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INPS MCA ENTRANCE EXAM
Definition. d [ ( x )] f (x ) . Then the dx
Let (x ) be the primitive or anti-derivative of a function f(x) defined on [a, b] i.e., definite integral of f(x) over [a, b] is denoted by
b
f (x )dx a
and is defined as [ (b) (a)] i.e.,
b
f (x )dx (b) (a) . a
This is also called Newton Leibnitz formula. The numbers a and b are called the limits of integration, ‘a’ is called the lower limit and ‘b’ the upper limit. The interval [a, b] is called the interval of integration. The interval [a, b] is also known as range of integration. Important Tips In the above definition it does not matter which anti-derivative is used to evaluate the definite integral, because if b
f x dx x c, then a f x d x x c a (b) c (a) c (b) (a). b
In other words, to evaluate the definite integral there is no need to keep the constant of integration.
Every definite integral has a unique value.
Example: 1
Solution: (b)
2 1 x 1 x 1 tan tan dx = 2 x 1 x 1
3
(a)
(b) 2
x 1 x 1 tan x 2 1 cot x 2 1 dx
I
Example: 2
sin
3
[UP MCA 2000]
(c) 3
(d) None of these
1
I 2
3
1
dx I [ x ]31 [3 1] 2 . 2 2 2
x dx is equal to
[MP MCA 1999]
0
(a) Solution: (b)
1 I 2
(b) /2
(c) 0
(d) None of these
1 2 sin 2 x dx [1 cos 2 x ]dx 2 0 0
I
1 sin 2 x 1 I [ ] . x 2 2 0 2 2
Definite Integral as the Limit of a Sum. Let f(x) be a single valued continuous function defined in the interval a x b, where a and b are both finite. Let this interval be divided into n equal sub-intervals, each of width h by inserting (n – 1) points a h, a 2 h, a 3 h...... a (n 1)h between a and b. Then nh b a . Now, we form the sum hf (a) hf (a h) hf (a 2 h) ........ hf (a rh) ...... hf [a (n 1)h] = h[ f (a) f (a h) f (a 2h) ..... f (a rh) .... f {a (n 1)h}]
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INPS MCA ENTRANCE EXAM n 1
=h
f (a rh) r 0
where, a nh b nh b a n 1
The lim h h 0
f (a rh), if it exists is called the definite integral of f(x) with respect to x between the limits a r 0
and b and we denote it by the symbol Thus,
b
f (x )dx . a
b
a
f (x )dx lim h[ f (a) f (a h) f (a 2h) ...... f {a (n 1)h}] h 0
b
a
f (x )dx lim h h 0
n 1
f (a rh) r 0
where, nh b a, a and b being the limits of integration. The process of evaluating a definite integral by using the above definition is called integration from the first principle or integration as the limit of a sum. Important Tips In finding the above sum, we have taken the left end points of the subintervals. We can take the right end points of the subintervals throughout. Then we have,
b
a f (x )dx
n b lim h[ f (a h) f (a 2h) ..... f (a nh )] , f (x )dx h f (a rh) a h0 r 1
where, nh = b – a.
b
b c
x x
a
dx
a c
x a dx b a 2 bx f x c dx
b
a
f x dx or
b
x x f x dx
a
b c
a c
f x c dx
b
1 n
nb
dx
2 8
f x dx
na
f x dx
a
Some useful results for evaluation of definite integrals as limit for sums n(n 1) n(n 1)(2n 1) (i) 1 + 2 + 3 + …….+ n = (ii) 1 2 2 2 3 2 ........ n 2 2 6 n(n 1) (iii) 1 3 2 3 3 3 ....... n 3 2 r 1
(v) a ar ar 2 .......... ..... ar n 1
2
(iv) a ar ar 2 ........ ar n 1
a(r n 1) ,r 1 , r 1
a (1 r n ) , r 1, r 1 1r
n 1 nh sin a h sin 2 2 [sin(a nh)] (vi) sin a sin(a h) ........ sin[a (n 1)h] h r 0 sin 2 n 1
n 1 nh cos a h sin 2 2 (vii) cos a cos(a h) cos(a 2h) ..... cos[ a (n 1)h] [cos(a nh)] h r 0 sin 2 n 1
2
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(viii) 1 (x) 1
INPS MCA ENTRANCE EXAM
1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 .......... .......... (ix) 1 2 2 2 2 2 .......... ...... 2 12 6 2 3 4 5 6 2 3 4 5 6 2
1 1 2 .......... .......... 8 32 52
(xii) cos
(xi)
e i e i e i e i and sin 2 2
1 22
1
42
(xiii) cosh
1 62
.......... ..........
2 24
e e e e and sinh 2 2
Evaluation of Definite Integral by Substitution. When the variable in a definite integral is changed, the substitutions in terms of new variable should be effected at three places. (i) In the integrand (ii) In the differential say, dx (iii) In the limits For example, if we put (x ) t in the integral
b
b
a
a
f { (x )} ' (x )dx , then
f { (x )} ' (x )dx
(b )
(a )
f (t) dt .
Important Tips
0
dx 2 1 sin x
/2
a
dx
0
a x
a
a
0
log tan x dx 0
0
/2
0
2
Example: 3
2
2
0
If h(a) h(b ), then
[ f (g[h(x )])] b
1
dx 2 log sin x cos x
2 1
dx a , where f a x f x f x 2 1e
dx x a 2 2a
2
a
a 2 x 2 dx
0
f ' (g[h(x )]) g ' [h(x )] h' ( x ) dx is equal to
a 2 4
[JNU 2001]
a
(a) 0 Solution: (a)
)]) t 1dt log( t)ff (( gg[[hh((ba)]) 0
f ( g [h(b )])
f ( g [h(a )])
The value of the integral
e2
e 1
(a) 3/2 Solution: (b)
f [g(a)] f [g(b)]
(c)
(d) None of these
Put f (g[h(x )]) t f ' (g[h( x )]) g ' [h(x )] h' ( x )dx dt
Example: 4
f (a) f (b)
(b)
[ h(a) h(b )]
log e x dx is x
[IIT 2000]
(b) 5/2
(c) 3
(d) 5
Put log e x t e t x dx e t dt and limits are adjusted as –1 to 2 I
Example: 5
/2
2
1
2 t t e dt | t | dt I t e 1
/ 2
/2
0
I
0
2
[MNR 1983; Rajasthan PET 1990; Kurukshetra CEE 1997]
(a) 0 I
dx equals 1 sin x
0
Solution: (b)
0
2 t2 t2 tdt tdt I I 5 /2 0 1 2 1 2 0 0
(b) 1
(c) –1
dx sin 2 x /2 cos 2 x /2 2 sin x /2 cos x /2
dx (sin x /2 cos x /2) 2
/2
0
sec 2 x /2 (1 tan x /2) 2
dx
(d) 2
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INPS MCA ENTRANCE EXAM
1 Put (1 tan x /2) t sec 2 x /2 dx dt 2
I2
2
1
dt t2
2
1 1 1 2 2 1 2 1 t 1
Properties of Definite Integral. (1)
b
b
a
a
f (x )dx f (t) dt
i.e., The value of a definite integral remains unchanged if its variable is replaced by
any other symbol. Example: 6
6
3
1 dx is equal to x 1
(a) Solution: (c)
I
log( x 1) 63
6
1 dx [log(x 1)]63 , x 1
3
(2)
b
(b) [log(t 1]63
I
6
3
(c) Both (a) and (b)
(d) None of these
1 dt [log(t 1)]63 t 1
a
f(x)dx f(x)dx i.e., by the interchange in the limits of definite integral, the sign of the integral is
a
b
changed. Example: 7
Suppose f is such that f(–x) = –f(x) for every real x and
1
f ( x ) dx 5 , then
0
(a) 10 Solution: (d)
Given,
(b) 5
(c) 0
f (t) dt 0
[UP MCA 2000]
1
(d) –5
f (x )dx 5 1
0
Put x t dx dt I
1
(3) or
b
f ( x )dx
a
0
f ( t)dt f (t)dt I 5 1
0
c
b
a
c
f (x )dx f (x )dx , (where a < c < b)
b
c1
a
a
f (x )dx
f (x )dx
c2
c1
f ( x )dx .....
b
cn
f ( x )dx ; (where a c 1 c 2 ........ c n b )
Generally this property is used when the integrand has two or more rules in the integration interval. Important Tips
b
( x a x b ) dx (b a)2
a
Note into intervals.
: Property (3) is useful when f (x ) is not continuous in [a, b] because we can break up the integral several integrals at the points of discontinuity so that the function is continuous in the sub The expression for f (x ) changes at the end points of each of the sub-interval. You might at times be puzzled about the end points as limits of integration. You may not be sure for x = 0 as the limit of the first integral or the next one. In fact, it is immaterial, as the area of the line is always zero. Therefore, even if you write
0
1
(1 2 x ) dx , whereas 0 is not included in its domain it is deemed to
be understood that you are approaching x = 0 from the left in the first integral and from right in the second integral. Similarly for x = 1. 4
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| 1 x 2
Example: 8
INPS MCA ENTRANCE EXAM
| dx is equal to
2
2
[IIT 1989; BIT Ranchi 1996; Kurukshetra CEE 1998]
(a) 2
| 1x
I
Solution: (b)
(b) 4
2
1
I (1 x 2 )dx 2
[x 1 .5
Example: 9
2
(c) 6
2
1
(d) 8
2
| dx | 1 x 2 | dx | 1 x 2 | dx | 1 x 2 | dx
2
2
1
1
1
1
1
2
(1 x 2 )dx (1 x 2 )dx I 1
4 4 4 4. 3 3 3
]dx , where [.] denotes the greatest integer function, equals
[DU 2000, 2001; IIT 1988; AMU 1998]
0
(a) 2 2
I
Solution: (b)
1. 5
(b) 2 2
1
[ x 2 ]dx [ x 2 ]dx 0
0
2
e cos x . sin x , | x | 2 If f (x ) , then 2, otherwise
Example: 10
(a) 0
a
f (x )dx
0
3
2
[ x 2 ]dx I 0
2
2 1
(d)
1dx
1
1. 5
2dx 2 1 3 2 2 I 2 2
2
f (x )dx
[IIT 2000]
(b) 1
I
2
(c) 2
(d) 3
| x | 2 2 x 2 and f (x ) e cos x sin x is an odd function.
Solution: (c)
(4)
[ x 2 ]dx
1
(c) 1 2
1 .5
3
2
f (x )dx
2
f (x )dx
2
a
f (a x )dx
3
f (x )dx I 0
2
2dx [2 x ] 3
2
3 2
2
f (x ) 0 if f(x) is odd and in (2, 3) f(x) is 2] a
[
a
: This property can be used only when lower limit is zero. It is generally used for
0
those complicated integrals whose denominators are unchanged when x is replaced by (a – x). Following integrals can be obtained with the help of above property. (i)
(ii)
sin x cos x n
0
(v)
1 tan x 2
0
2
0
(vii)
tan x n
0
/2
0
/2
(xii)
dx
0
/2
0
Example: 11
log tan xdx
log sin xdx
1 cot x
0
4
4
f (sin 2 x ) cos xdx
0
(iii)
(vi)
log cot xdx
(x)
0
/2
1 log 2 log 2 2 2 a sec x b cosec x dx sec x cosec x
/2
0
(viii)
/2
/2
0
f (cot x )dx
1 dx 1 tan n x
2
0
1 dx n 4 1 cot x
1
0
/4
0
f (sin x )dx
f (log x )dx
/2
0
1
0
f (cos x )dx
f [log(1 x )]dx
log(1 tan x )dx
log 2 8
log cos xdx
0
a sin x b cos x dx sin x cos x
dx
dx
cosec n x dx n n 4 cosec x sec x
/2
/2
/2
n
n
0
cos x sin x n
cot n x
/2
n
f (tan x )dx
cos n x
/2
0
0
f (sin 2 x ) sin xdx
/2
(xi)
dx
sec x dx n sec x cosec n x
0
n
n
/2
(iv)
(ix)
sin n x
/2
/2
0
/2
0
a tan x b cot x dx (a b ) tan x cot x 4
e sin
2
x
cos 3 xdx =
0
OR For any integer n,
0
98]
e sin
2
x
cos 3 (2n 1)xdx =
[MPMCA2002;
MNR
1992,
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INPS MCA ENTRANCE EXAM
(a) –1 Solution: (b)
(d) None of these
f 2 ( x ) cos 3 (2n 1)x f ( x )
I = 0.
f (x ) dx is equal to f ( x ) f (2a x )
2a
0
(a) a I
Solution: (a)
(c) 1
Let, f1 ( x ) cos 3 x f ( x ) and
Example: 12
(b) 0
(b) a/2
2a
f (x ) dx f ( x ) f (2 a x )
0
2I
[IIT 1988; Kurukshetra CEE 1999;BHU 2000]
0
f ( x ) f (2 a x ) dx f ( x ) f (2 a x )
2a
0
(c) 2a
(d) 0
f (2 a x ) dx f (2 a x ) f ( x )
2a
dx [x ] 2a
2a 0
0
2a
I a. /2
Example: 13
tan x
1 tan x
0
/2
sin x sin x cos x
0
(a) /4 Solution: (a)
dx is equal to
[MNR 1989, 90JNU 2005]
(b)
We know,
/2
0
(c) –1
(d) 1
tan xdx for any value of n 1 tan n x 4 n
I /4 .
(5)
a
f (x ) dx
a
a
0
f (x ) f ( x ) dx .
2 a f ( x ) dx , if f (x ) is even function or f ( x ) f (x ) In special case : f (x ) dx 0 a 0 , if f (x ) is odd is odd function or f ( x ) f (x ) This property is generally used when integrand is either even or odd function of x.
Example: 14
The integral
1 x [ x ] ln dx equal to 1 x
1 / 2
(b) 0
1/2
[ x ]dx 0 I
1 / 2
Example: 16
0
1 / 2
1
1
[ x ]dx
1/2
[ x ]dx I
0
log x x 2 1 dx is (b) log 2
0
1dx 0 [ x ] 01 / 2
1 / 2
[MP PET 2001]
(c) log 1/2
The value of
(1 x
2
) sin x cos 2 xdx is
[Roorkee 1992; MNR 1998]
(b) 1
(c) 2
Let, f1 (x ) (1 x ) , f2 (x ) sin x and f3 (x ) cos x 2
I
2
f ( x )dx [ f1 ( x ). f2 (x ). f3 (x )]dx = [ f1 ( x ). f2 ( x ). f3 ( x )] dx
I0
6
2a
0
(d) None of these
f (x ) log[ x x 1] is a odd function i.e. f ( x ) f (x ) f (x ) f ( x ) 0 I = 0.
Now, f1 ( x ) f1 ( x ) , f2 ( x ) f2 ( x ) and f3 ( x ) f3 ( x )
(6)
1 . 2
2
(a) 0 Solution: (a)
The value of the integral (a) 0
Solution: (a)
1 (d) 2 ln 2
(c) 1
1 x log is an odd function of x as f ( x ) f ( x ) 1 x
I Example: 15
1/2
1 2
(a) Solution: (a)
a
f (x )dx
a
0
f ( x )dx
a
0
f (2 a x ) dx
(d) 3
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INPS MCA ENTRANCE EXAM
, if f (2 a x ) f (x ) 0 2a In particular, f (x ) dx a 0 2 0 f (x ) dx , if f (2 a x ) f (x ) It is generally used to make half the upper limit.
Example: 17
If n is any integer, then
e cos
2
x
cos 3 (2n 1)x dx is equal to
[IIT 1985; Rajasthan PET
0
1995]
(a) x I
Solution: (c)
(b) 1
e cos
2
( x )
(c) 0
(d) None of these
. cos 3 (2n 1)( x ) dx
0
I
2
e cos
x
. cos 3 (2n 1)x dx I I
0
2I 0 I 0. Example: 18
If I1
3
f (cos
f (cos
x ) dx and I 2
2
0
(a) I1 I 2
2
x ) dx then
0
(b) I1 2I 2
(c)
I1 3I 2
(d) I1 4 I 2
f (cos x ) f (cos (3 x ))
Solution: (c)
2
2
I1 3 f (cos 2 x ) dx I1 3I 2 0
(7)
b
b
a
a
f (x ) dx f (a b x )dx
Note : a
f (x ) dx 1 (b a) f (x ) f (a b x ) 2
b
Example: 19
3 / 4
/4
dx is equal to 1 cos x
[IIT 1999]
(a) 2 Solution: (a)
I
(b) –2
3 / 4
3 / 4
Example: 20
2 dx 1 cos 2 x
/4
2I 2
3 / 4
/4
The value of
co sec 2 xdx 2 I 2[cot x ]3/4/4 4 I = 2. 3
x
2
5x x
(a) 1 Solution: (d)
I
2
dx is (b) 0
3
(d) –1/2
3 x cos x [ cos 4 4
1 dx 1 cos x
/4
2I
(c) 1/2
x 5x
(c) –1
(d) 1/2
dx x
Put x 2 3 t dx dt
I
5t
2
5t t
3
Example: 21
(dt )
3
5x
2
5x x
dx and 2 I
3
x 5x
2
5x x
3
dx 1 dx 2
2 I [ x ]22 1 I 1 / 2
If f (a b x ) f ( x ) then
x f (x )dx is equal to b
[Kurukshetra CEE 1993; AIEEE 2003]
a
(a) Solution: (b)
I
ab 2
b
a
(b)
a
x f (x )dx b
f (b x )dx and I
b
a
ab 2
b
f (x )dx
a
(a b x ) f (a b x )dx
(c)
b a 2
b
a
f (x )dx
(d) None of these
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INPS MCA ENTRANCE EXAM
(a b) f (x )dx x f (x )dx
b
I (a b x ) f ( x )dx I a
(8)
a
0
x f (x )dx
Example: 22
If
1 a 2
a
x f (sin x )dx k
a
f (x )dx b
a
f (sin x )dx , then the value of k will be (b) /2
Given,
x f (sin x )dx k
0
[IIT 1982]
f (sin x )dx
0
0
(d) 1
0
(c) /4
( x ) f (sin( x ))dx k
a
ab 2 I f ( x )dx (a b ) I 2 a b
0
(a)
b
f (x )dx if f (a x ) f (x )
0
0
Solution: (b)
b
f (sin x )dx 2k
0
0
0
f (sin x )dx
f (sin( x ))dx
x f (sin x )dx k
f (sin x )dx
0
f (sin x )dx 0
f (sin x )dx 0 ( 2k )
0
0
2 k 0 k /2 .
(9) If f (x ) is a periodic function with period T, then
nT
0
f (x )dx n
T
0
f (x )dx ,
Deduction : If f (x ) is a periodic function with period T and a R , then
a nT
nT
f ( x )dx
a
0
f (x )dx
(10) (i) If f (x ) is a periodic function with period T, then
a nT
a
f (x ) dx n
T
0
where n I
f (x ) dx
(a) In particular, if a = 0
nT
0
(b) If n = 1, (i)
(ii)
nT
a T
0
b nT
a nT
f (x ) dx
f ( x ) dx
b
T
T
0
T
0
where n I
f (x ) dx ,
0
f (x ) dx (n m )
mT
f ( x ) dx n
f ( x ) dx ,
where n, m I
f (x ) dx ,
where
f (x ) dx ,
a
n I
(11) If f (x ) is a periodic function with period T, then (12)
b
a
f (x ) dx (b a)
f (x ) is independent of a.
f (b a) x a dx 0
x
(x ) f (t) dt is an even function a
(14) If f (x ) is an even function, then (x )
x
0
f (t) dt is on odd function.
: If f(t) is an even function, then for a non zero ‘a’, will be odd function if
Example: 23
For n 0,
2
0
(a) 2 Solution: (a)
I
2
0
8
a T
a
1
(13) If f (t) is an odd function, then
Note
sin
x
0
f (t) dt is not necessarily an odd function. It
a
f (t) dt 0 0
x sin 2 n x 2n
x cos 2 n x
dx is equal to
(b) 2 2
x sin 2 n xdx and I sin 2 n x cos 2 n x
2
0
[IIT 1996]
(c)
3 2
(2 x ) sin 2 n (2 x )dx sin 2 n (2 x ) cos 2 n (2 x )
(d) 4 2
a
0
f (x )
f (a x ) a
0
www.inpsmcalucknow.com 2 I 2
0
using
nT
INPS MCA ENTRANCE EXAM
2
sin 2 n dx I sin 2 n x cos 2 n x
f (x ) n
0
2
0
sin 2 n x dx sin 2 n x cos 2 n x
T
f (x )dx
0
/2
I 4
0
sin 2 n x dx I 4 ( /4 ) 2 . sin x cos 2 n x 2n
If f (x ) is a continuous periodic function with period T, then the integral I
Example: 24
aT
f (x )dx
is
a
(b) Equal to 3a
(a) Equal to 2a Consider the function g(a)
Solution: (c)
(c) Independent of a
aT
0
T
a
a
0
(d) None of these
aT
f (x ) dx = f (x )dx f (x )dx f (x )dx
Putting x T y in last integral, we get
aT
f ( x )dx
T
T
a
f (y T ) dy
0
f (y) dy a
0
f (x )dx f (x ) dx f (x )dx = f (x )dx
g(a)
0
T
a
T
a
0
0
0
Hence g(a) is independent of a.
Important Tips
Every continuous function defined on [a, b] is integrable over [a, b]. Every monotonic function defined on [a, b] is integrable over [a, b].
If f(x) is a continuous function defined on [a, b], then there exists c (a, b ) such that
b
f (x )dx f (c).(b a) .
a
The number f (c)
1 (b a)
b
f ( x )dx is called the mean value of the function f (x ) on the interval [a, b].
a
If f is continuous on [a, b], then the integral function g defined by g(x) =
f (t)dt x
for x [a, b] is derivable on [a, b] and
a
g ( x ) f (x ) for all x [a, b ] .
If m and M are the smallest and greatest values of a function f(x) on an interval [a, b], then m (b a)
f (x )dx M (b a) b
a
If the function (x ) and (x ), are defined on [a, b] and differentiable at a point x (a, b) and f(t) is continuous for
(x ) (a) t (b ) , then f (t)dt f ( ( x )) ' ( x ) f (( x ))' ( x ) ( x )
b
b
f ( x )dx | f (x ) | dx
a
a
f ( x ) g( x ) dx a
b
f 2 (x ) dx a
b
1/2
b 2 g (x ) dx a
1/2
If f 2 ( x ) and g 2 (x ) are integrable on [a, b], then
Change of variables : If the function f(x) is continuous on [a, b] and the function x (t) is continuously differentiable on the interval [t1 , t 2 ] and a (t1 ), b (t 2 ), then
b
a
f ( x )dx
f ((t))' (t)dt . t2
t1
Let a function f (x , ) be continuous for a x b and c d . Then for any [c, d ] , if I( )
f (x, )dx , b
a
I' ( )
f '(x , )dx , b
a
Where I' ( ) is the derivative of I( ) w.r.t. and f ' ( x , ) is the derivative of f (x , ) w.r.t. , keeping x constant.
then
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INPS MCA ENTRANCE EXAM
For a given function f (x ) continuous on [a, b] if you are able to find two continuous function f1 (x ) and f 2 (x ) on [a, b] such that f1 (x ) f (x ) f2 ( x ) x [a, b ], then
b
f1 ( x ) dx
a
b
f ( x ) dx
a
b
a
f2 (x ) dx
Summation of Series by Integration. We know that
b
a
f (x )dx lim h n
n
f (a rh) , where nh b a r 1
Now, put a = 0, b = 1, nh 1 or h
Note
1 . Hence n
: Express the given series in the form
1
0
Example: 25
1
1
n
0
r 1 r . Replace by x, by dx and the limit of the sum is n h n n
1
f
f (x ) dx .
If S n
1 1 n
1
2 2n
........
(a) log 2 Solution: (b)
r
f (x )dx lim n f n
1 n n2
then lim S n is equal to
(b) 2 log 2
lim r
1
n
rn
lim S n n
lim
1
n
(c)
(d) 4 log 2
3 log 2
1 r r n n n
1 x (1 x )
0
[Roorkee 2000]
n
dx
= 2[log(1 x )]10 2 log 2 1/n
Example: 26
(n! )1 / n n! or lim n n n n n lim
(a) e Solution: (b)
is equal to
[WB JEE 1989 BHU 1998]
(b) e–1
Let A lim
n
(d) None of these
(n! ) n
1/n
1 . 2 . 3 ...... n log A lim log n nn
log A
(c) 1
1/n
1/n
n 1 2 3 log A lim log . . .... n n n n n
log xdx [x log x x ] 1
0
1 0
log A lim
n
1 n
n
r 1
log A 1 A e 1
Gamma Function. If m and n are non-negative integers, then
/2
0
sin m
m 1 n 1 2 2 n x cos xdx m n 2 2 2
where (n) is called gamma function which satisfied the following properties (n 1) n (n) n!
i.e.
(1) 1 and (1 / 2)
In place of gamma function, we can also use the following formula :
10
r
log n
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INPS MCA ENTRANCE EXAM
/ 2
0
(m 1)(m 3).....( 2 or 1)(n 1)(n 3).....( 2 or 1) sin m x cos n xdx = (m n)(m n 2)....( 2 or 1)
It is important to note that we multiply by (/2); when both m and n are even. Example: 27
/2
The value of
sin 4 x cos 6 xdx =
[PUNE 1999]
0
(b) 5 / 512
(a) 3 /312 Solution: (c)
(c)
3 / 512
(d) 5 / 312
(4 1).(4 3).(6 1).(6 3).(6 5) 3 .1.5.3 .1 3 . . (4 6 )(4 6 2)(4 6 4 )(4 6 6)(4 6 8 ) 2 10 . 8 . 6 . 4 . 2 . 2 512
I
Reduction formulae Definite Integration. (1)
0
b
e ax sin bxdx
Example: 28
If In
e
a b 2
x
(2)
2
x n 1dx , then
0
e
x
0
e ax cos bxdx
a a b 2
(3)
2
0
e ax x n dx
n! a 1 n
x n 1 dx is equal to
0
(a) I n Solution: (c)
1 In
(b)
(c)
In n
(d) n In
Put, x t, dx dt , we get,
e
x
1 n
x n 1dx
0
e
t n 1
t
dt =
0
1
n
e
x
x n 1 dx
0
In
n
Walli's Formula.
/2
0
/2
0
sin xdx n
sin m x cos n dx
/2
0
2 n 1 n 3 n 5 n . n 2 . n 4 ...... 3 , cos xdx n 1 n 3 n 5 3 1 . . ....... . . , 4 2 2 n n2 n4
when n is even
(m 1)(m 3)......... .(n 1)(n 3)........ [If m, n are both odd +ve integers or one odd +ve integer] (m n)(m n 2)
(m 1) (m 3)......... ...( n 1) (n 3) . (m n) (m n 2) 2
Example: 29
when n is odd
n
/2
[If m, n are both +ve integers]
sin7 xdx has value
[BIT Ranchi 1999]
0
(a) Solution: (c)
37 184
(b)
Using Walli’s formula, I
17 45
(c)
16 35
(d)
16 45
7 1 7 3 7 5 6 . 4 .2 16 . . 7 7 2 7 4 7 .5 . 3 35
6.9 Leibnitz’s Rule. (1) If f(x) is continuous and u(x), v(x) are differentiable functions in the interval [a, b], then, d v(x ) d d {v(x )} f {u(x )} {u(x )} f (t)dt f {v(x )} u ( x ) dx dx dx (2) If the function (x ) and (x ) are defined on [a,b] and differentiable at a point x (a, b), and f (x , t) is
continuous, then, Example: 30
d dx
(x )
Let f (x )
(x )
x
f (x , t) dt
(x )
(x )
d d (x ) d (x ) f (x , t) dt f (x , (x )) f ( x , (x )) dx dx dx
2 t 2 dt . Then the real roots of the equation x 2 f ' (x ) 0 are
[IIT 2002]
1
(a) 1
(b)
1 2
(c)
1 2
(d) 0 and 1
www.inpsmcalucknow.com Solution: (a)
f (x )
x
INPS MCA ENTRANCE EXAM
2 t 2 dt f ' ( x ) 2 x 2 .1 2 1 . 0 2 x 2
1
x 2 f ' (x ) 2 x 2 x 4 x 2 2 0 ( x 2 2)(x 2 1) 0 x 1 (only real). Example: 31
Let f : (0, ) R and f (x )
f (t)dt . If f (x ) x (1 x ), then f (4) equals x
2
2
[IIT 2001]
0
(a) 5/4 Solution: (c)
(b) 7
By definition of f(x) we have f (x 2 )
(c) 4
f (t)dt x x
(d) 2
2
2
x3
(given)
0
Differentiate both sides, f (x 2 ).2 x 0 2 x 3 x 2 Put, x 2 4 f (4 ) 16 f (4 ) 4
6.10 Integrals with Infinite Limits (Improper Integral). A definite integral
b
f (x )dx a
is called an improper integral, if
The range of integration is finite and the integrand is unbounded and/or the range of integration is infinite and the integrand is bounded. 1 1 e.g., The integral dx is an improper integral, because the integrand is unbounded on [0, 1]. Infact, 0 x2 1 1 as x 0 . The integral dx is an improper integral, because the range of integration is not 2 0 1 x2 x finite. There are following two kinds of improper definite integrals:
(1) Improper integral of first kind : A definite integral
b
f (x )dx a
is called an improper integral of first
kind if the range of integration is not finite (i.e., either a or b or a and b ) and the integrand f(x) is bounded on [a, b].
1
1 x
2
dx ,
1 1 x
0
2
dx ,
1 1 x
2
dx ,
1
3x (1 2 x ) 3
dx are improper integrals of first kind.
Important Tips
In an improper integral of first kind, the interval of integration is one of the following types [a, ), (–, b], (–, ).
The improper integral
f (x )dx is said to be convergent, if a
improper integral. If lim
k
k a
The improper integral
exists finitely and this limit is called the value of the
f (x ) dx is either + or –, then the integral is said to be divergent.
f (x )dx
f (x )dx k
lim
k a
is said to be convergent, if both the limits on the right-hand side exist finitely and are
independent of each other. The improper integral
f (x )dx is said to be divergent if the right hand side is + or –
(2) Improper integral of second kind : A definite integral
b
f (x )dx a
is called an improper integral of
second kind if the range of integration [a, b] is finite and the integrand is unbounded at one or more points of [a, b].
12
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If
b
f (x )dx a
INPS MCA ENTRANCE EXAM
is an improper integral of second kind, then a, b are finite real numbers and there exists at least
one point c [a, b ] such that f (x ) or f (x ) as x c i.e., f(x) has at least one point of finite discontinuity in [a, b]. For example :
(i) The integral
1 1 dx , is an improper integral of second kind, because lim . x 2 x 2 x 2
3
1 1
log xdx ; is an improper integral of second kind, because log x as x 0 .
(ii) The integral
0
2
1 1 becomes dx , is an improper integral of second kind since integrand 0 1 cos x 1 cos x infinite at x [0, 2 ] .
(iii) The integral
(iv)
1
0
sin x sin x dx , is a proper integral since lim 1. x 0 x x
Important Tips Let f(x) be bounded function defined on (a, b] such that a is the only point of infinite discontinuity of f(x) i.e., f(x) as x a. Then the improper integral of f(x) on (a, b] is denoted by
b
f ( x )dx and is defined as
a
b
a
f ( x )dx lim 0
limit on right hand side exists. If l denotes the limit on right hand side, then the improper integral
b
b
a
f ( x )dx . Provided that the
f ( x )dx is said to converge to
a
l, when l is finite. If l = + or l = –, then the integral is said to be a divergent integral.
Let f(x) be bounded function defined on [a, b) such that b is the only point of infinite discontinuity of f(x) i.e. f(x) as x b.
f (x )dx and is defined as b
b
b
f (x )dx lim f (x )dx a a 0 a Provided that the limit on right hand side exists finitely. If l denotes the limit on right hand side, then the improper integral
Then the improper integral of f(x) on [a, b) is denoted by
b
f ( x )dx is said to converge to l, when l is finite.
a
If l or l , then the integral is said to be a divergent integral.
Let f(x) be a bounded function defined on (a, b) such that a and b are only two points of infinite discontinuity of f(x) i.e., f(a) , f(b) . Then the improper integral of f(x) on (a, b) is denoted by
b
f ( x )dx and is defined as
a
b
c
b
f (x )dx lim f (x )dx lim f ( x )dx , a c b a 0 a 0 a Provided that both the limits on right hand side exist. Let f(x) be a bounded function defined [a, b]-{c}, c[a, b] and c is the only point of infinite discontinuity of f(x) i.e. f(c). Then
the
improper
integral
of
f(x)
on
[a,
b]
–
{c}
is
denoted
by
b
f ( x )dx
and
is
defined
as
a
b
a
f (x )dx lim x 0
c x
a
f (x )dx lim 0
b
c
f (x )dx
Provided that both the limits on right hand side exist finitely. The improper integral
b
f ( x )dx is said to be convergent if both the
a
limits on the right hand side exist finitely. If either of the two or both the limits on RHS are , then the integral is said to be divergent.
Example: 32
The improper integral
e
x
dx is …… and the value is….
0
(a) Convergent, 1
(b) Divergent, 1
(c) Convergent, 0
(d) Divergent, 0
www.inpsmcalucknow.com I
Solution: (a)
e
e x dx lim
x
k 0
0
Thus, lim
k
k 0
Example: 33
INPS MCA ENTRANCE EXAM k
dx I lim [e x ]k0 lim [e k e 0 ] I lim (1 e k ) 1 0 1 [ lim e k e 0 ] k
k
I
Solution: (b)
0
0
dx lim a x 2 k - 2
-
k
e x dx exists and is finite. Hence the given integral is convergent.
1 dx , a 0 is a2 x 2 (a) Convergent and equal to a (c) Divergent and equal to a The integral
k
0
k
(b) Convergent and equal to (d) Divergent and equal to
2a
2a
dx a2 x 2
0
x k 1 1 1 1 1 I lim tan 1 lim tan 1 0 tan 1 I 0 tan 1 () k a a k k a a a a a 2 2a Hence integral is convergent.
Example: 34
1 dx is e x e x (a) Convergent and equal to /6 (c) Convergent and equal to /3 The integral
I
Solution: (d)
1 dx ex e x
(b) Convergent and equal to /4 (d) Convergent and equal to /2
ex dx 2x 1 e
Put e t e dx dt x
I
x
0
Example: 35
1 dt I [tan 1 t]0 [tan 1 tan 1 0] I /2 , which is finite so convergent. 1 t2
x 1
2
x 1
1
dx is
(a) Convergent and equal to
14 3
(b) Divergent and equal to
(c) Convergent and equal to
I
Solution: (a)
2
x 1 dx
1
Example: 36
2
1
2
1
dx x 5x 4 2
(d) Divergent and equal to 2
2
2 = (x 1)3 / 2 [4 x 1 ]12 = 14 /3 which is finite so convergent. 3 1
dx
x 1
dx is
(a) Convergent and equal to
1 log 2 3
(b) Convergent and equal to 3/log2
(c) Divergent I
Solution: (c)
2
1
1 dx = (x 1)( x 4 ) 3
(d) None of these
x 4 x 1 dx 2
1
1
1
=
1 [log 2 ] 3
So the given integral is not convergent.
6.11 Some Important results of Definite Integral. (1) If I n (2) If I n (3) If I n 14
/4
0
0
/4
tan n xdx then I n I n 2
/4
0
3 14
cot n xdx then I n I n 2
sec n xdx then I n
1 n 1 1 1n
( 2 )n 2 n 2 In 2 n 1 n 1
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(4) If In
(5) If In
(6) If In
/4
0
/2
0
/2
0
INPS MCA ENTRANCE EXAM
cosec n xdx then In
n2
( 2) n2 In 2 n 1 n 1
x n sin xdx then In n(n 1)In 2 n( / 2)n 1 x n cos xdx then In n(n 1)In 2 ( /2)n
(7) If a b 0, then
(8) If 0 a b then
(9) If a b 0 then
/2
0
/2
0
/2
(10) If 0 a b , then
0
dx a b cos x
a b
dx a b cos x
b a
dx a b sin x
a b
/2
0
dx a b sin x
(11) If a b, a 2 b 2 c 2 , then
(12) If a b, a 2 b 2 c 2 , then
(13) If a b, a 2 b 2 c 2 then
/2
0
/2
0
/2
0
2 2
2
1 2
2
log
2 2
2
b a
b a b a b a b a
tan 1
1 2
ab ab
tan 1
2
log
ab ab ba ba ba b a
dx a b cos x c sin x
dx a b cos x c sin x dx a b cos x c sin x
2 a b c 2
2
2
tan 1
1 b 2 c2 a2 1 b 2 c2 a2
log
log
ab c a2 b 2 c 2
a b c b 2 c2 a2 a b c b 2 c2 a2 b a c b 2 c2 a2 b a c b 2 c2 a2
Important Tips x
lim
x 0
b
f (x ) dx f (0) 0
x
f ( x )dx (b a)
a
1
f (b a) t a dt
0
Integration of Piecewise Continuous Functions. Any function f (x ) which is discontinuous at finite number of points in an interval [a, b] can be made continuous in sub-intervals by breaking the intervals into these subintervals. If f (x ) is discontinuous at points x 1 , x 2 , x 3 .......... x n in (a, b), then we can define subintervals (a, x 1 ), (x 1 , x 2 )......... ....( x n 1 , x n ), (x n , b ) such that f (x ) is continuous in each of these subintervals. Such functions are called piecewise continuous functions. For
integration of Piecewise continuous function. We integrate f (x ) in these sub-intervals and finally add all the values. Example: 37
20
10
[cot 1 x] dx , where [.] denotes greatest integer function
(a) 30 cot 1 cot 3
(b) 30 cot 1 cot 2 cot 3
(c) 8 30 cot 1 cot 2
(d) None of these
www.inpsmcalucknow.com Solution: (b)
Let I
20
10
INPS MCA ENTRANCE EXAM
[cot 1 x ] dx ,
we know cot 1 x (0, ) x R
thus,
[cot
I
Hence, Example: 38
[x 2
2
1
3, 2, x] 1 0
cot 3
10
3 dx
x (, cot 3) x (cot 3, cot 2) x (cot 2, cot 1) x (cot 1, )
cot 2
2 dx
cot 3
cot 1
1 dx
cot 2
20
0 dx 30 cot 1 cot 2 cot 3
cot 1
x 1] dx , where [.] denotes greatest integer function
0
(a)
Solution: (a)
7 5 2
Let I
2
0
16
[ x 2 x 1] dx
(b)
1 5 2
0
7 5 2
[ x 2 x 1] dx
(c)
2
1 5 2
5 3 2
[ x 2 x 1] dx
(d) None of these
1 5 2
0
1 dx
2
1 5 2
2dx
7 5 2
