 x 1 tan6000 x
C C
! Z D
!
dx
=2
dx
1 dx
D =2:
0
a
100. Use substitution to prove that
Z
f.x/dx a
We assume that f that f is is continuous. If f f .x/ .x/ is an odd function, then f then f .x/
SOLUTION
du
D 0 if f is f is an odd function.
D dx or dx or du D dx . Accordingly, a
Z
0
f.x/dx a
Z D Z D
a
f.x/dx a
a
b=a b=a
0
f.x/dx
f.u/du
0
f .u/du
a
a
f.x/dx
a
Z
D
0
0
b
Z C Z
D f.x/. f.x/. Let u Let u D x . Then x Then x D u and
Z C
f.x/dx
0
D 0:
1 101. Prove that a x1 d x for a,, b > 0. 0 . Then show that the regions under the hyperbola over the intervals Œ1 intervals Œ1;; 2, Œ2; Œ2; 4, 1 x d x for a Œ4; Œ4; 8; : : : all : all have the same area (Figure 2).
R
D
R
y
1 y = x 1
Equal area
1 2
1 4
1 8
x
1
2
4
8
D x1 over Œ2 over Œ2n ; 2nC1 is the same for all n all n D 0 ; 1 ; 2 ; : : : : Let u Let u D xa . Then au Then au D x and du and du D a1 d x or a or a du D dx . Hence FIGURE 2 The area under under y
SOLUTION
b
Z a
b=a
1 d x x
b=a
Z D 1
a d u au
b=a
Z D 1
1 d u: u
b=a 1 1 d u d x after the substitution x substitution x u. u x 1 1 2 The area under the hyperbola over the interval Œ1 interval Œ1;; 2 is given by the definite integral 1 x1 d x . Denote this definite integral by A by A.. Using the result from part (a), we find the area under the hyperbola over the interval Œ2 interval Œ2;; 4 is
Note that
Z
D
Z
D
R
4
Z 2
1 d x x
4=2
Z D 1
1 d x x
2
Z D 1
1 d x x
D A:
1 d x x
D A:
Similarly, the area under the hyperbola over the interval Œ interval Œ4; 4; 8 is 8
Z 4
1 d x x
8=4
D
Z 1
1 d x x
2
D
Z 1
In general, the area under the hyperbola over the int erval Œ2 erval Œ2n ; 2nC1 is 2nC1
Z
2n
1 d x x
2nC1 =2n
D
Z 1
1 d x x
2
D
Z 1
1 d x x
D A:
102. Show that the two regions in Figure 3 have the same area. Then use the identity cos 2 u second area. y
1
y y = 1 − x 2
1
y = cos2 u
x
u
1
1
(A)
(B)
FIGURE 3
D
1 2 .1
C cos 2u/ to 2u/ to compute the
SECTION
SOLUTION
p
1 x2
R p 1 0
The area of the region region in Figure 3(A) is given by
p D
2
1 sin u
D cos u. Hence, 1
Z p
1 x 2 dx
0
Further Further Transce Transcendenta ndentall Functions Functions
5.7
1 x 2 dx. dx . Let x
=2
=2
Z D
cos u cos u du
0
Z D
D
sin u. Then dx
D
637
cos u du and
cos2 udu:
0
This last integral represents the area of t he region in Figure 3(B). The two regions in Figure 3 therefore have the same area. =2
=2
Z
2
cos u du
D
0
103. Area of of an Ellipse Ellipse
1 2
cos2 u du. du . Using the trigonometric identity cos2 u
R Z
Let’s now focus on the definite integral
0
=2
1
1 2
C cos 2udu D
0
u
C
ˇˇˇ
1 sin 2u 2
ˇ
=2
0
D 12 .1 C cos 2u/, 2u/, we have
D 12 2 0 D 4 :
D ab for ab for the area of the ellipse with equation (Figure 4)
Prove the formula A formula A
x2 a2
2
C yb2 D 1
A is equal to ab to ab times times the area of the unit circle. Hint: Use a change of variables to show that A is y b
a
a
!
x
b
!
FIGURE 4 Graph of
SOLUTION
half-plane, y half-plane, y
Consider the ellipse with equation
D f.x/
r D
x2 a2
b2 1
x2 a2
x2 a2
2
C yb 2 D 1.
2
C yb D 1; here a;b here a;b > 0. 0 . The area between the part of the ellipse in t he upper 2
, and the x the x-axis -axis is
a f.x/dx . By symmetry, the part of the elliptical region in the l ower a f.x/dx.
R
half-plane has the same area. Accordingly, the area enclosed by the ellipse is a
2
Z
a
f.x/dx a
Now, let u let u
Z s
D2
b2
x2 a2
1
a
a
dx
D 2b
Z q
1 .x=a/2 dx
a
D x=a. x=a . Then x Then x D au and au and a a du D dx . Accordingly, a
2b
Z r
1
a
Here we recognized that 2 .
R p 1 1
x a
2
1
dx
D 2ab
Z p
1 u2 du
1
D 2ab
Preliminary Questions b
1. Find b Find b such such that
1
(a) ln 3 SOLUTION
dx is equal equal to x (b) 3
For b For b > 0, 0, b
Z 1
dx x
b
ˇj jˇ D ˇ
D ln x
ln b ln 1
1
(a) For the value of the definite integral to equal ln 3, we must have b have b (b) For the value of the definite integral to equal 3, we must have have b b
2. Find b Find b such such that
Z 0
2
ab
1 u2 du represents du represents the area of the upper unit semicircular disk, which by Exercise 102 is 2. 4 /
5.7 Further Transcendental Functions
Z
D
D . 3 C
dx 1 x2
D 3.
D e3 .
D ln b:
D
638
THE INTEGRAL
CHAPTER 5
In general,
SOLUTION
b
Z
dx 1 x2
1
C D tan
0
b
x
ˇˇ D ˇ
1
tan
1
b tan
0
0
1
D tan
b:
For the value of the definite integral to equal 3 , we must have 1
tan
b
D 3
or
3. Which integral should be evaluated evaluated using substitution? 9 dx (a) 1 x2
Z
D tan 3 D
b
Z
(b)
C
p
3:
dx 1 9x 2
C
Use the substitution u substitution u
D 3x on 3x on the integral in (b) . p p 4. Which relation between x between x and and u u yields yields 16 C x 2 D 4 1 C u2 ? p p SOLUTION To transform transfo rm 16 C x 2 into 4 into 4 1 C u2 , make the substitution x substitution x D 4u. 4u. SOLUTION
Exercises In Exercises Exercises 1–10, 1–10, evaluate evaluate the definite definite integr integral. al. 9
1.
dx x
Z 1
9
1 d x x
Z
SOLUTION
1
20
2.
Z 4
dx x 20
Z
SOLUTION
4
e3
3.
Z 1
Z
SOLUTION
1
e
4.
e2
1 d x x
1 dt t
e
SOLUTION
e2
12
5.
Z
ˇj jˇˇ
dt
C 4
SOLUTION
Let u Let u
e3
e
e2
D 3t C 4. Then d Then du u D 3 dt and 2
e
6.
e
D ln j ej ln j e2 j D ln ee2 D ln.1=e/ ln.1=e/ D 1.
12
3
D ln e3 ln 1 D 3.
1
ˇj j ˇˇ
Z Z
D ln 20 ln 4 D ln 5.
4
D ln t
3t
2
ˇj jˇ
D ln t
1 dt t
1
20
1 dt t
Z
9
D ln x
1 dt t e3
Z
ˇˇ ˇ
D ln jx j D ln 9 ln 1 D ln 9.
dt
D 3t C 4
Let u Let u
10
1 dt t ln t
Z e
tan 8
tan 1
du u
D
ˇj jˇ ˇ
1 ln u 3
40 10
3
Z D 1
du u
D
j j 31 D ln 3 ln 1 D ln 3:
ln u
dx x2
SOLUTION
Z C
D 13 .ln 40 ln 10/ D 13 ln 4:
D ln t . Then du Then du D .1=t/dt and .1=t/dt and e3
7.
40
Z
dt t ln t
SOLUTION
Z
1 3
ˇˇ
1
tan 8
tan 1
dx 1 x2
1
C D tan
tan 8
x
ˇˇ ˇ
tan 1
1
D tan
1
.tan 8/ tan
.tan 1/
D 8 1 D 7.
SECTION
7
8.
x dx x2 1
Z
C
2
D x 2 C 1. Then du Then du D 2x dx and dx and
Let u Let u
SOLUTION
7
x dx x2 1
Z
du u
5
1=2
Z
SOLUTION
p
2=
3
1
1 x2
0
p
dx
D sin
1=2
x
D
2
2=
Z
p
3
dx
p D sec 2 jx j x 2 1 11. Use the substitution u substitution u D x=3 to x=3 to prove SOLUTION
1 1
D sin
0
p dx2 jx j x 1
ˇˇ ˇ
1
2
2=
x
ˇˇˇ ˇ
Let u Let u
1
3
D sec
2
D 12 .ln 50 ln 5/ D 12 ln 10:
5
0
D 6 .
p 2
1
Z SOLUTION
sin
p
50
ˇj jˇ ˇ
1 ln u 2
1 x2
0
10.
50
Z
p dx
Z Z
1 2
C D
2
1=2
9.
Further Further Transce Transcendenta ndentall Functions Functions
5.7
dx 9 x2
3
1 D tan C 3
1 x
3
1
sec
D 56 23 D 6 .
.2/
C C
D x=3. x=3. Then, x Then, x D 3u, 3u, dx D 3 du, du , 9 C x 2 D 9.1 C u2 /, and
Z
dx x2
9
3 du
Z D
u2 /
D
1 3
Z
du
1 D tan 3 1C
1
u2
u
C 9.1 C dx 12. Use the substitution u substitution u D 2x to 2x to evaluate . 4x 2 C 1 SOLUTION Let u Let u D 2x. 2x . Then, x Then, x D u=2, u=2, dx D 12 d u, 4x 2 C 1 D u2 C 1, and
Z
Z
dx
D 12 4x 2 C 1
Z
du u2 1
1 C D 2 tan
1
u
C C D D 13 tan
C C D D 12 tan
1
2x
1 x
3
C C:
C C:
In Exercises Exercises 13–32, 13–32, calculate calculate the the integral. integral. 3
13.
Z
dx x2
0
C3 Let x Let x
SOLUTION
D 3
Z 0
4
14.
Z
3u. 3u. Then dx Then dx
dx x2
D
C9 Let t Let t
SOLUTION
C3
3 du and du and
p Z D p
3
1
3
0
4
0
Z p
dt
1
1
D p tan u2 C 1 3
ˇˇˇp D p 3
1
u
3
0
dt
D 4t 2 C 9
1 6
8=3
Z
du
D u2 C 1
0
1 tan 6
Let u Let u
3
1
1
.tan
p
1
3 tan
0/
ˇˇ ˇ
8=3
u
0
D 16 tan
1 8
3
:
D 4t . Then d Then du u D 4 dt , and
Z p
dt
1 16t 2
Z p p
1
1 16t 2
SOLUTION
16.
du
D .3=2/u. .3=2/u. Then d Then dtt D .3=2/du .3=2/du,, 4t 2 C 9 D 9t 2 C 9 D 9.t 2 C 1/, 1/, and
Z 15.
p
dt 4t 2
0
p
dx
4 25x 2
du
Z D p
1 D sin 2 4
4 1u
1
u
C C D D 14 sin
1
.4t/
C C:
D p :
3 3
639
640
THE INTEGRAL
CHAPTER 5
Note that the domain of the function is from 2=5 to 2=5 to 2=5, 2=5, so we will integrate over Œ over Œ2=5; 2=5; 2=5. Let x Let x
SOLUTION
dx
D 25 d u;
4 25x 2
D 4.1 u2 /;
and 2=5
Z
p dx
4 25x
2=5
1
2 D 2 5
D
1 sin 5
Z p
Let t Let t
D
p
5=3u. 5=3u. Then d Then dtt
1=2
Z
1=4
p dx2
x 16x
Let x Let x
SOLUTION
1
sin
1
.1/ sin
: 5
D
.1/
1
Z D p p 3
du
1
1
D p sin 2 3
1u
u
1
1
C C D D p sin 3
r
3 t 5
1
D u=4. u=4. Then dx Then dx D du=4, du=4, 16x 2 1 D u2 1 and 2
p dx2
x 16x
1=4
1
Z
D
p du2
u u
1
1
D sec
2
ˇˇ D
1
u
dx
1
sec
1
1
2 sec
1
D 3 :
x 12x 2 3 Let u Let u
SOLUTION
D 2x. 2x . Then du Then du D 2 dx and dx and
Z p x
x dx x4 1
Z
C
SOLUTION
Let u Let u
dx
D p 1
12x 2 3
3
du
Z p
1
u2 1
u
D p 1 sec
C C D D p 1 sec
3
1
u
3
.2x/
C C:
D x 2 . Then du dx and Then du D 2x dx and
Z
dx
Z p
x dx x4
C 1 D
1 2
Z
du u2
1 D C 1 2 tan
1
u
C C D D 12 tan
1
x2
C C:
x x4 1 Let u Let u
SOLUTION
D x 2 . Then du dx , and Then du D 2x dx, dx
Z p
x4 1
x
0
22.
1
5=3du and 5=3du and
5 1 t2
1=2
21.
p
5=3du
Z Z p
D
p Z D p p
dt
5 3t 2
20.
u
5 3t 2
Z p
19.
1
1
dt
SOLUTION
18.
du
4.1 u2 /
1
D 15 17.
1
Z p ˇˇ ˇ
Z
Z D
p du2
2u u
1
D 12 sec
1
C C D D 12 sec
u
1
x2
C C:
.x
1=2
SOLUTION
p C 1/dx 2 1x
Observe that .x
Z p C
1/dx
x dx
1 x2
In the first integral on the right, we let u let u .x
Z p C
1/dx
1 x2
dx
Z D p
1 x2
Z C p
1 x2
:
D 1 x 2, du d u D 2x dx. dx . Thus D
1 2
Z
du
1 dx
u1=2
Z C p
1x
p
D 2
1 x2
1
C sin
x
C C:
Finally, 0
Z
1=2
.x
p C 1/dx D . 2 1x
p
1 x2
1
C sin
0
x/
ˇˇ ˇ
1=2
D 1 C
p
3 2
C 6 :
C C:
D 2u=5. 2u=5. Then
SECTION
0
23.
Z
e x dx 1 e 2x
C
ln 2
D ex . Then du Then du D e x dx, dx , and
Let u Let u
SOLUTION
0
Z Z
ln. ln.cos
Z
1=2
1
D 4 tan
1
u
1=2
.1=2/:
1 x . Then du Then du
D ln cos cos
ln. ln.cos
1 x/dx
1 x2
1 x/
.cos
D cos1 1 x p 1
p
1x
D 2
Z
u du
, and
D 12 u2 C C D D 12 .lncos
1
x/ 2
C C:
1 x dx
C x2 D tan
Then du Then du
.tan
C x 2/ 1 x.
Let u Let u
SOLUTION
D tan
Then du Then du
Z p
3
1
C x2
Z D
1
u du
2
D 12 u2 C C D D .tan 2 x/ C C:
D 1 Cdxx2 , and
dx .tan 1 x/.1
1
Z
1 x dx
dx 1 x/.1
1
D 1 Cdxx2 , and
tan 1
Z
3
Z
1 x.
Let u Let u
p
27.
C D tan
ˇˇˇ ˇ
tan 1
SOLUTION
26.
C
1
1x
Z 25.
du 1 u2
p x/dx 2
Let u Let u
SOLUTION
1
Z D
1
1 x/
.cos
e x dx 1 e2x
ln 2
24.
Further Further Transce Transcendenta ndentall Functions Functions
5.7
=3
C x2/
Z D
=4
1 d u u
ˇj jˇ ˇ
D ln u
3x dx
=3 =4
D ln 3 ln 4 D ln 43 :
0
1
Z
SOLUTION
x
3 dx
0
1
28.
Z
x
3
D
3x ln 3
dx
1
ˇˇˇ D 0
1 .3 1/ ln 3
D ln23 .
0
SOLUTION
Let u Let u
D x. Then du dx and Then du D dx and 1
Z
x
3
dx
0
log4 .3/
29.
Z
1
D
Z
u
3 du
0
D
3u ln 3
4x dx
1
ˇˇ ˇ
D
0
1 ln 3
1 3
C 1 D 3 ln2 3 :
0
log4 .3/
SOLUTION
Z
x
4 dx
0
1
30.
Z
D
4x ln 4
2
t 5t dt
log4 3
ˇˇ ˇ
0
D ln14 .3 1/ D ln24 D ln12 .
0
SOLUTION
Let u Let u
D t 2 . Then du Then du D 2t dt and dt and 1
Z
t2
t5
dt
0
31.
Z
D
1 2
1
Z
u
5 du
0
D
9x sin.9 sin.9x / dx
SOLUTION
Let u Let u
5u 2 ln 5
1
ˇˇ D ˇ 0
5 2 ln 5
1 2 ln 5
D ln25 :
D 9x . Then du Then du D 9x ln 9 dx and dx and
Z
9x sin.9 sin.9x / dx
D ln19
Z
sin u du
D ln19 cos u C C D D ln19 cos.9 cos.9x / C C:
641
642
32.
THE INTEGRAL
CHAPTER 5
dx
Z p
52x 1 First, rewrite
SOLUTION
Z p
dx
52x
x . Then du Then du
Now, let u let u
D5
Z p
x
D 5 dx
52x 1
dx
Z D
1
p x
5
5
15
Z D p
2x
x
dx
15
2x
:
ln 5 dx and dx and
D
1 ln 5
du
D ln15 sin
Z p
1 u2
1
u
C C D D ln15 sin
1
x
.5
In Exercises Exercises 33–70, 33–70, evaluate evaluate the integral integral using using the methods methods covered covered in the text text so far far. 33.
Z
2
ye y dy
34.
Z
Z
dx 3x 5
C
Let u Let u
SOLUTION
35.
D y 2; du D 2y dy. dy . Then
Use the substitution u substitution u
SOLUTION
4x 2
Z
C9
Let u Let u
SOLUTION
.x x
SOLUTION
37.
Z
7
x
SOLUTION
38.
Z
SOLUTION
39.
Z
40.
D 12 eu C C D D 12 ey C C: 2
Z
du u
D 13 ln juj C C D D 13 ln j3x C 5j C C:
x
dx
C9
D
1 8
Z
u
1=2
du
D 14 u1=2 C C D D 14
p
4x 2
C 9 C C:
/ dx
Z
.x x
2 2
/ dx
Z D
.x 2 2x
1
Cx
/ dx
D 13 x 3 2 ln jx j 13 x 3 C C . C .
7u du
D ln7 7 C C D D 7ln 7 C C:
e u du
D 121 eu C C D D 121 e9
4
dx Let u Let u
D x . Then du dx and Then du D dx and x
7
dx
D
Z
u
x
dt
Let u Let u
D 9 12t . Then du 12dt and Then du D 12dt and
Z
e
912t
dt
D
1 12
Z
12t
C C:
sec2 tan tan7 d d
SOLUTION
Z
e u du
2 2
12t
1 3
C D
Z e9
D
dx 3x 5
4x 2
36.
dy
Z
D 4x 2 C 9. Then du Then du D 8x dx and dx and
Z p
Z
1 2
D 3x C 5. Then d u D 3 dx and dx and Then du
x dx
Z p
ye
y2
Let u Let u
cos. cos.ln t / d t t
SOLUTION
Let u Let u
D tan . Then d and Then du u D sec2 d d and
Z
2
Z D D
7
sec tan tan d d
u7 du
D 18 u8 C C D D 18 tan8 C C:
D ln t . Then du Then du D dt=t and dt=t and
Z
cos. cos.ln t / d t t
D
Z
cos u du
D sin u C C D D sin. sin.ln t / C C:
/
C C:
SECTION
41.
t dt
Z p
7 t2 Let u Let u
SOLUTION
D 7 t 2 . Then d Then du u D 2t dt and dt and t dt
Z p
7t
42.
Further Further Transce Transcendenta ndentall Functions Functions
5.7
Z
D 2
1 2
Z
u
1=2
p
D u1=2 C C D D
du
7 t2
C C:
2x e4x dx First, note that
SOLUTION
2x
D ex ln 2
dx
Z D
2x e 4x
so
D e.4Cln 2/x :
Thus,
43.
Z
Z
.3x 2/dx x2 4
C C
x 4x
2 e
e .4Cln 2/x dx
D 4 C1ln 2 e.4Cln 2/x C C:
Write
SOLUTION
Z In the first integral, let u let u
.3x 2/dx x2 4
C C
Z D
3x dx x2 4
C
2 dx : x2 4
Z C
C
D x 2 C 4. Then du Then du D 2x dx and dx and
du 3 3 D ln.x 2 C 4/ C C 1 : ln juj C C 1 D ln.x C u 2 2 For the second integral, let x let x D 2u. 2u. Then dx Then dx D 2 du and du and
Z
Z
3x dx x2 4
2 dx x2 4
C
Combining these two results yields
44.
Z
Z tan.4x tan.4x
dx
Z p
du u2
C C
1
C 1 D tan
.3x 2/dx x2 4
u
C C 2 D tan
D 32 ln.x ln.x 2 C 4/ C tan
1
1
.x=2/
.x=2/
C C 2:
C C:
sin.4x C1/ C 1/dx as 1/dx as cos dx d x . Let u Let u D cos.4x cos.4x C 1/. 1/. Then d Then du u D 4 sin.4x sin.4x C 1/dx, 1/dx , and .4x C1/ sin.4x sin.4x C 1/ 1 du 1 ln j cos.4x D dx D cos.4x C 1/j C C: cos.4x cos.4x C 1/ 4 u 4
R
Z
Z
1 16x 2 Let u Let u
D 4x. 4x . Then d u D 4 dx and dx and Then du dx
Z p
1 16x 2
Z p C et
et
SOLUTION
Z
.e
x
SOLUTION
D
1 4
Use the substitution u substitution u
1 u2
D 14 sin
1
u
C C D D 14 sin
1
.4x/
C C:
D et C 1;du D et dt . Then p u du D 2 u3=2 C C D D 2 .et C 1/3=2 C C: et C 1 dt D
Z p
du
Z p
1 dt
e
47.
Z D
C 1/dx
R
SOLUTION
46.
Z
First we rewrite tan.4x tan .4x
SOLUTION
45.
3 2
t
Z
3
3
4x/dx
First, observe that
Z
.e
x
4x/dx
Z D
e
x
dx
Z
Z D
4x dx
e
x
dx
2x 2 :
643
644
CHAPTER 5
THE INTEGRAL
In the remaining integral, use the substitution u substitution u
Z
e
x
D x;du D dx. dx . Then
dx
D
Z
e u du
D eu C C D D e x C C:
Finally,
Z
.e
48.
Z
x
4x/dx
D e
x
2x 2
C C:
.7 e 10x / dx First, observe that
SOLUTION
Z
.7 e 10x / dx
In the remaining integral, use the substitution u substitution u
Z
e
10x
dx
D
Z
7 dx
Z
e 10x dx
Z
D 7x
e 10x dx :
D 10x;du D 10dx. 10dx . Then D
1 10
Z
e u du
D 101 eu C C D D 101 e10x C C:
Finally,
Z
.7 e10x / dx
49.
Z
e 2x
e
4x
ex
D 7x 101 e10x C C:
dx
SOLUTION
Z 50.
Z p
e 2x
e4x
ex
.e x
d x
3x
D ex e 3 C C:
e3x / dx
dx
x 25x 2 1 Let u Let u
SOLUTION
D 5x. 5x . Then du Then du D 5 dx and dx and
Z p
dx
du
x 25x 2 1
51.
! Z D
.x
Z p C
5/dx
Z D p
D sec
5/dx
x dx
u u2 1
1
u
1
C C D D sec
.5x/
C C:
4 x2
SOLUTION
Write .x
Z p C
4 x2
In the first integral, let u let u
5 dx
Z D p
4 x2
Z C p
4 x2
:
D 4 x 2. Then d Then du u D 2x dx and dx and x dx
Z p
4x
In the second integral, let x let x
D 2
1 2
Z
u
1=2
du
p
D u1=2 C C 1 D
4 x2
C C 1 :
D 2u. 2u. Then dx Then dx D 2 du and du and 5 dx
Z p
4x
D5 2
du
Z p
D 5 sin
1
p
C 5 sin
1 u2
u
1
C C 2 D 5 sin
.x=2/
C C 2 :
Combining these two results yields .x
Z p C
5/dx
4 x2
52.
Z C .t
SOLUTION
p
C 1 dt Let u Let u D t C 1. Then du Then du D dt and p .t C 1/ t C 1 dt D
D
4 x2
1
.x=2/
C C:
1/ t
Z
Z
u3=2 du
D 25 u5=2 C C D D 25 .t C 1/5=2 C C:
SECTION
53.
Z
e x cos.e cos.e x / dx Use the substitution u substitution u
SOLUTION
54.
Z p
ex
D ex ; du D ex dx . Then
Z
ex
C 1 dx
Use the substitution u substitution u
SOLUTION
e x cos.e cos.e x / dx
Z p
dx
Z p
Z
D
ex
dx
C1
Z D p D p C du
u
D D 2p ex C 1 C C:
C
First rewrite dx
Z p
D
9 16x 2
Now, let u let u
dx
9 16x
Z
1 4
D 2
du
Z p
Let u Let u
Z r
dx
1
2
:
4 3x
SOLUTION
1 sin 4
1
u
C C D D
1 sin 4
1
4x 3
dx .4x 1/ ln.8x ln.8x 2/
1 4
D
Z
du u
D 14 ln juj C C D D 14 ln j ln.8x ln.8x 2/j C C:
C 1/3 dx
Use the substitution u substitution u
Z
x
e .e
2x
3
C 1/
D ex ; du D ex dx . Then
dx
Z D
2
u
3
C 1
du
Z D
u6
C 3u4 C 3u2 C 1
d u
D 17 u7 C 35 u5 C u3 C u C C D D 17 .ex /7 C 35 .ex /5 C .ex /3 C ex C C 7x
5x
D e 7 C 3e5 C e3x C ex C C: 58.
Z
dx x.ln x. ln x/ 5
SOLUTION
59.
Z
60.
D ln x. Then du dx=x and Then du D dx=x and
Z
C2
SOLUTION
Z
Let u Let u
x 2 dx x3
Let u Let u
dx x.ln x. ln x/ 5
C
Let u Let u
Z D
u
5
du
1 D 14 u 4 C C D D 4.ln C C: 4.ln x/ 4
D x 3 C 2. Then du dx , and Then du D 3x 2 dx,
Z
.3x 1/dx 9 2x 3x 2
SOLUTION
C C:
8 4 D ln.8x ln.8x 2/. 2/. Then du Then du D dx D dx , and 8x 2 4x 1
Z e x .e 2x
D 2
1u
dx .4x 1/ ln.8x ln.8x 2/
SOLUTION
Z
1 3
D 43 x. Then du Then du D 43 dx d x and
Z p
57.
2 u
9 16x 2
SOLUTION
56.
D sin u C C D D sin.e sin.e x / C C:
cos u du
D ex C 1;du D ex dx . Then ex
55.
Further Further Transce Transcendenta ndentall Functions Functions
5.7
x 2 dx x3 2
C D
1 3
Z
du u
D 13 ln jx 3 C 2j C C:
D 9 2x C 3x2 . Then du Then du D .2 C 6x/dx D 2.3x 1/dx, 1/dx , and
Z
.3x 1/dx 9 2x 3x 2
C
D 12
Z
du u
D 12 ln.9 ln.9 2x C 3x 2 / C C:
645
646
61.
CHAPTER 5
Z
cot x dx
R
We rewrite cot x dx as dx as
SOLUTION
62.
Z
THE INTEGRAL
cos x sin x
R
cos x dx d x 2 sin x 3
C
Let u Let u
SOLUTION
where we have used the fact that 2 that 2 sin x 63.
4 ln x x
C 5 dx Let u Let u
SOLUTION
64.
Z
D sin x . Then du Then du D cos x dx , and
cos x dx sin x
Z D
cos x dx d x 2 sin x 3
C
D
1 2
Z
du u
D 12 ln.2 ln.2 sin x C 3/ C C;
Z
.sec tan /5sec d tan /5
Let u Let u
D sec . Then du and Then du D sec tan tan d d and
Z
Z
Z
Z
5u du
u
sec
D ln5 5 C C D D 5ln 5 C C:
2
Let u Let u
D x 2 . Then du Then du D 2x dx, dx , and
Z
ln. ln.ln x/ dx x ln x Let u Let u
SOLUTION
Z
D D
x3 x dx
SOLUTION
67.
Z
D 4 ln x C 5. Then du Then du D .4=x/dx, .4=x/dx , and 4 ln x C 5 1 1 D 18 .4 ln x C 5/2 C C: dx D u du D u2 C C D x 4 8
.sec tan tan /5sec d
66.
D ln j sin x j C C:
C 3 1 to drop the absolute value.
Z
SOLUTION
65.
du u
D 2 sin x C 3. Then d Then du u D 2 cos x dx , and
Z Z
dx . Let u Let u
x3
x2
dx
D
1 2
Z
u
3 du
D
1 3u 2 ln 3
2
C C D D
3x 2 ln 3
C C:
1 1 D ln. ln.ln x/. x/ . Then du Then du D dx and dx and ln x x
Z
ln. ln.ln x/ dx x ln x
Z D
2
u du
2
ln.ln x// D u2 C C D D .ln. C C: 2
cot x ln. ln.sin x/dx Let u Let u
SOLUTION
D ln. x/ . Then ln.sin x/. du
D sin1 x cos x dx D cot x dx;
and
68.
Z
tdt
Z p
cot x ln. ln.sin x/dx
Z
2
u du
2
ln.sin x// D u2 C C D D .ln. C C: 2
1 t4
SOLUTION
Let u Let u
D t 2. Then du Then du D 2t dt and dt and t dt
Z p
1t
69.
D
Z
p
t2 t
3 dt
D 4
1 2
Z p
du
1 D sin 2 2
1u
1
u
C C D D 12 sin
1 2
t
C C:
SECTION
Let u Let u
SOLUTION
Further Further Transce Transcendenta ndentall Functions Functions
5.7
647
D t 3. Then t Then t D u C 3, du D dt and p p t 2 t 3 dt D .u C 3/2 u du
Z
Z Z D
.u
2
C 6u C 9/p u du D
Z
.u5=2
C 6u3=2 C 9u1=2 / du
D 27 u7=2 C 125 u5=2 C 6u3=2 C C D 27 .t 3/7=2 C 125 .t 3/5=2 C 6.t 3/3=2 C C: 70.
Z
cos x5
SOLUTION
2 sin x
Let u Let u
dx
D 2 sin x . Then du dx and Then du D 2 cos x dx and
Z
cos x5
2 sin x
dx
D
1 2
Z
5u du
u
2 sin x
5 D 2ln5 C C D D 5 2 ln 5 C C:
71. Use Figure 1 to prove prove x
Z p
1 t 2 dt
0
D 12 x
p
1 x2
C 12 sin
1
x
y
1
x
x
FIGURE 1 x
SOLUTION
The definite integral
Z p
1 t 2 dt represents the area of the region under the upper half of the unit circle from 0
0
to x to x . The region consists of a sector of the circle and a right triangle. The sector has a central angle of Hence, the sector has an area of 1 2 .1/ 2 2
1
cos
x
p
D
1 sin 2
1
2
D D x .
, where cos
x:
p
The right triangle has a base of length x length x,, a height of 1 x 2 , and hence an area of 12 x 1 x 2 . Thus, x
Z p
1 t 2 dt
0
72. Use the substitution u substitution u
D 12 x
D tan x to evaluate
p
1 x2
C 12 sin
1
x:
dx
Z
1
C sin2 x :
2
du D C 1 2u2 x
Hint: Show that
dx 1 SOLUTION
If u u
C sin
D tan x, then du then du D sec2 x dx and dx and du 1 2u2
C
2
x dx dx dx dx D 1sec D D D 2 2 2 2 C 2 tan x cos2 x C 2 sin x cos2 x C sin x C sin x 1 C sin2 x :
Thus
Z
dx 1
2
C sin
x
Z D
du 1
C
2u2
Z D
du
1 p D p tan 1 C . 2u/2 2
1
p
. 2u/
C C D D p 1 tan
1
2
p
..tan ..tan x/ 2/
C C:
648
THE INTEGRAL
CHAPTER 5
73. Prove:
Z Let G.t/ Let G.t/
SOLUTION
D
p
1 t2
d dt
G 0 .t / D
1
C t sin
p
1 t2
1
sin
t dt
This proves the formula 74. (a) Verify for r for r
Z
1
sin
t dt
D
¤ 0:
1 t2
1
C t sin
t:
t . Then d dt
C
t sin
1 t2
t
t
D p 1
C t sin T
Z
1
t ert dt
0
C
1 t2
C sin
1 t2
p
1
D p t 2 C p t 1t
p
D
t
1
D sin
d t sin dt
1
t
1
C sin
t
t:
t.
D e
rT .rT 1/
r2
C1
6
T / be the value of the integral on the left. By FTC II, F 0 .t/ .t / t e rt and F .0/ 0. Show that the same is Hint: For fixed r fixed r , let F. let F.T and F .0/ true of the function on the right. opital’s Rule to show that for fixed T fixed T ,, the limit as r as r 0 of 0 of the right-hand side of Eq. (6) is equal to the value of the (b) Use L’Hˆopital’s integral for r for r 0.
D
D
!
D
SOLUTION
(a) Let rt
f.t/
D er 2 .r t 1/ C r12 :
Then f 0 .t /
D r12 ert r C .rt .r t 1/.re rt / D t e rt
and f.0/
D r12 C r12 D 0;
as required. (b) Using L’Hˆopital’s opital’s Rule, lim
r
!0
e rT .rT 1/ r2
T
If r r
D 0 then,
Z 0
te
rt
r
T
dt
Z D
T e rT
C 1 D lim tdt
0
D
2r
!0
t2 2
ˇˇˇ ˇ
T 0
C .rT 1/.TerT / D lim r
!0
rT 2 e rT 2r
2 rT
2
T e T D rlim D : !0 2 2
2
D T 2 .
Further Insights and Challenges f.t/ g.t/ for g.t/ for t t 75. Recall that if f.t/
0, then for all x all x
0,
x
Z
x
f.t/dt
0
Z
g.t/dt
7
0
The inequality e inequality e t 1 holds 1 holds for t for t 0 because 0 because e e > 1. Use Eq. (7) to prove that e that e x integration, the following inequalities (for x (for x 0): ex SOLUTION
C x C 12 x 2;
1
Integrating both sides of the inequality inequality e e t x
Z 0
et dt
ex
1
1
C x C 12 x 2 C 16 x 3
1 yields
D ex 1 x
C x for x for x 0. 0 . Then prove, by successive successive
or
ex
1
C x:
SECTION
Further Further Transce Transcendenta ndentall Functions Functions
5.7
649
Integrating both sides of thi s new inequality then gives x
Z
et dt
D ex 1 x C x2 =2
or
ex
1
C x C x 2=2:
D ex 1 x C x2 =2 C x 3=6
or
ex
1
C x C x 2 =2 C x3 =6
0
Finally, integrating both sides again gives x
Z
et dt
0
as requested. 76. Generalize Exercise 75; 75; that is, use induction (if you are familiar with this method of proof) to prove that for all n all n
ex SOLUTION
true for n for n
1
C x C 12 x 2 C 16 x3 C C n1Š x n
.x
D
C
ex
D k . We need to prove the statement is
C x C x 2=2 C C x k = kŠ:
1
0,
0/
1 , e x 1 x by Exercise 75. Assume the statement is true for n For n For n 1, k 1. By the Induction Hypothesis,
D C
Integrating both sides of this inequality yields x
Z
e t dt
0
or
ex
D ex 1 x C x 2 =2 C C xkC1 =.k C 1/Š 1
C x C x 2 =2 C C xkC1=.k C 1/Š
as required. 77. Use Exerc Exercise ise 75 to show show that that e x =x 2
that that lim e x =x n x
!1
SOLUTION
x=6 and conclude conclude that lim e x =x 2 x
D 1 for all n all n..
By Exercise 75, e 75, e x
1
2
D 1. Then Then use Exerc Exercise ise 76 to prove prove more more genera generally lly
3
C x C x2 C x6 . Thus ex x2
1 x2
C x1 C 12 C x6 x6 :
D 1, xlim lim e x =x 2 D 1. More generally, by Exercise 76, !1
Since Since lim x=6 x
!1
!1
2
ex
1 xn
1
n 1
C
C x2 C C .nx C 1/Š :
Thus ex xn Since Since lim
x .n 1/Š
x !1 C
x n
C C .n Cx 1/Š .n Cx 1/Š :
D 1, xl!1 im xe D 1.
Exercises Exercises 78–80 78–80 develop develop an elega elegant nt approach approach to the exponen exponential tial and logarithm logarithm functions functions.. Define a function function G.x/ for x > 0 : x
G.x/
Z D 1
1 dt t
78. Defining ln x as an Integral This exercise exercise proceeds as if we didn’t know know that G.x/ that G.x/ all the basic properties of the logarithm. Prove the following statements. ab ab
D b D C D D D D
D ln x and shows directly that G.x/ that G.x/ has has
1 0. Hint: Use the substitution u =a. (a) a 1t dt all a;b > 0. substitution u t =a. 1 t dt for all a;b (b) G.ab/ G.a/ G.b/. G.b/. Hint: Break up the integral from 1 from 1 to to ab ab into into two integrals and use (a). (c) G.1/ 0 and G.a and G.a 1 / G.a/ for G.a/ for a a > 0. 0. n (d) G.a / nG.a/ for nG.a/ for all a all a > 0 and 0 and integers n integers n.. 1 (e) G.a 1=n / G.a/ for G.a/ for all a all a > 0 and 0 and integers n integers n 0. n (f) G.a r / rG.a/ for rG.a/ for all a all a > 0 and 0 and rational numbers r numbers r . G.x/ is increasing. Hint: Use FTC II. (g) G.x/ is number a such such that G.a/ that G.a/ > 1. 1 . Hint: Show that G.2/ that G.2/ > 0 and 0 and take a take a 2m for m for m > 1=G.2/. 1=G.2/. (h) There exists a number a (i) lim G.x/ and and lim lim G.x/
R
x
R
D1
!1
D
D
¤
x
!0C
(j) There exists a unique number E number E such such that G.E/ that G.E/ r (k) G.E / r for every rational number r number r .
D
D
D 1
D 1.
650
THE INTEGRAL
CHAPTER 5
SOLUTION
Let u (a) Let u
D t =a. =a. Then du Then du D dt=a, dt=a, u.a/ D 1, u.ab/ D b and ab
1 dt t
Z a
ab
D
Z
1 dt t
C
a dt at
a
b
D
1 d u u
Z 1
b
D
1 dt : t
Z 1
(b) Using part (a), ab
G.ab/
D
Z 1
1 dt t
a
Z
D
1
ab
1 dt t
Z a
a
1 dt t
Z
D
1
b
C
Z 1
1 dt t
D G.a/ C G.b/:
(c) First, 1
G.1/
Z D 1
1 dt t
D 0:
Next, G.a
1
/
DG
1 a
1=a
Z Z D D
1
a
1
1
1 dt t
D
1 dt t
1 dt t
D G.a/:
Z a
by part (a) with b with b
D a1
(d) Using part (a),
G.an /
an
Z Z D D
1
a
1
1 dt t
1 dt t
a
Z Z C D
1
a
1
1 dt t
1 dt t
a2
C
Z a
C
1 dt t a
Z C 1
an
C C
1 dt t
Z
an1
1 dt t
D nG.a/:
1 D G..a1=n /n D nG.a1=n /. Thus, G.a G.a/. Thus, G.a 1=n / D G.a/. n (f) Let r Let r D m=n where m=n where m m and and n n are are integers. Then G.ar / D G.am=n / D G..am /1=n / D n1 G.am / by part (e)
(e) G.a/
D mn G.a/ D rG.a/:
by part d
G.x/ is continuous on .0; (g) By the Fundamental Theorem of Calculus, G.x/ is increasing and one-to-one for x for x > 0. 0. (h) First note that 2
Z D
G.2/
1
because
1 1 > for t for t t 2
1/ and G 0 .x/ D
1 x
> 0 for x > 0. Thus, G.x/ is G.x/ is
1 1 dt > > 0 t 2
2 .1; .1; 2/. 2/. Now, let a let a D 2m for m for m an an integer greater than 1=G.2/. 1=G.2/. Then G.a/
1 D G.2m / D mG.2/ > G.2/ G.2/ D 1:
(i) First, let a let a be be the value from part ( h) for which G.a/ which G.a/ > 1 (note 1 (note that a that a itself itself is greater t han 1). Now,
lim G.x/
x
For the other limit, let t let t
!1
m D mlim !1 G.a / D G.a/ mlim !1 m D 1:
D 1=x and 1=x and note x
lim G.x/
!0C
D t !1 lim G
1 t
D t !1 lim G.t/ D 1:
0 and by part (h) there exists an a 1. the Intermediate Value (j) By part (c), G.1/ (c), G.1/ 0 and an a such such that G.a/ that G.a/ > 1. Value Theorem then guarantees there exists a number E such E such that 1 that 1 < E < a and a and G.E/ G.E/ 1. We know that E that E is is unique because G because G is is one-to-one. (k) Using part (f) and then part (j),
D
D
G.E r /
D rG.E/ D r 1 D r:
SECTION
Further Further Transce Transcendenta ndentall Functions Functions
5.7
651
79. Defining e x
Use Exercise 78 to prove the following statements. G.x/ has an inverse with domain R and range x x > 0 . Denote the inverse by F.x/ by F.x/.. (a) G.x/ has F.x/F.y/ for all x; all x; y . Hint: It suffices to show that G.F.x/F.y// that G.F.x/F.y// G.F.x (b) F .x y/ E r for all numbers. In particular, F.0/ particular, F.0/ 1. (c) F.r/ 0 F.x/. F.x/. Hint: Use the formula for the derivative derivative of an inverse function. (d) F .x/
C D D D
f W
g
D
D
This shows that E that E
C y//. y// .
D e and F.x/ and F.x/ is is the function e function e x as defined in the text.
SOLUTION
(a) The domain of G.x/ is G.x/ is x x > 0 and, 0 and, by part (i) of the previous exercise, the range of G.x/ is G.x/ is R. Now,
G 0 .x/
D x1 > 0
for all x all x > 0. Thus, G.x/ Thus, G.x/ is is increasing on its domain, which implies that G.x/ that G.x/ has has an i nverse. nverse. The domain of the inverse is R and the range is x x > 0 . Let F.x/ Let F.x/ denote denote the i nverse of G.x/. G.x/. (b) Let x Let x and and y y be be real numbers and suppose t hat x hat x G.w/ and G.w/ and y y G.z/ for G.z/ for some positive real numbers w numbers w and and z z.. Then, using part (b) of the previous exercise
f W
g
D
D
F .x
C y/ D F.G.w/ C G.z// D F.G.wz// D wz D F.x/ C F.y/: (c) Let r Let r be be any real number. By part (k) of the previous exercise, exercise, G.E r / D r . By definition of an inverse function, it then follows that F.r/ that F.r/ D E r . (d) By the formula for t he derivative derivative of an inverse function
1 1 D G0 .F.x// D 1=F.x/ D F.x/:
F 0 .x/
80. Defining b x Let b Let b > 0 and let f let f .x/ .x/ F.xG.b// with F.xG.b// with F F as as in Exercise 79. Use Exercise 78 (f) to prove that f .r/ br x x x for every rational number r number r . This gives us a way of defining b for irrational x irrational x , namely b namely b f .x/. x/. With this definition, b definition, b is a differentiable function of x (because x (because F F i iss differentiable).
D
SOLUTION
D
D
By Exercise Exercise 78 (f),
D F.rG.b// D F.G.br // D b r ;
f.r/ for every rational number r number r . 81. The formula
Z
x n dx
n 1
C
D nx C 1 C C is is valid for n for n ¤ 1. Show that the exceptional case n case n D 1 is a limit of the general case
by applying L’Hˆopital’s opital’s Rule to the limit on the left. x
n
lim
!
1 1
Note that the integral on the left is equal to
Z
t
x
n
dt D
Z
t
1
dt
.for fixed x fixed x > 0/
1
x nC1 1 . n 1
C
SOLUTION
x
n
lim
!
Z
1 1
t n dt
D n! lim 1
D
x
ˇˇ C ˇˇ D !
t nC1 n 1
n
1
x nC1 1 lim n! 1 n 1
1
C
D n! lim .x 1
C
lim
x nC1 n 1 n 1
C
1nC1 n 1
C
! x
/ ln x
D ln x
Z D
t
1
dt
1
Note that when using L’Hˆopital’s opital’s Rule in the second line, we need to differentiate with respect to n to n.. x nC1 1 . Investigate Investigate the limit graphically by plotting n 1
82.
The integral on the left in Exercise 81 is equal to f to f n .x/
f n .x/ for .x/ for n n
D 0, 0:3, 0:3, 0:6, 0:6, and 0:9 together 0:9 together with ln x on a single plot.
D
C
SOLUTION y n = 0
2 1
n = !0.3 n = !0.6 n = ! 0.9 y = ln x x
1 !
1
2
3
4
5
652
CHAPTER 5
THE INTEGRAL ln ln a y e
R
(a) Explain why the shaded region in Figure 2 has area 0 a ln ln a (a) Prove the formula 1 ln x dx a ln a 0 e y dy. dy. a (b) Conclude that 1 ln x dx a ln a a 1. (c) Use the result of (a) to find an antiderivative antiderivative of ln x .
83.
R R
D
R
D
dy. dy.
C
y = ln x
y
ln a
1
x
a
FIGURE 2 SOLUTION
(a) Interpreting the graph with y with y as as the independent variable, we see that the function is x ln ln a 0
D e y . Integrating in y in y then then gives the
area of the shaded region as e y dy of y ln x from x from x 1 to x to x a by computing the area of the rectangle extending (b) We can obtain the area under the graph of y from x from x 0 to x to x a horizontally and from y from y 0 to y to y ln a vertically and then subtracting the area of the shaded region. This yields
D
D
R
D D
D
D
D
a
Z
ln a
ln x dx
1
D a ln a
Z
e y dy:
0
(c) By direct calculation ln a
Z
y
e dy
0
De
y
Thus, a
Z 1
ln x dx
ln a
ˇˇ ˇ
0
D a 1:
D a ln a .a 1/ D a ln a a C 1:
(d) Based on these results results it appears that
Z
ln x dx
D x ln x x C C:
5.8 Exponential Growth and Decay Preliminary Questions 1. Two Two quantities increase exponentially with growth constants k constants k rapidly? SOLUTION
D 1:2 and 1:2 and k k D 3:4, 3:4, respectively. respectively. Which quantity doubles more
Doubling time is inversely inversely proportional to the growth constant. Consequently Consequently,, the quantity with k with k
more rapidly.
D 3:4 doubles 3:4 doubles
2. A cell population grows exponentially beginning with one cell. Which takes longer: increasing from one to two cells or increasing from 15 million to 20 million cells?
It takes longer for the population to increase from one cell to two cells, cells, because this requires doubling the population. Increasing from 15 million to 20 million is less than doubling the population. SOLUTION
3. Referring to his popular book A Brief History of Time , the renowned physicist Stephen Hawking said, “Someone told me that each equation I included in the book would halve its sales.” Find a differential equation satisfied by the function S.n/ function S.n/,, the number of copies sold if the book has n has n equations. equations. SOLUTION
Let S.0/ Let S.0/ denote denote the sales with no equations in the book. Translating Hawking’s Hawking’s observation into an equation yields S.n/
D S.0/ : 2n
Differentiating with respect to n to n then then yields dS dn
D S.0/ dd n 2 n D ln 2S.0/2 n D ln 2S.n/:
SECTION
5.8
Exponentia Exponentiall Growth Growth and Decay Decay
653
4. The PV of N N dollars received at time T time T is (choose the correct answer): (a) The value at at time T time T of of N N dollars invested today (b) The amount you would have to invest invest today in order to receive N receive N dollars dollars at time T time T SOLUTION The correct response response is (b) : the PV of N N dollars received received at time T time T is the amount you would have to invest today in order to receive N receive N dollars dollars at time T time T ..
5. In one year, you will be paid $1. Will the PV increase or decrease decrease if the interest rate goes up? SOLUTION
If the interest rate goes up, the present value of of $1 a year from now now will decrease.
Exercises 1. A certain population P population P of of bacteria obeys the exponential exponential growth law P.t/ law P.t/ (a) How many bacteria are present present initially?
D 2000e1:3t (t in hours).
(b) At what time will there be 10,000 bacteria? SOLUTION
D 2000e0 D 2000 bacteria 2000 bacteria initi ally. 1:3t (b) We solve 2000e D 10;000 for solve 2000e 10;000 for t t . Thus, e Thus, e 1:3t D 5 or (a) P.0/
1 D 1:3 1:24 hours:: ln 5 1:24 hours 2. A quantity P quantity P obeys the exponential exponential growth law P.t/ law P.t/ D e5t (t in years). D 10? time t is is P P D 10? (a) At what time t t
(b) What is the doubling time for P for P ?? SOLUTION
(a) e 5t
D 10 when 10 when t t D 15 ln 10 0:46 years. 0:46 years.
(b) The doubling time is 1 5 ln 2
0:14 years. 0:14 years.
D 5.7/t in the form f form f .t/ D P 0 ekt for some P some P 0 and k and k.. SOLUTION Because 7 Because 7 D e ln 7 , it follows that f.t/ D 5.7/t D 5.eln 7 /t D 5et ln 7 : Thus, P Thus, P 0 D 5 and k and k D ln 7. 4. Write f Write f .t/ D 9e 1:4t in the form f form f .t/ D P 0 b t for some P some P 0 and b and b.. Write f .t/ 3. Write f
SOLUTION
Observe that f.t/
so P so P 0
D 9e 1:4t D 9
t
e 1:4
;
D 9 and b and b D e 1:4 4:0552. 4:0552.
5. A certain RNA molecule replicates every every 3 minutes. Find the differential equation for the number N.t/ number N.t/ of of molecules present at time t time t (i (in n minutes). How many molecules will be present after one hour if there is one molecule at t 0? SOLUTION
The doubling time is
ln 2 so k so k k
D
D
ln 2 . Thus, the differential equation is N 0 .t / doubling time
D kN.t/ D
ln 2 N.t/. N.t/. 3
With one molecule initiall y, N.t/
D e.ln 2=3/t D 2t =3:
Thus, after one hour, there are N.60/
D 260=3 D 1;048;576
molecules present. P obeys the exponential growth law P.t/ 6. A quantity P obeys law P.t/ doubling time is 7 years and P and P .0/ .0/ 100. 100.
D
The doubling time is 7 years, years, so 7 so 7 100e .
SOLUTION
P.t/
D
0:099t
D
C e kt (t in years). Find the formula for P.t/, P.t/, assuming that the
D ln 2= k, or k D ln 2=7 D 0:099 years 0:099 years
1.
With P.0/ With P.0/
D 100, 100, it follows that
654
THE INTEGRAL
CHAPTER 5
7. Find all solutions to the differential equation equation y y 0 SOLUTION
y.t/
D 3:4e
y 0 D 5y, 5y, so y.t/ so y.t/ D C e
5t .
to y 0 8. Find the solution to y
0 p y D
SOLUTION
y.t/
D 20e
p
D
D 5y. 5y. Which solution satisfies the initial condition y.0/ condition y.0/ D 3:4? 3:4? 5t for some constant C D 3:4. constant C .. The initial condition y.0/ condition y.0/ D 3:4 determines 3:4 determines C C D 3:4. Therefore,
p
2y satisfying 2y satisfying y.0/ y.0/
2y, 2y , so y.t/ so y.t/
2t .
D Ce
p
2t
D 20. 20.
for some constant C constant C .. The initial condition y.0/ condition y.0/
D 20 determines D 20. 20 determines C C D 20. Therefore,
9. Find the solution to y to y 0
D 3y satisfying 3y satisfying y.2/ y.2/ D 1000. 1000. D 1000 SOLUTION y 0 D 3y, 3y, so y.t/ 1000 determines C C D so y.t/ D C e 3t for some constant C constant C .. The initial condition y.2/ condition y.2/ D 1000 determines . Therefore, e6 1000 y.t/ D 6 e 3t D 1000e3.t 2/ . e 10. Find the function y function y D f.t/ that f.t/ that satisfies the differential equation y equation y 0 D 0:7y and 0:7y and the initial condition y.0/ condition y.0/ D 10. 10. SOLUTION Given that y that y 0 D 0:7y and 0:7y and y.0/ y.0/ D 10, 10, then f then f .t/ D 10e 0:7t .
11. The decay constant constant of cobalt-60 is 0:13 is 0:13 year year1 . Find its half-life. SOLUTION
Half-life
ln 2 D 0:13 5:33 years. 5:33 years.
12. The half-life radium-226 is 1622 years. Find its decay constant. SOLUTION
Half-life
l n 2 ln 2 ln 2 D ln D D 4:27 10 so k so k D k half-life 1622
4
1.
years
13. One of the world’s smallest smallest flowering plants, Wolffia globosa (Figure 1), has a doubling time of approximately 30 hours. Find the growth constant k constant k and and determine the initial population if the population grew to 1000 after 48 hours.
FIGURE 1 The tiny plants are Wolffia , with plant bodies smaller than the head of a pin.
SOLUTION
By the formula formula for the doubling time, 30 time, 30 k
l n 2 D ln . Therefore, k
D ln302 0:023 hours 0:023 hours
1
:
D P 0e0:023t . If P P .48/ .48/ D 1000, 1000, then P 0 e .0:023/48 D 1000 ) P 0 D 1000e .0:023/48 332
The plant population after t after t hours hours is P.t/ is P.t/
14. A 10-kg quantity of a radioactive isotope decays to 3 kg after 17 years. Find the decay constant of the isotope. SOLUTION
P.t/
D 10e
kt .
Thus P Thus P .17/ .17/
D 3 D 10e
17k ,
so k so k
l n.3=10/ 0:071 years D ln 0:071 years 17
1.
2 e 0:06t (in millions), where t where t is measured in years. Calculate the time it takes for the 15. The population of a city is P.t/ population to double, to triple, and to increase seven-fold.
D
SOLUTION
Since k Since k
D 0:06, 0:06, the doubling time is ln 2 k
11:55 years 11:55 years::
The tripling time is calculated in the same way as the doubling time. Solve for in the equation P .t
C / D 3P.t/ 2 e 0:06.t C / D 3.2e0:06t / 2 e0:06t e 0:06 D 3.2e0:06t / e 0:06 D 3
SECTION
0:06 or ln 3=0:06 seven-fold is
D
18:31 years. 18:31 years.
Exponentia Exponentiall Growth Growth and Decay Decay
5.8
655
D ln 3;
Working Working in a similar fashion, fashion, we find that the time required required for the population population to increase
ln 7 ln 7 32:43 years:: 32:43 years k 0:06 16. What is the differential equation satisfied by P.t/ by P.t/,, the number of infected computer hosts in Example 4? Over which time interval would P.t/ would P.t/ increase increase one hundred-fold?
D
SOLUTION
Because the rate constant is k is k
D 0:0815 s 0:0815 s
1
, the differential equation for P.t/ for P.t/ is is
dP dt
D 0:0815P:
The time for the number of infected computers to increase one hundred-fold is ln 100 k
ln 100 D 0:0815 56:51 s:: 56:51 s
17. The decay constant constant for a certain drug is k is k 0:35 day 0:35 day1 . Calculate the time it takes for the quantity present in the bloodstream to decrease by half, by one-third, and by one-tenth.
D
SOLUTION
The time required for the quantity quantity present in the bloodstream to decrease decrease by half is ln 2 k
ln 2 1:98 days D 0:35 1:98 days::
ln 3 k
ln 3 3:14 days D 0:35 3:14 days::
ln 10 k
l n 10 D ln 6:58 days 6:58 days:: 0:35
To decay decay by one-third, the time is
Finally, to decay by one-tenth, the time is
The intensity of light passing through an absorbing medium medium decreases exponentially exponentially with the distance trav1 eled. Suppose the decay constant for a certain plastic block is k 4 m . How thick must the block be to reduce the intensity by a factor of one-third? 18. Light Intensit Intensity y
D
SOLUTION
Since intensity decreases exponentially exponentially,, it can be modeled by an exponential decay equation I.d/ equation I.d/ kd .
suming I.0/ suming I.0/
D 1; I.d/ D e Since the decay constant is k is k D 4, we have I.d/ have I.d/ D e ln l n.1=3/ 1 4d D 3 or when d D 4 0:275 m. of one-third when e when e when d D
4d .
D I 0 e
kd .
As-
Intensity will be reduced by a factor
19. Assuming that population growth is approximately exponential, which of the following two sets of data is most likely to represent the population (in millions) of a city over a 5-year period?
Year
2000
2001
2002
2003
2004
Set I Set II
3.14 3.14
3.36 3.24
3.60 3.54
3.85 4.04
4.11 4.74
SOLUTION If the population growth is approximately exponential, exponential, then the ratio between successive years’ years’ data needs to be approximately the same.
Year
2000
2001
2002
2003
2004
Data I Ratio tios
3.14 3.36 3.60 3.85 4.11 1.07006 1.07143 1.06944 1.06753
Data II Ratio tios
3. 3.14 3.24 3.54 4.04 4.74 1.03185 1.09259 1.14124 1.17327
As you can see, the rati o of successive years in the data from “Data I” is very close to 1:07. 1:07. Therefore, we would expect exponential t growth of about P.t/ about P.t/ .3:14/.1:07 /. 20. The atmospheric atmospheric pressure pressure P.h/ (in kilopascal kilopascals) s) at a height height h meters meters above above sea level level satisfies satisfies a differentia differentiall equation equation P 0 for some positive constant k constant k.. (a) Barometric measurements measurements show that P that P .0/ .0/ (b) Determine the atmospheric pressure at h at h
D 101:3 and 101:3 and P.30; P.30; 900/ 900/ D 1:013. 1:013. What is the decay constant k constant k?? D 500. 500.
D kP
656
THE INTEGRAL
CHAPTER 5
SOLUTION
Because P 0 constant k,, P.h/ C e kh where C where C (a) Because P kP for some positive constant k 30;900k We know that P that P .30;900/ .30;900/ 101:3e 1:013. 1:013. Solving for k for k yields yields
D
D
D
k (b) P.500/
D 101:3e
0:000149.500/
D
1 ln 30;900
1:013 101:3
0:000149 meters 0:000149 meters
kh .
D D P.0/ D 101:3. 101:3. Therefore, P.h/ Therefore, P.h/ D 101:3e
D
1
:
94:03 kilopascals. 94:03 kilopascals.
21. Degrees Degrees in Physics One study suggests that from 1955 to 1970, the number of bachelor’s bachelor’s degrees degrees in physics awarded awarded per year by U.S. universities grew exponentially, with growth constant k constant k 0:1. 0:1.
D
awarded per year to increase 14-fold? (a) If exponential growth continues, how long will it take for the number of degrees awarded (b) If 2500 degrees were awarded awarded in 1955, in which year were 10;000 were 10;000 degrees degrees awarded? SOLUTION
(a) The time required for the number of degrees to increase 14-fold is
ln 14 k
l n 14 D ln 26:39 years:: 26:39 years 0:1
(b) The doubling time is . is .ln ln 2/=0:1 0:693=0:1 6:93 years. 6:93 years. Since degrees are usually awarded once a year, we round off the doubling time to 7 years. The number quadruples after 14 after 14 years, years, so 10;000 so 10;000 degrees degrees would be awarded in 1969 in 1969..
D
22. The Beer–Lambert Law is used in spectroscopy to determine the molar absorptivity ˛ absorptivity ˛ or or the concentration c concentration c of of a compound dissolved in a solution at low concentrations (Figure 2). The law states that the intensity I intensity I of of light as it passes through the solution ˛cx , where I 0 is the initial intensity and x is the distance traveled by the light. Show that I I satisfies a satisfies ln.I=I ln.I=I 0 / kI for differential equation dI equation dI =dx =dx f or some constant k constant k..
D
D
Intensity I
x
Distance
Solution
0
0
x
FIGURE 2 Light of intensity I intensity I 0 passing through a solution.
SOLUTION
ln
I I 0
I D ˛cI D e˛cI or I D I 0 e˛cI . Therefore, ˛c I so so or I D I 0
dI dx where k where k
D I 0e˛cI .˛c/ D I.˛c/ D kI; kI ;
D ˛c is ˛c is a constant.
23. A sample of sheepskin parchment discovered by archaeologists had a C 14 -to-C12 ratio equal to 40% of that found in the atmosphere. Approximately how old is the parchment? SOLUTION
C 14 to C The ratio of C to C 12 is Re is Re
0:000121t
D 0:4R so 0:4R so 0:000121t D ln.0:4/ t D 7572:65 7600 years. 7600 years. ln.0:4/ or or t
In 1994, three French speleologists (geologists specializing in caves) caves) discovered discovered a cave in southern France 14 containing prehistoric cave paintings. A C analysis carried out by archeologist Helene Valladas showed the paintings to be between 29,700 and 32,400 years old, much older than any previously known human art. Given that the C 14 -to-C12 ratio of the atmosphere is R is R 10 12 , what range of C 14 -to-C12 ratios did Valladas find in the charcoal specimens? 24. Chauvet Chauvet Caves Caves
D
SOLUTION
The C14 -C12 ratio found in the specimens ranged from 12
10
e
0:000121.32;400/
14
1:98 10
2:75 10
to 12
10
e
0:000121.29;700/
14
:
25. A paleontologist discovers remains of animals that appear to have died at the onset of the Holocene ice age, between 10,000 and 12,000 years ago. What range of C 14 -to-C12 ratio would the scientist expect to find in the animal remains?
SECTION
SOLUTION
Exponentia Exponentiall Growth Growth and Decay Decay
5.8
657
The scientist scientist would expect expect to find C14 -C12 ratios ranging from 12
10
e
0:000121.12;000/
13
2:34 10
2:98 10
to 12
10
e
0:000121.10;000/
13
:
26. Inversion Inversion of Sugar When cane sugar is dissolved in water, water, it converts to invert invert sugar over a period of several several hours. The percentage f.t/ percentage f.t/ of of unconverted cane sugar at time t time t (in (in hours) satisfies f satisfies f 0 . What percentage of cane sugar remains after 0:2f . 5 hours? After 10 hours?
D
SOLUTION
Thus f.5/ Thus f.5/
f 0
D 0:2f , , so f.t/ D C e
D 100e
0:2.5/
0:2t . Since Since f is is a percentag percentage, e, at t
36:79 percent 36:79 percent and f and f .10/ .10/
D 100e
0:2.10/
D 0, C D D 100 percent. Therefore. f.t/ D 100e
0:2t .
13:53 percent. 13:53 percent.
27. Continuing with Exercise 26, suppose that 50 that 50 grams grams of sugar are dissolved in a container of water. water. After how many hours will 20 grams 20 grams of invert sugar be present? SOLUTION
If there are 20 grams of invert sugar present, then there are 30 grams of unconverted unconverted sugar. This means means that f
Solving 100e
0:2t
D 60. 60.
D 60
for t for t yields yields t
1 D 0:2 ln 0:6 2:55 hours 2:55 hours::
28. Two bacteria colonies are cultivated in a laboratory. laboratory. The first colony has a doubling time of 2 hours and the second a doubling time of 3 hours. Initially, the first colony contains 1000 bacteria and the second colony 3000 bacteria. At what time t time t will will the sizes of the colonies be equal? SOLUTION
solve e solve e k1 t
D 1000ek t and P and P 2 .t / D 3000e k t . Knowing that k that k 1 D
P 1 .t /
1
2
D 3ek t for t for t . Thus
ln 2 hours 2
1
and k and k 2
D
ln 2 hours 3
1
, we need to
2
k1 t
D ln.3e ln.3ek t / D ln 3 C ln.e ln.e k t / D ln 3 C k2 t; 2
2
so t
D k ln3k D 6lnln23 9:51 hours 9:51 hours:: 1
29. Moore’s Moore’s Law Law tially.
2
In 1965, Gordon Moore Moore predicted that that the number number N of N of transistors on a microchip would increase exponen-
table of data below below confirm Moore’ Moore’ss prediction prediction for the period period from 1971 1971 to 2000? If so, so, estimate estimate the growth growth constant constant k k.. (a) Does the table (b)
Plot the data in the table.
(c) Let N.t/ Let N.t/ be be the number of transistors t transistors t years years after 1971. Find an approximate formula N.t/ years after 1971. (d) Estimate the doubling time in Moore’s Law for the period from 1971 to 2000. (e) How many transistors will a chip contain in 2015 if Moore’s Law Law continues to hold? (f) Can Moore have expected expected his prediction to hold indefinitely? Processor
Year
No. Transistors
4004 8008 8080 8086 286 386 processor 486 DX processor Pentium processor Pentium II processor Pen Pentium tium III III proc rocesso ssor Pentiu tium 4 proc rocessor
1971 1972 1974 1978 1982 1985 1989 1993 1997 1999 1999 2000
2250 2500 5000 29,000 120,000 275,000 1,180,000 3,100,000 7,500,000 24, 24,000 000,000 ,000 42,000,000
Xeon processor
2008
1,900,000,000
C e kt , where t where t is is the number of
658
THE INTEGRAL
CHAPTER 5
SOLUTION
(a) Yes, the graph looks like an exponential graph especially towards the latter years. We We estimate the growth constant by setting 1971 as our starting point, so P 0 2250. 2250. Therefore, P.t/ 2250e kt . In 2008, t 37. 37. Therefore, P.37/ 2250e 37k ln 844;444:444 1;900;000;000, 1;900;000;000, so k 0:369. Note: A better estimate can be found by calculating k for each time period and 0:369. 37 then averaging the k the k values. values. (b)
D
D
D
D
D
D
y
4 " 10 7 3 " 10 7 2 " 10 7 1 " 10 7 x
1980 1985 1990 1995 2000
(c) N.t/
D 2250e0:369t
(d) The doubling time is ln 2=0:369 1:88 years. 1:88 years. (e) In 2015, t 2015, t
D 44 years. 44 years. Therefore, N.44/ Therefore, N.44/ D 2250e0:369.44/ 2:53 1010 .
(f) No, you can’t make a microchip microchip smaller than an atom.
30. Assume that in a certain country, the rate at which jobs are created is proportional to the number of people who already have jobs. If there are 15 million million jobs at t at t 0 and 15.1 million jobs 3 months later, how many jobs will t here be after 2 years?
D
Let J.t/ Let J.t/ denote denote the number of people, in millions, who have jobs at time t , in months. Because the rate at which jobs are created is proportional to the number of people who already have j obs, J obs, J 0 .t / kJ.t/, kJ.t/, for some constant k . Given that kt J.0/ 15, 15, it then follows that J.t/ that J.t/ 15e . To determine k determine k,, we use J.3/ use J.3/ 15:1; 15:1; therefore, SOLUTION
D
D
D
D
k
D
1 ln 3
15:1 15
3
2:215 10
1
months
:
Finally, after two years, there are J.24/
D 15e0:002215.24/ 15:8 million 15:8 million
jobs. doubling t ime are the exponential functions P functions P 0 ek t with k with k > 0. 0 . Show that the doubling time 31. The only functions with a constant doubling of linear function f function f .t/ at b at time t time t 0 is t is t 0 b=a (which b=a (which increases with t with t 0 ). Compute the doubling times of f f .t/ 3t 12 at t at t 0 10 and 10 and t t 0 20. 20.
D
SOLUTION
D C
D
Let f Let f .t/
equation
C
D C
D at C b and suppose f suppose f .t0 / D P 0 . The time it takes for the value of f of f to to double is the solution of the 2P 0
D 2.at0 C b/ D at C b or t D 2t0 C b=a: For the function f.t/ function f.t/ D 3t C 12, 12, a D 3, b D 12 and 12 and b b=a =a D 4. With t With t 0 D 10, 10, the doubling time is then 24; with t with t 0 D 20, 20, the
doubling time is 44.
32. Verify that the half-life of a quantity that decays exponentially with decay constant k constant k is is equal to . to .ln ln 2/=k. 2/=k .
Let y Let y C e kt be an exponential decay function. Let t be the half-life of the quantity y quantity y , that is, the time t time t when C C y C e k t for t . Solving for t we we get ln 2 k t , so t so t ln 2= k . 2 2 33. Compute the balance after 10 years if $2000 is deposited in an account paying 9% interest and int erest is compounded (a) quarterly, (b) monthly, and (c) continuously. continuously.
D D
SOLUTION
D
D
D
SOLUTION
D 2000.1 C 0:09=4/4.10/ D $4870:38 (b) P.10/ D 2000.1 C 0:09=12/12.10/ D $4902:71 (c) P.10/ D 2000e0:09.10/ D $4919:21 (a) P.10/
34. Suppose $500 is deposited into an account paying interest at a rate of 7%, continuously compounded. Find a formula for the value of the account at time t time t . What is the value of the account after 3 years? SOLUTION Let P.t/ Let P.t/ denote denote the value of the account at time t time t . Because the initial deposit is $500 and the account pays interest at a rate of 7%, compounded continuously, continuously, it foll ows that P.t/ that P.t/ 500e0:07t . After three years, the value of the account is P.3/ 500e0:07.3/ $616:84. 616:84.
D
D
D
SECTION
Exponentia Exponentiall Growth Growth and Decay Decay
5.8
659
35. A bank pays interest at a rate of 5%. What is the yearly multiplier if interest is compounded (a) three times a year? year?
(b) continuously?
SOLUTION
(a) P.t/
D (b) P.t/ D
0:05 3t 0:05 P 0 1 , so the yearly multiplier is 1 3 3 P 0e 0:05t , so the yearly multiplier is e is e 0:05 1:0513. 1:0513.
C
C
3
1:0508. 1:0508.
36. How long will it take for $4000 to double in value value if it is deposited deposited in an account account bearing 7% interest, continuousl continuously y compounded? ln 2 SOLUTION The doubling time is 9:9 years. 9:9 years. 0:7 37. How much must one invest today in order to receive $20,000 after 5 years if interest is compounded continuously at the rate r 9%?
D
SOLUTION
Solving 20;000 Solving 20;000
D P 0e0:09.5/ for P for P 0 yields P 0
D 20;000 $12;752:56: e0:45
38. An investment investment increases in value at a continuously compounded rate of 9%. 9 %. How large must the initial i nvestment nvestment be in order to build up a value of $50,000 over a 7-year period? SOLUTION
Solving 50;000 Solving 50;000
D P 0e0:09.7/ for P for P 0 yields P 0
$26;629:59: D 50;000 e0:63
39. Compute the PV of $5000 received in 3 years if the interest rate is (a) 6% and (b) 11%. What is the PV in these two cases if the sum is instead received in 5 years? SOLUTION
In 3 years: 0:06.3/
D D 5000e D D 5000e
D D D D
(a) P V (b) P V
In 5 years: (a) P V 5000e 5000e (b) P V
0:11.3/
0:06.5/ 0:11.5/
D $4176:35 D $3594:62 D $3704:09 D $2884:75
40. Is it better to receive $1000 today or $1300 in 4 years? Consider r Consider r
D 0:08 and 0:08 and r r D 0:03. 0:03.
0:08, 0:08, then the present value of $1300 four years from now is continuous continuous compound compounding, ing, if r 1300e $943:99. 943:99. It is better to get $1000 now. On the other hand, if r 0:03, 0:03, the present value of $1300 four 0:03.4/ years from now is 1300e is 1300e $1153:00, 1153:00, so it is better to get the $1,300 in four years. SOLUTION Assuming Assuming 0:08.4/
D
D
D
D
41. Find the interest rate r rate r if if the PV of $8000 to be received in 1 year is $7300. SOLUTION
Solving 7300 Solving 7300
D 8000e
r.1/
for r for r yields yields r
D ln
7300 8000
D 0:0916;
or 9.16%. 42. A company can earn additional profits of $500;000 $ 500;000/year /year for 5 years by investing $2 million to upgrade its factory. Is the investment investment worthwhile if the interest rate is 6 is 6%? %? (Assume the savings are received as a lump sum at the end of each year.) SOLUTION
The present value of the stream of additional additional profits is 500;000.e
0:06
Ce
0:12
Ce
0:18
Ce
0:24
Ce
0:3
/
D $2;095;700:63:
This is more than the $2 million cost of the upgrade, so the upgrade should be made. 43. A new computer system costing $25,000 will reduce labor costs by $7000 $7000/year /year for 5 years.
investment if r r 8%? (a) Is it a good investment (b) How much money will the company actually save?
D
SOLUTION
(a) The present value of the reduced labor costs is
7000.e
0:08
Ce
0:16
Ce
0:24
Ce
0:32
Ce
0:4
/
D $27;708:50:
This is more than the $25,000 cost of the computer system, so t he computer system should be purchased. (b) The present value of the savings is $27;708:50 $25;000
D $2708:50:
660
THE INTEGRAL
CHAPTER 5
44. After winning $25 million i n the state lottery, Jessica learns that she will receive five five yearly payments of $5 million beginning immediately. (a) What is the PV of Jessica’s Jessica’s prize if r r
D 6%?
(b) How much more would the prize be worth if the entire amount were paid today? SOLUTION
(a) The present value of the prize is 0:24
5;000;000.e
Ce
0:18
Ce
0:12
Ce
0:06
Ce
0:06.0/
/
D $22;252;915:21:
(b) If the entire amount were paid today, the present value would be $25 million, or $ 2;747;084:79 more 2;747;084:79 more than the stream of payments made over five years. 45. Use Eq. (3) to compute the PV of an income stream paying out R.t/ r 0:05. 0:05.
D
10
Z D D
P V
SOLUTION
5000e
0:05t
dt
0
0:05t
D 100;000e
ˇˇ ˇ
D $5000/year 5000/year continuously for 10 years, assuming
10
D $39;346:93. 39;346:93.
0
46. Find the PV of an investment that pays out continuously at a rate of $800 $800/year /year for 5 years, assuming r assuming r 5
Z D D
P V
SOLUTION
800e
0:08t
dt
0
D 10;000e
0:08t
5
ˇˇ D ˇ
$3296:80. 3296:80.
0
rate R.t/ 47. Find the PV of an income stream that pays out continuously at a rate R.t/ 7
Z D D
P V
SOLUTION
5000e
0:1t
e
0:05t
0
7
dt
Z D
D 0:08. 0:08.
0:05t
5000e
dt
0
D $5000e0:1t /year for 7 years, assuming r assuming r D 0:05. 0:05. 0:05t
D 100;000e
7
ˇˇ D ˇ
$41;906:75. 41;906:75.
0
48. A commercial property generates income at the rate R.t/. R.t/. Suppose that R.0/ $70;000/year 70;000/year and that R.t/ increases R.t/ increases at a continuously compounded rate of 5%. Find the PV of the income generated in the first 4 years if r r 6%.
D
4
Z D D
P V
SOLUTION
70;000e0:05t e
0:06t
dt
0
D 70;000 e 0:01
0:01t
4
ˇˇ D ˇ
D
$274;473:93:
0
49. Show that an investment investment that pays out R out R dollars dollars per year continuously for T for T years has a PV of R.1 R.1 e rT /=r .
The present value of an an investment investment that pays out R out R dollars dollars= =year continuously for T for T years years is
SOLUTION
T
Z D D
P V
Re
rt
dt :
0
Let u Let u
D rt; du D r d t . Then P V
D D
1 r
rT
Z
u
Re du
0
D
R u e r
ˇˇˇ ˇ
rT
0
D Rr .e
rT
1/
D Rr .1 e
rT
/:
50. R=r R= r .
Explain this statement: If T T is very large, then the PV of the income stream described in Exercise 49 is approximately
SOLUTION
Because lim e
rT
T
!1
1 D T lim D 0; !1 e rt
it follows that R .1 T !1 r lim
e
rT
/
D Rr :
51. Suppose that r that r 0:06. 0:06. Use the result of Exercise 50 to estimate t he payout rate R rate R needed needed to produce an income stream whose PV is $20,000, assuming that the stream continues for a large number of years.
D
SOLUTION
R D D Rr so 20;000 1200. so 20;000 D or R or R D $1200. 0:06
From Exercise 50, P 50, P V
52. Verify by differentiation:
Z
te
rt
dt
D e
rt .1
C r t / C C
r2
Use Eq. (5) to compute the PV of an investment that pays out income continuously at a rate R.t/ year for 5 years, assuming r assuming r 0:05. 0:05.
D
5
D .5000 C 1000t/ dollars 1000t/ dollars per
SECTION
Exponentia Exponentiall Growth Growth and Decay Decay
5.8
661
SOLUTION
d dt
e
rt .1
C r t / D 1 e
r2
rt
.r/
r2
C .1 C rt/.r e
rt
1
D
/
r
e
rt
e
rt
rt e
rt
D
te
rt
Therefore 5
Z
P V
D D
.5000
0
D
5000 .e 0:05
C 1000t/e 0:05.5/
0:05t
5
dt
Z
D
1/ 1000
5000e
0:05t
5
dt
0
e
C
0:05.5/ .1
Z !
0:05t
1000te
dt
0
C 0:05.5// C 1000 1 .0:05/2 .0:05/2
D 22;119:92 389;400:39 C 400;000 $32;719:53: 53. Use Eq. (5) to compute the PV of an investment that pays out income continuously at a rate R.t/ dollars per year for 10 years, assuming r assuming r 0:08. 0:08.
D
D .5000 C 1000t/e0:02t
SOLUTION
10
Z D D
P V
.5000
0
D
5000 .e 0:06
C 1000t/.e
0:06.10/
0:02t
/e
10
0:08t
dt
Z D
5000e
0:06t
10
dt
0
1/ 1000
e
0:06.10/
Z C
0:06t
1000te
dt
0
.1 0:06.10// .0:06/2
C
!
1 C 1000 .0:06/ 2
D 37;599:03 243;916:28 C 277;777:78 $71;460:53: 54. Banker’s Rule of 70 If you earn earn an interest rate of R percent, R percent, continuously compounded, your money doubles after approximately 70=R approximately 70=R years. years. For example, at R 5%, your money doubles after 70=5 after 70=5 or or 14 years. Use the concept of doubling time Note: Sometimes, the rule 72=R to justify the Banker’s Rule. ( Note: rule 72=R is is used. It is less accurate but easier to apply because 72 is divisible by more numbers than 70.)
D
SOLUTION
The doubling time is t
70 l n 2 ln l n 2 100 69:93 D ln D D : r r% r% r%
Let y.t/ be be the drug concentration (in mg/kg) in a patient’s body at time t . The initial con55. Drug Dosing ing Interval Let y.t/ centration is y.0/ is y.0/ L. Additional doses that increase the concentration by an amount d amount d are are administered at regular time intervals of length T length T .. In between doses, y.t/ decays y.t/ decays exponentially—that is, y is, y 0 ky . Find the value of T of T (in (in terms of k and k and d d )) for which ky. the the concentration varies between L between L and and L L d as as in Figure 3.
D
D
Exponential decay y (mcg/ml) L
L − d
Dose administered administered T
2T
t
3T
FIGURE 3 Drug concentration with periodic doses. SOLUTION
Because y Because y 0
D ky and ky and y.0/ y.0/ D L, it follows that y.t/ that y.t/ D Le Le
kT
D L d
or
kt . We want y. want y.T T/
D D k1 ln
T
1
D L d , d , thus
d : L
Exercises Exercises 56 and and 57: The Gompertz The Gompertz differential equation
dy dt
D ky ln ky ln
y M
6
(where M and k are constants) was introduced in 1825 by the English mathematician Benjamin Gompertz and is still used today to model aging and mortality. 56. Show that y that y
D M eae
kt
satisfies Eq. (6) for any constant a constant a..
662
CHAPTER 5
Let y Let y
SOLUTION
THE INTEGRAL
D M eae
kt
. Then dy dt
D M.kaekt /eae
kt
and, since ln.y=M/ ln.y=M/
D aekt ;
we have ky ln ky ln.y=M/ .y=M/
D Mkae kt eae D dy : dt kt
57. To model mortality in a population of 200 laboratory rats, a scientist assumes that the number P.t/ of P.t/ of rats alive at time t time t (in months) satisfies Eq. (6) with M with M 204 and 204 and k k 0:15 month 0:15 month1 (Figure 4). Find P.t/ Find P.t/ [note [note that P.0/ that P.0/ 200] 200] and determine the population after 20 months.
D
D
D
Rat population P(t ) 200
100
t (mo)
10
20
30
40
FIGURE 4
The solution to the Gompertz Gompertz equation with M with M
D D 204 and 204 and k k D 0:15 is 0:15 is of the form:
SOLUTION
P.t/
D 204eae
0:15t
Applying the initial condition allows us to solve for a for a::
D 204ea 200 D ea 204 200
ln so that a that a
0:02: After 0:02: After t t
200 204
Da
D 20 months, 20 months, P.20/
D 204e
0:02e0:15.20/
D 136:51;
so there are 136 rats. 58. Isotop topes for Dating ing Which of the following would be be most suitable for dating extremely extremely old rocks: carbon-14 (halflife 5570 years), lead-210 (half-life 22.26 years), or potassium-49 (half-life 1.3 billi on years)? Explain why. SOLUTION For extremely old rocks, you need to have an isotope that decays very slowly. slowly. In other words, you want a very large half-life such as Potassium-49; otherwise, the amount of undecayed isotope in the rock sample would be too small to accurately measure.
P.t/ be a quantity that obeys an exponential growth law with growth constant k m-fold 59. Let P Let P P.t/ be constant k.. Show that P that P increases increases m -fold after an interval of .ln .ln m/=k years. m/=k years.
D D
SOLUTION
For m For m-fold -fold growth, P.t/ growth, P.t/
ln m D mP 0 for some t some t . Solving mP Solving mP 0 D P 0 e kt for t for t , we find t find t D k
Further Insights and Challenges 60.
Average Time ime of Decay Physicists use the radioactive radioactive decay decay law R law R
M unti unti l an atom decays. Let F.t/ Let F.t/ R=R0 (a) Find the inverse inverse function t.F/ function t.F/..
D
De
kt
D R0 e
kt
to compute the average or mean time
be the fraction of atoms that have survived to time t time t without without decaying.
SECTION
5.8
Exponentia Exponentiall Growth Growth and Decay Decay
663
(b) By definition of t.F/, t.F/, a fraction 1=N fraction 1=N of of atoms decays in the time interval
j ;t N
t
Use this to justify the approximation M approximation M
1 N
N
X t
j 1
D
j
1
N
j . Then argue, by passing to the l imit as N as N N
! ! 1, that M D that M D
1 t.F/dF . 0 t.F/dF .
R
Strictly speaking, this is an improper integral because t.0/ because t.0/ is is infinite (it takes an infinite amount of time for all atoms to decay). Therefore, we define M define M as as a limit 1
M
D D clim !0
R
(c) Verify the formula ln x dx
Z
t.F/dF
c
D x ln x x by differentiation and use it to show that for c 0, for c > 0,
M
1 k
D D clim C !0
(d) Show that M that M
1 .c ln .c ln c c/ k
D D 1=k 1= k by evaluating the limit (use L’Hˆopital’s opital’s Rule to compute lim c ln c ). c !0
(e) What is the mean time to decay for radon (with a half-life of 3.825 days)? SOLUTION
(a) F
D D e
kt
ln l n F D D k t and t.F/ and t.F/ D k
so ln F
1 N t.j=N/. t.j=N/. For the interval Œ0 interval Œ0;; 1, from the approximation given, the subinterval length is 1=N and 1=N and thus the N j D1 right-hand endpoints have x have x-coordinate -coordinate .j=N/ .j=N/.. Thus we have a Riemann sum and by definition,
(b) M
P
1 lim N !1 N d (c) .x ln .x ln x x/ dx
Dx
1 x
N
X
1
Z D
t.j=N/
t.F /dF: /dF:
0
j 1
D
C ln x 1 D ln x . Thus 1
M
D D clim !0
Z
t .F /df /df
c
D clim !0
D ! lim
c
0
1
ˇˇ ! D ˇ !
1 .F ln ln F F / k
lim
c
c
1 .1 1 ln 1 .c c ln c// k 1 k
D clim C !0
1 .c ln .c ln c c/ k
0
1 .F F ln ln F / k
1
ˇˇ ! ˇ c
(d) By L’Hˆopital’s opital’s Rule, 1
c 1
Thus, M Thus, M
D D clim !0
Z
t.F /dF /dF
c
lim c ln c
!0C
1 k
D
61. Modify the proof of the relation e 1
C xn
SOLUTION
1 1 .c ln c c/ . k k ln 2 1 and 5:52. 5:52. Thus, M Thus, M 3:825 k
D D clim C !0
(e) Since the half-life is 3:825 is 3:825 days, days, k k
ln.1 ln.1
D c!lim0C cln c1 D c!lim0C cc 2 D c!lim0C c D 0: D
D D D 5:52 days. 5:52 days. 1 C n1 n given in the text to prove ex D D nlim !1
/ as an integral and estimate above and below by rectangles. Start by expressing
ln 1
C
1 x= n
x n
D Z C
ln 1
C xn
1
dt : t
Following the proof in the t ext, we note that x n
Cx
x n
C
x n
provided x provided x > 0, 0 , while x n
ln 1
x n
Cx
lim
n
!1
x n n . Hint: Express
C 1
664
CHAPTER 5
THE INTEGRAL
when x when x < 0. 0 . Multiplying both sets of inequalities by n by n and and passing to the limit as n as n lim
n
!1
n
x n
C ln 1
! 1, the squeeze theorem guarantees that
D x:
Finally, lim
n
!1
62. Prove that, for n for n > 0, 0,
1
1 n
C
n
x n
C 1
n
e
D ex :
1
C
1 n
n 1
C
Hint: Take logarithms and use Eq. (4). SOLUTION
Taking logarithms throughout the desired inequality, inequality, we find the equivalent equivalent inequality
n ln 1
C
1 n
1 .n
C 1/ ln 1 C
1 n
:
Multiplying Eq. (4) by n by n yields yields n
n
1 n
C 1 n ln 1 C
1;
which establishes the left-hand side of t he desired inequality. On the other hand, multiplying Eq. (4) by n by n 1 .n
C 1/ ln 1 C n1
1
C 1 yields
C n1 ;
which establishes the right-hand side of the desired inequality. 63. A bank pays interest at the rate r rate r , compounded M compounded M times yearly. yearly. The effective interest rate r e is the rate at which interest, if compounded annually, annually, would have to be paid to produce the same yearly return.
Find re if r r (a) Find r
D 9% compounded monthly. (b) Show that r that re D .1 C r=M/M 1 and that r that re D e r 1 if interest is compounded continuously. continuously. Find re if r r D 11% 11% compounded continuously. continuously. (c) Find r (d) Find the rate r rate r that, that, compounded weekly, would yield an effective rate of 20%. SOLUTION
(a) Compounded monthly, monthly, P.t/
D P 0.1 C r=12/12t . By the definition of r re , P 0 .1 C 0:09=12/12t D P 0 .1 C re /t
so .1
C 0:09=12/12t D .1 C re /t
or
re
D .1 C 0:09=12/12 1 D 0:0938;
or 9:38 or 9:38% % (b) In general,
P 0 .1
C r=M/M t D P 0 .1 C re /t ;
so .1 so .1 r=M/M t .1 re /t or r or re rt t r e .1 re / or r e 1. or re
D C D .1 C r=M/M 1. If interest is compounded continuously, continuously, then P then P 0 e rt D P 0 .1 C re /t so D (c) Using part (b), r (b), re D e 0:11 1 0:1163 or 0:1163 or 11.63%. C D C
(d) Solving
0:20 for r for r yields yields r r
D 52.1:21=52 1/ D 0:1826 or 0:1826 or 18.26%.
D 1 C 52r
52
1
Chapter Chapter Review Review Exercise Exercises s
665
CHAPTER REVIEW EXERCISES In Exercises Exercises 1–4, 1–4, refer refer to the function function f .x/ whose graph is shown in Figure 1.
y
3 2 1 x
1
2
3
4
FIGURE 1
1. Estimate L Estimate L4 and M and M 4 on Œ0 on Œ0;; 4. SOLUTION
With n With n
D 4 and an interval of Œ0 of Œ0;; 4, x D 4 4 0 D 1. Then,
L4
D x.f.0/ C f.1/ C f.2/ C f .3// C 1 C C 2 D 234 .3// D 1
1 4
5 2
and M 4
D x
1 2
3 2
5 2
7 2
D 1 12 C 2 C 94 C 94 D 7:
f
C f
C f
C f
2. Estimate R Estimate R4 , L4 , and M and M 4 on Œ1 on Œ1;; 3. SOLUTION
With n With n
D 4 and an interval of Œ1 of Œ1;; 3, x D 3 4 1 D 12 . Then,
R4
D x
L4
D x
M 4
D x
3 2
5 2
f.3/
1 2
2
5 2
9 4
2
5 2
1 2
1
2
5 2
9 4
35 8
C C C D C C C D I C C C D C C C D I f
f.1/
5 4
f
f.2/
f
3 2
f
f.2/
7 4
C f
f
9 4
C f
C f
11 4
31 and 8
D 12 32 C 94 C 52 C 178 D 67 : 16
b
3. Find an interval Œa interval Œa;; b on which R which R4 is larger than SOLUTION
In general, R general, R N is larger than
Z a
f.x/dx. f.x/dx . Do the same for L4 .
b f.x/dx on a f.x/dx on
R
any interval Œa; Œa; b over which f which f .x/ .x/ is increasing. Given the graph of
f.x/, f.x/, we may take Œa; Œa; b Œ0;2. In order for L for L 4 to be larger than We may therefore take Œa; Œa; b Œ2;3.
D
4. Justify SOLUTION
3 2
2
Z 1
D
f.x/dx
b f.x/dx , f .x/ .x/ must a f.x/dx,
R
be decreasing over the i nterval Œa nterval Œa;; b .
9 . 4
.x/ is increasing on Œ1 Because f Because f .x/ on Œ1;; 2, we know that 2
LN
Z
f.x/dx
1
RN
for any N any N .. Now, L2
D
1 .1 2
C 2/ D
3 2
and
R2
D
so 3 2
2
Z 1
f.x/dx
9 : 4
1 2
2
C
5 2
D 94 ;
666
THE INTEGRAL
CHAPTER 5
In Exercises Exercises 5–8, 5–8, let f .x/ .x/
D x 2 C 3x.
5. Calculate R Calculate R6 , M 6 , and L and L6 for f for f .x/ .x/ on the interval Œ interval Œ2; 2; 5. Sketch the graph of f of f .x/ and x/ and the corresponding rectangles for each approximation. SOLUTION
Let f Let f .x/ .x/
D x 2 C 3x. D 6 subintervals has 3x . A uniform partition of Œ2; Œ2; 5 with N with N D x
D 5 6 2 D 12 ;
D a C jx D 2 C j2 ;
xj
and xj
f .xj /
D 12
C 18 C
91 4
D a C
j
1 2
x
D 74 C j2 :
Now, 6
R6
D x D
1 2
X
j 1
D 55 4
5 2
f
C 28 C
C f.3/ C f
135 4
7 2
C f.4/ C f
9 2
C f.5/
C 40 D 625 : 8
The rectangles corresponding to this approximation are shown below. below. y
35 30 25 20 15 10
x
2.0
2.5
3.0
3.5
4.0
11 4
C f
4.5
Next, 6
M 6
D x D
1 2
X
f .xj /
189 16
253 16
j 1
D
C
D 12 C
9 4
13 4
15 4
17 4
f
325 16
C
C f
405 16
C
493 16
C
589 16
C f
C f
C f
1127 D 2254 D : 32 16
The rectangles corresponding to this approximation are shown below. below. y
35 30 25 20 15 10
x
2.0
2.5
3.0
3.5
4.0
4.5
Finally, 5
L6
D x D
1 2
X
f .xj /
j 0
D 10
C
55 4
19 4
D
1 2
C 18 C
f.2/
91 4
C f
5 2
C f.3/ C f
135 4
D 505 : 8
C 28 C
The rectangles corresponding to this approximation are shown below. below. y
35 30 25 20 15 10
x
2.0
2.5
3.0
3.5
4.0
4.5
7 2
C f.4/ C f
9 2
Chapter Chapter Review Review Exercise Exercises s x
6. Use FTC I to evaluate evaluate A.x/ A.x/
Z D
f.t/dt . 2
SOLUTION
Let f Let f .x/
D x 2 C 3x. 3x . Then x
A.x/
Z D
.t
2
2
C 3t/dt D
1 3 t 3
C
3 2 t 2
x
ˇˇ
ˇ Z D 2
Let f Let f .x/
D x
8 3
C 6 D 13 x 3 C 32 x 2 103 :
f.x/dx by f.x/dx by taking the limit.
2
52 3 D x 2 C 3x on D 3x on the interval Œ interval Œ2; 2; 5. Then x D and a and a D 2. Hence, N N N
RN
C
3 2 x 2
5
7. Find a formula for R for RN for f .x/ on x/ on Œ2 Œ2;; 5 and compute SOLUTION
1 3 x 3
X
f .2
j 1
D
D 30 C N 632 D 30 C
63 N 2
C jx/ D
N
27 N 3
N
X C
3 N
2
j 1
D
j 1
N 2 2
C
N 2
2
C3 2C
3j N
!
N
X
3 N
D
10
j 1
D
C
21j N
N
j2
X C X D ! D j
3j N
j 1
C
N 3 3
27 N 3
N 2 2
C
C
N 6
!
45 9 D 141 C C 2 N 2N 2 and lim RN
D N lim !1
N
!1
Let f Let f .x/
C
45 N
9 2N 2
C
D 141 : 2
2
x/ on Œ0 Œ0;; 2 and compute 8. Find a formula for L for L N for f .x/ on SOLUTION
141 2
Z
f.x/dx by f.x/dx by taking the limit.
0
D x 2 C 3x and 3x and N N be be a positive integer. Then 0 D 2 N D N 2
x
and
D a C jx D 0 C 2jN D 2j N
xj for 0 for 0
j
N . N . Thus, N 1
LN
D x
X
f .xj /
j 0
D
D
2 N
N 1
X
j 0
D
4j 2
C N 2
6j N
!
8
D N 3
N 1
X
j
2
j 0
D
12
C N 2
N 1
X
j
j 0
D
1/.2N 1/ 6.N 1/ 4 D 4.N 3N C N D 263 10 C : 2 N 3N 2 Finally, 2
Z
f.x/dx
0
.x/ 9. Calculate R Calculate R5 , M 5 , and L and L5 for f for f .x/ SOLUTION
Let f Let f .x/
D .x 2 C 1/
1.
D N lim !1
D .x 2 C 1/
1
26 3
10 N
C 3N 4 2 D 263 :
on the interval Œ0 interval Œ0;; 1.
A uniform partition of Œ0; Œ0; 1 with N with N
D D 5 subintervals has
D 1 5 0 D 15 ;
D a C jx D j5 ;
x
xj
and
xj
D a C
j
1 2
x
D 2j10 1 :
Now, 5
R5
D x
X
j 1
D
f .xj /
D
1 5
1 5
2 5
3 5
4 5
f
C f
C f
C f
C f.1/
C
9j 2 N 2
!
667
668
CHAPTER 5
THE INTEGRAL
D
1 5
25 26
25 29
25 34
25 41
1 2
1 10
C f
C C C C
0:733732:
Next, 5
M 5
D x D
1 5
X
f .xj /
100 101
100 109
j 1
D
C
D 15
3 10
1 2
7 10
9 10
f
4 5
C C
100 149
C
C f
C f
C f
100 181
C f
0:786231:
Finally, 4
L5
D x D
X
j 0
D
1 5
f .xj /
1
25 26
1 5
D
f.0/
25 29
25 34
C C C C
25 41
1 5
3 5
C f
4 5
D x 3 on Œ0 on Œ0;; 4 (Figure 2).
2
C 1/ D 64.N N C 2
C f
0:833732:
10. Let R Let RN be the N the N th th right-endpoint approximation for f for f .x/ .x/ (a) Prove that R that RN
2 5
C f
.
(b) Prove that the area of the region within the right-endpoint rectangles above the graph is equal to
64.2N 1/ N 2
C C
y
4
2
x
1
2
3
4
Approximation RN for f for f .x/ .x/ FIGURE 2 Approximation R
D x3 on Œ0 on Œ0;; 4.
SOLUTION
(a) Let f Let f .x/ .x/
for 0 for 0
j
D x 3 and N and N be a positive integer. Then 40 4 D x D N N
D a C jx D 0 C 4jN D 4N j
xj
and
N . N . Thus, N
RN
D x
X
f .xj /
j 1
D
(b) The area between the graph of y y
D
4 N
N
X
64j 64j 3
j 1
D
N 3
256
D N 4
N
j3
X
j 1
D
2
2
2
C 1/ D 64.N C C 1/ N .N C D 256 4 4 N N 2
:
D x 3 and the x the x-axis -axis over Œ0 over Œ0;; 4 is 4
Z
3
x dx
0
D
1 4 x 4
4
ˇˇ D ˇ
64:
0
The area of the region below the right-endpoint rectangles and above the graph is therefore
C C 1/2 64 D 64.2N C C 1/ :
64.N
N 2 N 2 11. Which approximation to the area is represented by the shaded rectangles in Figure 3? Compute R Compute R 5 and L and L 5 . y
30 18 6 x
1
2
3
FIGURE 3
4
5
Chapter Chapter Review Review Exercise Exercises s
669
SOLUTION There are five five rectangles and the height of each each is given by the function value at at the right endpoint of the subinterval. Thus, the area represented by the shaded rectangles is R5 . From the figure, we see that x 1. Then
D R5 D 1.30 C 18 C 6 C 6 C 30/ D 90 L5 D 1.30 C 30 C 18 C 6 C 6/ D 90: and 12. Calculate any two Riemann sums for f for f .x/ .x/ D x 2 on the interval Œ2;5 interval Œ2;5, but choose partitions with at least five subintervals subintervals of
unequal widths and intermediate points that are neither endpoints nor midpoints. Let f Let f .x/
SOLUTION
D x 2. Riemann sums wil l, of course, vary. Here are two possibilit ies. Take D 5, Take N D D fx0 D 2; x1 D 2:7; P D 2:7; x2 D 3:1; 3:1; x3 D 3:6;x 3:6;x 4 D 4:2; 4:2; x5 D 5g
and C
D D fc1 D 2:5; 2:5; c2 D 3; c3 D 3:5;c 3:5;c 4 D 4; c5 D 4:5g:
Then, 5
R.f;P;C/
X D
xj f .cj /
j 1
D
D 0:7.6:25/ C 0:4.9/ C 0:5.12:25/ C 0:6.16/ C 0:8.20:25/ D 39:9:
Alternately, take N take N
D D 6, P
D D fx0 D 2; x1 D 2:5; 2:5; x2 D 3:5; 3:5; x3 D 4; x4 D 4:25; 4:25; x5 D 4:75; 4:75; x6 D 5g
and C
D D fc1 D 2:1; 2:1; c2 D 3; c3 D 3:7; 3:7; c4 D 4:2;c5 D 4:5; 4:5; c6 D 4:8g:
Then, 6
X D
R.f;P;C/
xj f .cj /
j 1
D
D 0:5.4:41/ C 1.9/ C 0:5.13:69/ C 0:25.17:64/ C 0:5.20:25/ C 0:25.23:04/ D 38:345: In Exercises Exercises 13–16, 13–16, express express the the limit as an an integral integral (or (or multiple of an integr integral) al) and evaluat evaluate. e. 13.
lim
N
X
!1 6N j D1
N
sin
Let f Let f .x/
SOLUTION
j C 6N 3
D sin x and N and N be a positive integer. integer. A uniform partition of the interval Œ interval Œ=3; =2 with N with N subintervals subintervals has x
for 0 for 0
j
D 6N
j D 3 C 6N
xj
and
N . N . Then N
6N
X
sin
j 1
D
C 3
j 6N
N
X
D x
f .xj /
j 1
D
D RN I
consequently, lim
N
X
!1 6N j D1
N
3 14. lim N !1 N SOLUTION
N 1
X
10
k 0
D
Let f Let f .x/
C
3k N
sin
C 6N 3
j
=2
=2
Z D
sin x dx
=3
D cos x
ˇˇˇ
=3
D 0 C 12 D 12 :
D x and N and N be a positive integer. A uniform partition of the int erval Œ10 erval Œ10;; 13 with N with N subintervals subintervals has x
for 0 for 0
j
D N 3
and
D 10 C 3j N
xj
N . N . Then 3 N
N 1
X
10
k 0
D
C
3k N
N 1
D x
X
j 0
D
f .xj /
D LN I
670
THE INTEGRAL
CHAPTER 5
consequently, N 1
3 lim N !1 N
X
10
k 0
D
C
3k N
13
Z D
x dx
10
D
1 2 x 2
13
ˇˇ ˇ
10
100 69 D 169 D : 2 2 2 15.
lim
N
!1
5 N
N
X p C 4
j 1
D
.x/ Let f Let f .x/
SOLUTION
5j=N
D p x and N and N be a positive integer. A uniform partition of the interval Œ4 interval Œ4;; 9 with N with N subintervals subintervals has D N 5
x
for 0 for 0
j
D 4 C 5jN
xj
and
N . N . Then 5 N
N
N
X p C 4
5j=N
D D x
j 1
D
X
f .xj /
j 1
D
D RN I
consequently, 5 lim N !1 N 16.
lim
1k
C 2k C C N k N k C1
N
!1
N
9
Z D D p
X p C 4
5j=N
x dx
4
j 1
D
0) (k > 0)
1k
Now, let f let f .x/ .x/
"
C 2k C 3k C C N k D 1 N N k C1
1 N
k
C
4
54 3
C
#
16 3
D 383 :
3 N
k
C
N N
k
D N 1
N
X
j 1
D
j N
D N j
xj
and
N . N . Then 1 N
N
X
j 1
D
j N
N
k
D x
X
f .xj /
j 1
D
D RN I
consequently, 1 lim N !1 N
N
X
j 1
D
j N
k
1
Z D
k
x dx
0
1
D k C 1x
k 1
C
1
ˇˇ D ˇ 0
1
k
C 1:
In Exercises Exercises 17–20, 17–20, use the given substitutio substitution n to evaluate evaluate the the integral. integral. 2
17.
Z 0
dt 4t
SOLUTION
D 4t C 12 Let u Let u D 4t C 12. 12. Then du Then du D 4dt 4d t , and the new limits of integration are u are u D 12 and 12 and u u D 20. 20. Thus,
C 12 ,
u
2
Z 0
18.
Z
.x 2 .x 3
SOLUTION
k
D xk and N and N be a positive integer. A uniform partition of the interval Œ0 interval Œ0;; 1 with N with N subintervals subintervals has D N 1
j
k
2 N
x
9
ˇˇˇ D ˇ
Observe that
SOLUTION
for 0 for 0
D
2 3=2 x 3
dt
D 4t C 12
1 4
20
Z
12
du u
D
20
ˇˇˇ
1 ln u 4
12
1 5 D 14 .ln 20 ln 12/ D 14 ln 20 D ln : 12 4 3
C 1/dx , u D x 3 C 3x C 3x/4 Let u Let u D x 3 C 3x. 3x . Then d Then du u D .3x 2 C 3/dx D 3.x 2 C 1/dx and 1/dx and .x 2 C 1/dx 1 1 1 3 3 D D C C: u 4 du D u 3 C C D .x C 3x/ 3 4 3 9 9 .x C 3x/
Z
Z
:
Chapter Chapter Review Review Exercise Exercises s =6
19.
Z
sin x cos4 x dx ,
D cos x p Let u Let u D cos x . Then du Then du D sin x dx and the new limits of integration are u are u D 1 and u and u D 3=2. 3=2. Thus,
0
SOLUTION
u
=6
Z
sin x cos4 x dx
0
3=2
D
D 15
Z
, sec2 .2 / tan.2 tan.2 / d
1
1
9 3 32
:
u
D D 12
Z
sec2 .2 / tan.2 tan.2 / d
In Exercises Exercises 21–70, 21–70, evaluate evaluate the the integral. integral.
Z
Z
u du
D 14 u2 C C D D 14 tan2 .2 / C C:
.20x 4 9x 3 2x/dx
Z
.20x 4 9x 3 2x/dx
SOLUTION
2
22.
3=2
D tan.2 tan.2 / and Let u Let u D tan.2 tan.2 /. Then du Then du D 2 sec2 .2 / d and
SOLUTION
21.
u4 du
1
1 5 u 5
D
20.
Z p ˇˇˇp ˇ p !
Z
D 4x5 94 x 4 x 2 C C . C .
.12x 3 3x 2 / dx
0
2
Z
SOLUTION
.12x
3
2
3x / dx
D .3x
0
23.
Z
3
Z
.2x
1
Z
2
2
ˇˇˇ D
x /
.4x 4 12x 3
3x/ dx
D 40. 40.
C 9x2 / dx D 45 x 5 3x4 C 3x3 C C . C .
Z D
2
.48 8/ 0
0
.2x 2 3x/ 2 dx
SOLUTION
24.
4
.x 7=3 2x 1=4 / dx
0
1
Z
SOLUTION
.x
7=3
2x
1=4
/ dx
D
0
25.
x
Z
5
C 3x
Z
3
26.
Z
4
r
3 10=3 8 5=4 x x 10 5
4
dx
x2
SOLUTION
x5
C 3x4 dx D x2
Z
.x 3
1
ˇˇ
3 10
ˇD 0
8 5
D 13 . 10
C 3x2 / dx D 14 x 4 C x 3 C C . C .
dr
1
3
SOLUTION
Z
r
4
1
3
27.
Z j
dr
1 r 3
D
x 2 4 dx
3
j
3
3
ˇˇ D ˇ 1
1 3
1 27
1
D 26 . 81
SOLUTION
3
Z j 3
x
2
2
4
j dx
Z D
.x
2
2
Z C
4/dx
3
D D
D 463 :
.4 x / dx 2
.x 2 4/dx
2
2
Z C
2
ˇˇ ˇˇ C C ˇ ˇ
1 3 x 4x 3
16 3
3
2
3
C
4x
3
16 3
1 3 x 3
C 163 C
3
2
C 163
1 3 x 4x 3
3
ˇˇ
ˇ
2
671
672
THE INTEGRAL
CHAPTER 5
4
28.
Z j
.x 1/.x 3/ dx
j
2
SOLUTION
4
1
Z j
.x 1/.x 3/ dx
j
2
Z D
.x
3
2
4x
C 3/dx
2
D
1 3 x 2x 2 3
D 43
50 3
Z C 1
1
C 3x
.x
ˇˇˇ C 2
C0
2
4 3
4
C 4x 3/dx
1 3 x 3
C 2x
2
Z C ˇˇˇ C
.x 2 4x
3
3
3x
1
C 43 0
C 3/dx
1 3 x 2x 2 3
C 3x
D 623 : 3
29.
Z
Œt dt
1
SOLUTION
3
Z
2
Œt dt
1
2
30.
Z
.t
Z D
3
Z C
Œt dt
1
2
Œt dt
2
2
3
Z Z D C dt
1
2 dt
2
ˇ ˇ D ˇ C t
3
2t
1
Œt /2 dt
ˇˇ D ˇ 2
.2 1/
C .6 4/ D 3:
0
SOLUTION
2
Z
.t Œt /2 dt
0
1
Z
D
t 2 dt
0
1 3 t 3
D
1
ˇˇ C ˇ 0
2
C
Z
1 .t 3
.t
31.
.10t
3
1/
.10t
3
32.
2 1
D 10t 7. Then du 10dt and Then du D 10dt and
Z
Z p
ˇˇ ˇ
7/14 dt
Let u Let u
SOLUTION
1/2 dt
1
D 13 C 13 D 23 :
Z
14
7/
dt
D
1 10
Z
u14 du
1 15 1 D 150 D 150 u C C D .10t 7/15 C C:
7y 5 dy
2
SOLUTION
Let u Let u
D 7y 5. Then d Then du u D 7dy and 7dy and when y when y D 2, u D 9 and when y when y D 3, u D 16. 16. Finally, 3
Z p
7y 5 dy
2
33.
.2x 3 .3x 4
Z
SOLUTION
1
3
1=2
u
du
9
D
1 2 3=2 u 7 3
16
ˇˇ ˇ
9
D 212 .64 27/ D 74 : 21
Z
x dx 5/2
.x 2
SOLUTION
16
Z
C 3x/dx C 9x2/5 Let u Let u D 3x 4 C 9x 2 . Then d Then du u D .12x 3 C 18x/dx D 6.2x 3 C 3x/dx and 3x/dx and .2x 3 C 3x/dx D 16 u 5 du D 241 u 4 C C D D 241 .3x4 C 9x2 / 4 C C: .3x 4 C 9x 2 /5
Z
34.
Z
D
1 7
C
Let u Let u
D x 2 C 5. Then du Then du D 2x dx and dx and 1
Z
3
x dx 2 .x 5/2
1 2
6
Z
2
1 u 2
1
C D 14 u du D D 12 16 141 D 211 :
ˇˇ ˇ
6 14
ˇˇˇ
ˇ
4 3
Chapter Chapter Review Review Exercise Exercises s 5
35.
Z
p
15x x
C 4 dx Let u Let u D x C 4. Then x Then x D u 4, du D dx and dx and the new limits of integration are u are u D 4 and u and u D 9. Thus, 5 9 p p 15x x C 4 d x D 15.u 4/ u du
0
SOLUTION
Z
Z
0
36.
Z
t2
4
Let u Let u
D 15
.u3=2 4u1=2 / du
4
2 5=2 u 5
D 15 D 506:
p t C 8 dt
SOLUTION
9
D 15
Z
486 5
D t C 8. Then du Then du D dt , t D u 8, and p p t 2 t C 8 dt D .u 8/2 u du D
Z
Z
ˇˇ ˇ
8 3=2 u 3
72
Z
.u5=2
9 4
64 5
3=2
16u
64 3
C 64u1=2 / du
D 27 u7=2 325 u5=2 C 128 u3=2 C C 3 D 27 .t C 8/7=2 325 .t C 8/5=2 C 128 .t C 8/3=2 C C: 3
1
37.
Z
cos
0
.t 3 1
SOLUTION
Z
cos
0
38.
Z
C 2/
sin
=2
SOLUTION
5 6
3
d t
.t
C 2/
d t
D
3 sin .t 3
ˇˇ
1
C 2/ D
ˇ
d
0
p
3 3 . 2
Let u
D 5 6
D 56 d :
du
so that
Then
Z
sin
=2
39.
Z
40.
5 6
d
D D
6 5
2=3
Z
sin u du
=4
D D
6 5
1 2
2
ˇˇ ˇ
6 cos u 5
3
=4
p
2 2
!
D 35 .1 C
p
2/:
t 2 sec2 .9t 3
SOLUTION
Z
C 1/dt Let u Let u D 9t 3 C 1. Then d Then du u D 27t 2 dt and
Z
2
2
t sec .9t
3
C 1/dt D
1 27
Z
sec2 u du
D 271 tan u C C D D 271 tan.9t tan .9t 3 C 1/ C C:
sin2 .3 / cos.3 cos.3 / d
SOLUTION
Let u Let u
D sin.3 and sin.3 /. Then du Then du D 3 cos.3 cos.3 /d /d and
Z
2
sin .3 / cos.3 cos.3 / d
D D
1 3
Z
u2 du
D 19 u3 C C D D 19 sin3 .3 / C C:
673
674
41.
Z
csc2 .9 2 / d
Let u Let u
SOLUTION
42.
Z
D 9 2 . Then du and Then du D 2 d and
Z
p
D D 12
csc2 .9 2 / d
Z
=3
sin cos2=3
0
Z
d
Z
Z
0
sec2 t d t .tan t
45.
2x
3
Z
1
sec2 t dt .tan t 1/2
Z D
2
u
du
D u 1 C C D D tan t1 1 C C:
dx
D 9 2x. 2x . Then du Then du D 2 dx, dx , and
Z e4x
D tan t 1. Then du Then du D sec2 t d t and
Let u Let u
SOLUTION
46.
1
Z e9
3
ˇˇ ˇ
1/2
Let u Let u
SOLUTION
Z
D 12 cot u C C D D 12 cot.9 cot.9 2 / C C:
D cos . Then du D 0, u D 1 and when D D 3 , u and Then du D sin d d and when D , u D 12 . Finally, p 1=2 =3 sin 1=2 3 4 2=3 1=3 1=3 D D 3.2 1/ D 3 2 : D d u du D 3u cos2=3
Let u Let u
SOLUTION
44.
csc2 u du
D 4 cos . Then du and Then du D sin d d and p D u1=2 du D 23 u3=2 C C D D 23 .4 cos /3=2 C C: D sin 4 cos d d
Let u Let u
Z Z
Z
sin 4 cos d d
SOLUTION
43.
THE INTEGRAL
CHAPTER 5
3
e
92x
dx
D
1 2
Z
e u du
D 12 eu C C D D 12 e9
2x
C C:
dx
1
3
Z
SOLUTION
4x 3
e
dx
1
47.
Z
x2e
x3
Z
ex
ex
3
ˇˇ D ˇ
1 9 .e e/. e/. 4
1
D x 3 . Then du Then du D 3x 2 dx, dx , and
Z
ln 3
48.
3
dx
Let u Let u
SOLUTION
D
1 4x e 4
2 x3
x e
dx
1 3
D
Z
e u du
D 13 eu C C D D 13 ex C C: 3
dx
0
SOLUTION ln 3
u
De D
Note e Note e x 3. Thus,
ex
ln 3
Z
e
D ex e
x e x
49.
dx
Z
Z D
x
e e
ex
0
D ex . Then d Then du u D e x dx , and the new limits of integration are u are u D e0 D 1 and 3
dx
Z D
e
u
du
1
D e
e x 10x dx
SOLUTION
50.
Now, let u let u ln 3
0
Z
ex .
e
2x
SOLUTION
Z
e x 10x dx
sin.e sin.e
2x
Let u Let u
D
Z
.10e/x dx
x
x
t
3
ˇˇˇ D ˇ
.e
3
e
1
/
1
De
1
x x
.10e/ .10e/ 10 e D ln.10e/ C D C D C C . C D C D C . ln.10e/ ln 10 C ln e ln 10 C 1
/ dx
De
2x . Then du Then du
Z
e
2x
D 2e
sin e
2x
2x dx , and
d x
D 12
Z
sin u du
D cos2 u C C D D 12 cos
C e
2x
C:
e
3
:
Chapter Chapter Review Review Exercise Exercises s
51.
Z
e x dx .e x 2/3
C
Let u Let u
SOLUTION
52.
Z
D e x C 2. Then du Then du D e
Z 2
sin cos cos e cos Let u Let u
SOLUTION
Z
e x dx .e x 2/3
dx and dx and
C D
Z
3
u
D 2u12 C C D D 2.e x1C 2/2 C C:
du
C1 d
D cos2 C 1. Then du and Then du D 2 sin cos cos d d and
Z
=6
53.
x
sin cos cos e
cos2
C1 d D 1 D 2
Z
D 12 eu C C D D 12 ecos
2
e u du
C1 C C:
d tan 2 d
0
=6
Z
SOLUTION
tan 2 d d
D D
0
2=3
54.
Z
cot
=3
ˇjˇ ˇ
1 ln sec 2 2
j
1 d 2
=6
0
D 12 ln 2.
SOLUTION
2=3
Z
cot
=3
55.
Z
t .1
C
Let u Let u
e
57.
1
Let u Let u
ln x dx x Let u Let u
SOLUTION
dt t .1
2
C .ln t / /
Z
Z D
cos. cos.ln x/ dx x
1
du
3 2
1
C u2 D tan
1
Z D
cos u du
ln
1 2
!
6
D ln 3:
u
1
C C D D tan
.ln t /
C C:
Z D 0
x ln x
SOLUTION
Let u Let u
Z p D Z x ln x
4x 2
C9
u du
D
1 2 u 2
1
ˇˇˇ D 0
1 : 2
D ln x. Then du Then du D x1 d x , and dx
dx
D sin u C C D D sin. sin.ln x/ C C:
1
ln x dx x
dx
Z p Z
ln
3
p
lnsin
D ln x. Then du Then du D dx new limits of integration are are u u D ln 1 D 0 and u and u D ln e D 1. Thus, x and the new e
59.
D2
lnsin
=3
D ln x. Then du Then du D dx x , and
Z 58.
D2
3
cos. cos.ln x/dx x
SOLUTION
Z
D D 2 ln sin 2
2
ˇˇˇ ˇˇ ˇ
D ln t . Then, d Then, du u D 1t dt and
Z 56.
dt .ln t /2 /
SOLUTION
Z
ˇˇˇ
1 d 2
u
1=2
du
p D 2p u C C D D 2 ln x C C:
675
676
THE INTEGRAL
CHAPTER 5
D 2x3 . Then x Then x D 32 u, dx D 32 d u, and
Let u Let u
SOLUTION
dx
Z 0:8
60.
0:8
Z
SOLUTION
p
12
p dx2
x x
3
62.
0
D sin 2
p 2
x x
1
C
Let u Let u
Z
3
1
C 1 D
u2
C9 Let u Let u
u
1 tan 6
1
C C D D
2x 3
C C:
3
0
dx
D sec
12
1
x
ˇˇˇ
1
D sec
4
Let u Let u
1
4:
12 sec
x dx x2 9
1 2
18
Z 9
du u
18
ˇˇˇ
1 ln u 2
D
dx
D x2 C 9
1 3
1
Z
dt
D t2 C 1
0
1 tan 3
1
1
D sin
0:8. 0:8.
D 12 .ln 18 ln 9/ D 12 ln 189 D 12 ln 2:
9
1
t
ˇˇ D ˇ 0
e 2x 1
SOLUTION
0
D x3 . Then du Then du D dx are u D 0 and u and u D 1. Thus, 3 , and the new limits of integration are u
Z Z p
1
0:8 sin
dx x2
SOLUTION
64.
1
D sin
0
C D
0
0
1 tan 6
D x 2 C 9. Then du Then du D 2x dx, dx , and the new limits of int egration are u are u D 9 and u and u D 18. 18. Thus, 3
63.
x
ˇˇˇ
x dx x2 9
SOLUTION
Z
0:8
1
dx
4
Z
du
1
12
Z
SOLUTION
dx
1x
0
4
C9 D
Z
1 x2
0
61.
C9
1 6
p dx
Z Z
4x 2
3 2 d u 9 2 4 4u
Z D
1 .tan 3
1
1
1 tan
0/
D 13
4
D
0
D ex . Then du
D ex dx )
du
D u dx )
u
1
du
D dx
By substitution, we obtain
Z p
dx
e 2x
D
1
du
Z p
u u2 1 1
65.
D sec
x dx
Z p
1 x4
SOLUTION
Let u Let u
D x 2 . Then du Then du D 2x dx, dx , and x dx
Z p
1x
1
66.
Z 0
D 4
Let x Let x
0
dx
D 25 x 2 D
0
.e x /
C C
p
1 u2 . Thus,
1 D sin 2 2
1
1u
u
C C D D 12 sin
1
.x 2 /
C C:
D 5u. 5u. Then dx Then dx D 5 du, du , and the new limits of i ntegration are u are u D 0 and u and u D 15 . Thus, 1
4
du
Z p
D
1
25 x 2
Z Z
1 x4
C C D D sec
dx
SOLUTION
67.
1 2
p
u
dx 2x 2 1
C
1 25
1=5
Z
D 1 u2
0
1 tanh 5
5 du
1
1=5
ˇˇˇ
u
0
D
1 5
5 25
1=5
Z
1 u2
0
11
tanh
du
5
1
tanh
0
D 15 tanh
11
5
:
12
:
Chapter Chapter Review Review Exercise Exercises s
Let u Let u
SOLUTION
p p D p 2x. 2x . Then du Then du D 2 dx , and the new limits of int egration are u are u D 0 and u and u D 4 2. Thus, 4
Z 0
2x 2
p p Z D 4
dx
C1
1 d u 2 u2 1
2
0
1
D p tan
1
2
8
68.
Z
p dx2
x x
5
8
Z 1
69.
Z
2
p
4
ˇˇˇ
4
2
dx
p
D
2
1 4
du
Z
p
u u2 1
5=4
1 sec 4
D
ˇˇ
1
C1 p
1
.4 2/ tan
ˇ
1 x/ 3 dx
C x2 Let u Let u D tan
2
u
.tan
1
tan
2
0
du u2
0
D p 1
u
2
1
D p
0
1
2
tan
p
.4 2/:
D 4u. 4u. Then dx Then dx D 4 du, du , and the new limits of integration are u are u D 54 and u and u D 2. Thus,
x x 2 16
5
C
p Z D p 1
16
Let x Let x
SOLUTION
677
5=4
D
1 4
1
sec
1 5
2 sec
4
1 4
D
3
1 5
sec
4
:
1
0
1 x.
SOLUTION
Then du
D 1 C1 x 2 d x
and 1
Z 0
1 tdt
=4
1 x/ 3 dx
.tan 1
Z D
C x2
3
u du
0
1 4 u 4
D
70.
cos
Z p
SOLUTION
1 t2 Let u Let u
D p 11 t
1 t . Then d Then du u
D cos
1t
1 t2 71. Combine to write as a single integral: integral:
dt
Z
D
4
C
1
0
f.x/dx
0
SOLUTION
0
4
D 12 u2 C C D D 12 .cos
u du
8
Z
D 14
4
D 1024 :
d t , and
cos
Z p
2
=4
ˇˇ ˇ
Z
2
t/
C C:
6
f.x/dx 2
C
Z
f.x/dx
8
First, rewrite 8
6
Z
f.x/dx
0
Z D
8
f.x/dx
0
Z C
f.x/dx
6
and observe that 6
Z
8
f.x/dx
8
D
Z
f.x/dx:
6
Thus, 8
Z
6
f.x/dx
0
Z C
6
f.x/dx
8
Z D
f.x/dx:
0
Finally, 8
Z D R
0
f.x/dx
C
0 x f.x/dx , 0 f.x/dx,
Z
6
f.x/dx 2
C
Z 8
6
f.x/dx
D
Z 0
0
f.x/dx
C
Z
6
f.x/dx 2
D
Z
f.x/dx: 2
.x/ is the function shown in Figure 4. Identify the location of the local minima, the local 72. Let A.x/ Let A.x/ where f where f .x/ maxima, and points of inflection of A.x/ on A.x/ on the interval Œ0; Œ0; E , as well as the intervals where A.x/ is A.x/ is increasing, decreasing, concave up, or concave down. Where does the absolute max of A.x/ of A.x/ occur? occur?
678
THE INTEGRAL
CHAPTER 5
y
x A
B
C
D
E
FIGURE 4 SOLUTION
Let f Let f .x/ .x/ be the function shown in Figure 4 and define x
A.x/
D
Z
f.x/dx:
0
Then A Then A 0 .x/ f.x/ and f.x/ and A A 00 .x/ f 0 .x/. .x/ . Hence, A.x/ Hence, A.x/ is is increasing when f when f .x/ .x/ is positive, is decreasing when f .x/ .x/ is negative, is concave up when f when f .x/ .x/ is increasing and is concave down when f when f .x/ is x/ is decreasing. Thus, A.x/ is A.x/ is increasing for 0 for 0 < x < B , is decreasing for B for B < x < D and for D for D < x < E , has a local maximum at x at x B and no local minima. Moreover, A.x/ Moreover, A.x/ is is concave up for 0 for 0 < x < A and A and for C for C < x < D, D , is concave down for A < x < C and and for D for D < x < E , and has a point of inflection at x A, x C and and x D . The absolute maximum value for A.x/ for A.x/ occurs occurs at x at x B.
D
D
D
D
D
D A.x/ D 73. Find the local minima, the local maxima, and the inflection points of A.x/
SOLUTION
D
x
Z 3
tdt t2
Let x
A.x/
Z D 3
C 1.
t dt t2
C 1:
Then A0 .x/
D x 2 xC 1
and A00 .x/
2 1 x2 x.2x/ D .x C.x1/.1/ D : 2 C 1/2 .x 2 C 1/2
Now, x Now, x 0 is the only critical point of A; A; because A because A 00 .0/ > 0, 0, it follows that A that A has has a local minimum at x at x 0. There are no local maxima. Moreover, A.x/ Moreover, A.x/ is is concave down for x > 1 and 1 and concave up for x < 1. 1 . A.x/ therefore A.x/ therefore has inflection points at x at x 1.
D
jj
74. A particle starts at the origin at time t time t
D
jj
D ˙
D 0 and moves with velocity v.t/ velocity v.t/ as as shown in Figure 5.
(a) How many times does does the particle return to the origin in t he first 12 seconds? (b) What is the particle’s maximum maximum distance from the origin? (c) What is particle’s maximum maximum distance to the left of the origin? v (t ) m/s
4 2 5 !
2
!
4
10
t (s)
FIGURE 5 SOLUTION
Because the particle starts at the origin, the position of the particle is given given by t
s.t/
Z D 0
v. / d
I
that is by the signed area between the graph of the velocity and the t -axis over the interval Œ0 interval Œ0;; t . Using the geometry in Figure 5, we see that s.t/ that s.t/ is is increasing for 0 for 0 < t < 4 and 4 and for 8 for 8 < t < 10 and 10 and is decreasing for 4 for 4 < t < 8 and 8 and for 10 for 10 < t < 12. 12 . Furthermore, s.0/
D 0 m; m;
s .4/
D 4 m; m;
s .8/
D 4 m; m;
s .1 .10/
D 2 m; m;
between t (a) In the first 12 seconds, the particle returns to the origin once, sometime between t (b) The particle’s maximum distance from the origin is 6 meters (to the left at t (c) The particle’s distance to the left of the origin is 6 meters.
and s.12/
D 6 m: m:
D 4 and t and t D 8 seconds.
D 12 seconds). 12 seconds).
Chapter Chapter Review Review Exercise Exercises s
679
75. On a typical day, a city consumes water at the rate of r.t/ of r.t/ 100 72t 3t 2 (in thousands of gallons per hour), where t is the number of hours past midnight. What is the daily water consumption? How much water is consumed between 6 P M and midnight?
D
SOLUTION
r.t/ With a consumption consumption rate of r.t/ 24
Z
.100
0
C 72t 3t
2
D 100 C 72t 3t 2 thousand gallons per hour, the daily consumption of water is
D
/ dt
C
100t
C 36t
2
t
3
24
ˇˇˇ
D 100.24/ C 36.24/2 .24/3 D 9312;
0
or 9.312 million gallons. From 6 PM to midnight, the water consumption is 24
Z
.100
18
C 72t 3t 2 / dt D
C 36t 2 t 3
100t
24
ˇˇ
18
D 100.24/ C 36.24/2 .24/3 100.18/ C 36.18/2 .18/3 D 9312 7632 D 1680;
or 1.68 million gallons. 12x 1=5 (in hours per bicycle), which means that it takes a bike 76. The learning curve in a certain bicycle factory is L.x/ mechanic L.n/ mechanic L.n/ hours hours to assemble the nth n th bicycle. If a mechanic has produced 24 bicycles, how long does it take her or him to produce the second batch of 12?
D
SOLUTION
The second batch of 12 bicycles consists of bicycles 13 through 24. The time it takes to produce these bicycles bicycles is 24
Z
12x
1=5
dx
13
D 15x
4=5
ˇˇˇ
24
D 15244=5 134=5 73:91 hours 73:91 hours::
13
P of major space projects. It has been found that the cost C of C of 77. Cost engineers at NASA have the task of projecting the cost P of 0:65 developing a projection increases with P with P at at the rate dC rate dC =dP =dP 21P , where C where C is in thousands of dollars and P and P in in millions of dollars. What is the cost of developing a projection for a project whose cost turns out to be P be P $35 million? 35 million?
D D
SOLUTION Assuming it costs nothing to develop develop a projection for a project with a cost of $0, the cost of developing developing a projection for a project whose cost turns out to be $35 million is
35
Z
0:65
21P
0
0:35
dP
D D 60P
or $208,245.
ˇˇˇ
35
D 60.35/0:35 208:245;
0
m , per degree-squared of sky, is 78. An astronomer estimates that in a certain constellation, the number of stars per magnitude m, 6 7:4 equal to A.m/ 2:4 10 m (fainter stars have higher magnitudes). Determine the t otal number of stars of magnitude between 6 and 15 and 15 in in a one-degree-squared one-degree-squared region of sky. sky.
D
SOLUTION
The total number number of stars of magnitude between 6 between 6 and and 15 15 in in a one-degree-squared region of sky is 15
Z
15
A.m/dm
6
Z D
8
79. Evaluate
2 7
6
10
8:4
m
2162
x 15 dx
ˇˇˇ
15 6
C cos2 x , using the properties of odd functions. Let f Let f .x/ D x and note that 3Ccos x 8
SOLUTION
m7:4 d m
6
D
Z
6
2:4 10
3
15
2
15
f .x/
Because f Because f .x/ .x/ is an odd function and the interval 8 x
8 is symmetric about x about x
8
Z
8
1
R
15
D 3 C.cosx/2 .x/ D cosx 2 x D f.x/: D 0, it follows that
x 15 dx 3
C cos2 x D 0:
f.x/dx , assuming that f .x/ is an even continuous function such that 80. Evaluate 0 f.x/dx, that f .x/ 2
Z 1
1
f.x/dx
D 5;
Z
f.x/dx 2
D 8
680
CHAPTER 5
SOLUTION
THE INTEGRAL
Using the given information 2
Z
1
f.x/dx 2
Z D
2
Z C
f.x/dx 2
f.x/dx
1
D 13:
.x/ is an even function, it follows that Because f Because f .x/ 0
Z
2
f.x/dx 2
f.x/dx;
0
so
Z D
2
Z
f.x/dx
0
D 132 :
Finally, 1
2
2
D 132 5 D 32 : 0 0 1 f .x/ D sin mx sin mx sin nx on nx on Œ0; Œ0; for the pairs .m;n/ .2;4/, .3; .3; 5/ and 5/ and in each case guess the value 81. Plot the graph of f pairs .m;n/ D .2;4/, D 0 f.x/dx. of I I D f.x/dx . Experiment with a few more values (including t wo cases with m with m D n) and formulate a conjecture for when I when I
Z
f.x/dx
Z D
f.x/dx
Z
f.x/dx
is zero.
R
The graphs of f f .x/ .x/ sin mx sin mx sin nx with nx with .m;n/ .m;n/ positive areas balance the negative areas, so we expect that
D .2;4/ and .2;4/ and .m;n/ .m;n/ D .3;5/ are .3;5/ are shown below. It appears as if the
D
SOLUTION
Z D D
I
f.x/dx
0
D0
in these cases. y
y
(2, 4)
(3, 5)
0.5
0.5 x
0.5
1
1.5
2
2.5
x
3
0.5
0.5
!
!
We arrive at the same conclusion for the cases .m;n/
1
1.5
2
2.5
3
0.5
D .4;1/ and .4;1/ and .m .m;; n/ D .5;2/. .5;2/.
y
y
(4, 1)
(5, 2)
0.5
0.5 x
0.5
1
1.5
2
2.5
x
3
0.5
!
However, when .m;n/ when .m;n/
!
0.5
1
1.5
2
2.5
3
0.5
1
1.5
2
2.5
3
0.5
D .3; .3; 3/ and 3/ and when .m when .m;; n/ D .5; .5; 5/, 5/, the value of
Z D D
I
f.x/dx
0
is clearly not zero as there is no negative area. y
y
0.5
0.5 x
0.5 !
1
1.5
2
2.5
0.5
x
3 !
(3, 3)
We therefore conjecture that I that I is is zero whenever m whenever m
0.5 (5, 5)
¤ n.
Chapter Chapter Review Review Exercise Exercises s
82. Show that
Z
x f .x/dx
D xF.x/ G.x/
Z
where F where F 0 .x/
D f.x/ and f.x/ and G G 0 .x/ D F.x/. F.x/. Use this to evaluate SOLUTION Suppose F Suppose F 0 .x/ D f.x/ and f.x/ and G G 0 .x/ D F.x/. F.x/. Then d .xF.x/ G.x// dx
x cos x dx .
D xF 0 .x/ C F.x/ G0 .x/ D xf.x/ C F.x/ F.x/ D xf .x/:
Therefore, xF.x/ Therefore, xF.x/ G.x/ is G.x/ is an antiderivative antiderivative of xf xf .x/ and .x/ and
Z
xf.x/dx
R
x cos x dx , note that f .x/ To evaluate x cos that f .x/
D xF.x/ G.x/ C C:
D cos x. Thus, we may take F.x/ take F.x/ D sin x and G.x/ and G.x/ D cos x . Finally,
Z
83. Prove 2
Z
2
x cos x dx
x
2 dx
D x sin x C cos x C C:
4
and
1
SOLUTION
The function f function f .x/
2
1 9
Z
x
3
dx
1
1 3
D 2x is increasing, so 1 so 1 x 2 implies that 2 that 2 D 21 2x 22 D 4. Consequently, 2
2
2
Z D
2 dx
1
On the other hand, the function f function f .x/
x
D3
Z
2
x
2 dx
Z
1
is decreasing, so 1 so 1 x 1 9
2
D 3
4 dx
1
x
3
2 implies that
1
3
D 4:
D 13 :
It then follows that 1 9 Plot the graph of f f .x/ .x/
84. SOLUTION
Let f Let f .x/
Dx
Dx
2 sin x .
2
D
Z 1
1 dx d x 9
2
Z
x
3
2
dx
1
Z 1
2 sin x , and show that 0:2 that 0:2
1 dx d x 3
D 13 :
2
Z
f.x/dx
0:9. 0:9.
1
From the figure below, we see that 0:2 f.x/ 0:9
for 1 for 1
x
2. Therefore, 2
0:2
D
Z
2
0:2 0:2 dx
1
Z
2
f.x/dx
1
Z
0:9 0:9 dx
1
y
1 0.8 0.6
x 2sin x !
0.4 0.2 x
0.5
1
1.5
1
85. Find upper and lower bounds for
Z 0
f.x/dx, f.x/dx , for f .x/ in Figure 6. for f .x/
2
D 0:9:
681
682
THE INTEGRAL
CHAPTER 5
y
f ( x x )
y = x 2 + 1
y = x 1/2 + 1
2
1
x
1
FIGURE 6
From the figure, we see that that the inequalities x inequalities x 2
SOLUTION
1
Z
.x
2
0
C 1 f.x/ p x C 1 hold for 0 for 0 x 1. Because
C 1/dx D
1 3 x 3
1
Cx
and 1
Z p C . x
1/dx
0
D
2 3=2 x 3
it follows that 4 3
f.x/dx
0
4 3
ˇD ˇˇ C ˇ D
1
Z
ˇˇˇ
0
1
x
0
5 ; 3
5 : 3
In Exercises Exercises 86–91, 86–91, find the the derivative. derivative. x
86.
Z Z D Z D Z D
A0 .x/, .x/ , where A.x/ where A.x/ D Let A.x/ Let A.x/
SOLUTION
3 x
sin.t sin.t 3 / dt
sin.t sin.t 3 / dt . Then A Then A0 .x/
3
x
87. A0 . /, where A.x/ where A.x/
Let A.x/ Let A.x/
SOLUTION
2 x
2
D sin.x sin.x 3/.
cos t dt 1 t
C
cos t dt . Then A Then A 0 .x/ 1 t
C
D 1cosC xx and
A0 . / d 88. dy
y
Z
3x dx
2
SOLUTION
89.
D 1cosC D 1 C1 :
d dy
y
Z
3x dx
D 3y :
2
sin x
Z Z D
G 0 .x/, .x/ , where G.x/ where G.x/ D
sin x
SOLUTION
Let G.x/ Let G.x/
t 3 dt
2
t 3 dt . Then
2
G 0 .x/
d D sin3 x dx sin x D sin3 x cos x:
x3
90.
Z p C Z D p C
G 0 .2/, .2/, where G.x/ where G.x/ D
t
1 dt
0
x3
SOLUTION
Let G.x/ Let G.x/
t
1 dt . Then
0
G 0 .x/ and G and G 0 .2/
D 3.2/2 p 8 C 1 D 36. 36.
D
p C x3
1
d 3 x dx
D 3x2
p C x3
1
Chapter Chapter Review Review Exercise Exercises s 9
91.
Z Z D
H 0 .1/, .1/, where H.x/ where H.x/ D
4x 2
9
SOLUTION
Let H.x/ Let H.x/
4x 2
1 dt t
1 dt t
4x 2
D
Z 9
1 dt . Then t
H 0 .x/ and H and H 0 .1/
683
d 8x 2 D 4x12 dx 4x 2 D 2 D x 4x
D 2.
92. Explain with a graph: If f f .x/ .x/ is increasing and concave up on Œa; Œa; b , then L then L N is more accurate than R N . Which is more accurate if f f .x/ is x/ is increasing and concave down? SOLUTION
Consider the figure below, below, which displays a portion of the graph of an increasing, concave up function. y
x
The shaded rectangles represent the differences between the right-endpoint approximation RN and the left-endpoint approximation LN . In particular, the portion of each rectangle that lies below the graph of y of y f.x/ is f.x/ is the amount by which L which LN underestimates the area under the graph, whereas the portion of each rectangle that li es above the graph of y y f.x/ is f.x/ is the amount by which R which RN overestimates the area. Because the graph of y of y f.x/ is f.x/ is increasing and concave up, the lower portion of each shaded rectangle is smaller than the upper portion. Therefore, L Therefore, L N is more accurate (introduces less error) than R N . By similar reasoning, if f if f .x/ .x/ is increasing and concave down, then R then RN is more accurate than L than LN .
D
D
D
b
93.
Explain with a graph: If f f .x/ .x/ is linear on Œa on Œa;; b , then the
Z
f.x/dx
a
SOLUTION
D 12 .RN C LN / for all N all N ..
Consider the figure below, below, which displays a portion of the graph of a linear linear function. y
x
The shaded rectangles represent the differences between the right-endpoint approximation RN and the left-endpoint approximation LN . In particular, the portion of each rectangle that lies below the graph of y of y f.x/ is f.x/ is the amount by which L which LN underestimates the area under the graph, whereas the portion of each rectangle that li es above the graph of y y f.x/ is f.x/ is the amount by which R which RN overestimates the area. Because the graph of y of y f.x/ is f.x/ is a line, the lower portion of each shaded rectangle is exactly the same size as the upper portion. Therefore, if we average L average L N and RN , the error in t he two approximations will exactly cancel, leaving
D
D
b
1 .RN 2
C LN / D
94. In this exercise, we prove
x x
(a) Show that ln.1 ln.1
C x/ D
that 1 t (b) Verify that 1
1
Z 0
x2 2
D
ln.1 ln.1
Z
f.x/dx:
a
C x/ x
.for x for x > 0/
dt 1
for x > 0. C t for x
all t > 0. 0. 1 for all t 1 t (c) Use (b) to prove Eq. Eq. (1). (d) Verify Eq. (1) for x for x 0:5, 0:5, 0.1, and 0.01.
C D
SOLUTION
(a) Let x Let x > 0. 0 . Then x
Z 0
dt
x
ˇˇˇ D ˇ
D ln.1 ln.1 C t / 1Ct 0
ln.1 ln.1
C x/ ln 1 D ln.1 ln.1 C x/:
1
684
THE INTEGRAL
CHAPTER 5
(b) For t For t > 0, 1 Hence,
C t > 1, so 1C1 t
< 1. Moreover, .1 Moreover, .1 t/.1
1t
1 C t / D 1 t 2 < 1. Because 1 Because 1 C t > 0, it follows that 1 that 1 t < 1C t .
1
C t 1: (c) Integrating each expression expression in the result from part (b) from t from t D 0 to t to t D x yields x (d) For x For x
x2 2
1
ln.1 ln.1
C x/ x:
D 0:5, 0:5, x D 0:1 and 0:1 and x x D 0:01, 0:01, we obtain the string of inequalities 0:375 0:405465
0:5
0:095 0:095310
0:1
0:01;
0:00995 0:00995033 respectively. 95. Let
F.x/ Prove that F.x/ that F.x/ and and cosh evaluating evaluating both at x at x 1.
1
p D
x2 1 2
x
x
Z p
t2 1 d t
1
x differ by a constant by showing that they have the same derivative. Then prove they are equal by
D
SOLUTION
Let F.x/
Dx
Then dF dx Also,
d .cosh 1 x/ dx
by a constant:
D p x1
D
2 1
p
x2 1
2
C p x2 x
1
x2 1 2
p
2 x2 1
; therefore, F.x/ therefore, F.x/ and and cosh
D 1. Because F.1/ Because F.1/ D 0 and cosh
t 2 1 dt:
1
1
F.x/
Now, let x let x
x
Z p
p
2
D p x2 x
p
x2 1
1
D p 21 x
1
:
x have the same derivative. We conclude that F.x/ and F.x/ and cosh 1
1
x differ
C C: 1 D 0. Therefore, 1 D 0, it follows that C that C D F.x/ D cosh 1 x: D cosh
x
96. Let f Let f .x/ be a positive increasing continuous function on Œa on Œa;; b , where 0 where 0 region has area
a < b as in Figure 7. Show that the shaded
b
Z
I
D D bf .b/ af .a/
f.x/dx
2
a
y
(b)
y = f ( x x)
(a) x a
b
FIGURE 7
We can construct the shaded shaded region in Figure 7 by taking a rectangle of length b length b and height f.b/ height f.b/ and and removing a rectangle of length a length a and and height f height f .a/ .a/ as well as the region between the graph of y y f.x/ and f.x/ and the x the x-axis -axis over the interval Œa interval Œa;; b . The area of the resulting region is then the area of the large rectangle minus the area of the small rectangle and minus the area under the curve y curve y f.x/; f.x/; that is, SOLUTION
D
D
b
I
D D bf .b/ af .a/
Z a
f.x/dx:
Chapter Chapter Review Review Exercise Exercises s
How can we interpret the quantity I quantity I in in Eq. (2) if a a < b
97.
685
0? Explain with a graph.
SOLUTION We will consider each term on the right-hand side of (2) separately. separately. For convenience, convenience, let I , II , III and IV denote the area of the similarly labeled region i n the diagram below. y f (b)
II I III
f (a) x
IV
a
b
Because b Because b < 0, 0 , the expression b expression bf f .b/ is .b/ is the opposite of the area of the rectangle along the right; that is, bf .b/
D II IV:
Similarly, b
af .a/
and
D III C IV
Z
f.x/dx
a
D I III:
Therefore, b
bf .b/ af .a/
Z
f.x/dx
a
D I III
that is, the opposite of the area of the shaded region shown below. below. y f (b)
f (a) x a
b
98. The isotope thorium-234 has a half-life half-life of 24:5 days. 24:5 days. (a) What is the differential equation equation satisfied by y.t/ by y.t/,, the amount of thorium-234 in a sample at time t time t ? (b) At t At t 0, a sample contains 2 contains 2 kg kg of t horium-234. How much remains after 40 days?
D
SOLUTION
(a) By the equation for half-life,
24:5
D lnk2 ;
k
so
ln 2 D 24:5 0:028 days 0:028 days
1
:
Therefore, the differential equation for y.t/ for y.t/ is is y0 at t (b) If there are 2 kg of thorium-234 at
D 0:028y:
D 0, then y.t/ then y.t/ D 2e 0:028t . After 40 days, the amount of thorium-234 is y.40/ D 2e 0:028.40/ D 0:653 kg 0:653 kg::
In Bat Cave, New New Mexico, archaeologists archaeologists found ancient human remains, remains, including cobs of popratio was approximately 48 approximately 48% % of that found in living matter. Estimate the age of the corn cobs.
99. The Oldest Snack Snack Food? Food?
ping corn whose C SOLUTION
14
12
-to-C
Let t Let t be be the age of the corn cobs. The C 14 to C to C 12 ratio decreased by a factor of e of e
That is: e
0:000121t
D 0:48;
so 0:000121t
D ln 0:48;
and t
1 D 0:000121 ln 0:48 6065:9:
We conclude conclude that the age of the corn cobs is approximately 6065:9 years. 6065:9 years.
0:000121t
which is equal to 0:48 to 0:48..
686
THE INTEGRAL
CHAPTER 5
100. The C 14 -to-C12 ratio of a sample is proportional to the disintegration rate (number of beta particles emitted per minute) that is measured directly with a Geiger counter. counter. The disintegration rate of carbon in a living organism is 15:3 is 15:3 beta beta particles per minute per gram. Find the age of a sample that emits 9.5 beta particles per minute per gram.
Let t Let t be the age of the sample in years. Because the disintegration rate for the sample has dropped from 15:3 beta 15:3 beta 14 12 particles= particles=min per gram to 9.5 beta particles= particles =min per gram and the C to C to C ratio is proportional to the disintegration rate, it follows that SOLUTION
e
0:000121t
9:5 D 15:3 ;
so t
1 9:5 D 0:000121 3938:5: ln 15:3
We conclude that the sample is approximately 3938.5 years old. 101. What is the interest rate if the PV of $50,000 to be delivered in 3 years is $43,000?
Let r denote the interest rate. The present value of $50,000 received received in 3 years with an interest rate rate of r of r is 50;000e Thus, we need to solve SOLUTION
43;000
D 50;000e
3r .
3r
for r for r . This yields
D 13 ln 43 D 0:0503: 50
r Thus, the interest rate is 5.03%.
102. An equipment upgrade costing $1 million will save a company $320,000 per year for 4 years. Is this a good investment if the interest rate is r is r 5%? What is the largest interest rate that would make the investment investment worthwhile? Assume that the savings are received received as a lump sum at the end of each year.
D
With an interest rate of r r
D 5%, the present value of the four payments is $320;000 e 0:05 C e 0:1 C e 0:15 C e 0:2 D $1;131;361:78:
SOLUTION
As this is greater t han the $1 mill ion cost of the upgrade, this is a good investment. To To determine the largest interest rate t hat would make the investment worthwhile, we must solve the equation
C e 2r C e for r for r . Using a computer algebra system, we find r D 10:13%. 10:13%.
320;000 e
r
3r
Ce
4r
D
1;000;000
5000e 0:1t dollars per year for 5 years, 5 years, assuming an 103. Find the PV of an income stream paying out continuously at a rate of of 5000e interest rate of r r 4%.
D D P V D
SOLUTION
5
Z
5000e
0:1t
e
0:04t
5000e
(b)
lim
n
!1
1
C
1 n
SOLUTION
(a)
(b)
(c)
n
1
4 n
4n
1
1 n
3n
1
4 n
!1 C !1 C
lim
n
lim
n
lim
n
!1
C
0:14t
0
C
dt
0
104. Calculate the limit: 4 n (a) lim 1 n!1 n
5
Z D
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AP-5
Chapter 5: The Integral Preparing for the AP Exam Solutions Multiple Choice Quesitons 1) B 6) B 11) C 16) E
2) D 7) A 12) B 17) C
3) E 8) C 13) C 18) D
4) D 9) C 14) C 19) E
5) C 10) A 15) E 20) C
Free Response Questions 1.
a) If v(t ) 0 , then x(t ) will be increasing, so set
5
t 2
6
1 ( sin t ) dt = 3 + 0 2 5 1 1 6 c) ( sin t ) dt 6 ( sin t ) dt + 0 2 2 6
b) 3 +
1 2
sin t 0 . Solution is 0 t
6
and
2
d) When t
, v(t )
1
2
2 2 4 v(t ) , or the speed, is increasing.
2
5
(
6
1 2
sin t ) dt = 2 3
0 and a(t ) cos t
2 2
3
0 . v(t ) is negative and decreasing, so
POINTS: (a) (3 pts) 1)
1 2
sin t 0 ; 1) 0 t
6
; 1)
5 6
t 2
(b) (1 pt) (c) (3 pts) 1) integrates integrates v(t ) over answer to part (a) ; 1) integrates v(t ) over complement; 1) answer (d) (2 pts) 1) v (
) 0 and a( ) 0 ; 1) Answer 4 4
2. a) We need t 64 defined on the interval whose endpoints are 0 and x 2 . Since x 0 for all x, the domain is the entire number line. 3
b) g ( x)
2
6 T hus, since g is continuous at 0, g( x x) is ( x 2 ) 3 64 (2 x) = 2 x x 64 > 0 for x > 0. Thus,
increasing on [0, ) . c)
g ( x) 2 x 6 64 2 x
1 2 x 64 6
6 x 5 . Thus g (0) 16
POINTS: 2 (a) (2 pts) 1) x 0 ; 1) Answer (b) (4 pts) 2) g ( x)
( x 2 ) 3 64 (2 x) Note: 1 pt for
g ( x) 0 ; 1) Answer (c) (3 pts) 2) g ( x) ; 1) g (0)
( x 2 ) 3 64 , 1 pt for chain rule; 1) Sets
AP-5
3. a) g has a local maximum when g ( x) f ( x) changes from positive to negative; this happens when x = 4. 1 b) The maximum maximum occurs occurs either at at a local local maximum, maximum, or at an an end point. point. g (4) 2 4 4 , the area of the 2 triangle; g decreases from 4 to 5, so we only need to check g ( – 3) 3) =
3
2
f ( x)dx =
2
3
f ( x)dx (
0
3
f ( x)dx +
2
0
f ( x)dx ) = (9 4) 5 . The maximum value
of g( x x) is 5. c) The graph of g is concave up when g (= f ) is increasing, that is on ( – 3, 3, 2). POINTS: (a) (3 pts) pts) 1) Identifies Identifies g ( x) f ( x) ; 1) x = 4; 1) justification (b) (4 pts) 1) Evaluates Evaluates g (4) ;1) deals with left end point 1) deals with right end point; 1) answer (c) (2 pts) 1) answer; 2) justification justification
4. a) g (0)
0
1
f (t )dt
1
0 f (t)dt 14
(1)2 4
b) g ( x) exists for all x because f is is continuous. c) g( x) fails to exist at x = 2 and 6 because g ( x) f ( x) and f is is not differentiable at 2 and 6. d) g (0) 4 ; g increases from 0 to 2.
g (2)
g (6)
4
4
; g decreases from 2 to 6.
12 (2)2 74 ; g increases from 6 to 10.
g (10) 74 ( 12 )(4)(4) 74 8 0
g ( x) 0 has three solutions, one each in (0, 2), (2, 6), and (6, 10). POINTS: (a) (1 pt) Answer (b) (2 pts) pts) 1) Answer; Answer; 1) f is is continuous (c) (3 pts) pts) 1) g f ; 1) x = 2 and 6; 1) f not not differentiable (d) (3 pts) 1) Finds g(2) and g(6); 1) Finds g(0) and g(10); 2) Uses sign changes of g
