DEFINITE INTEGRATION (Integration as a process of summation) FIRST LECTURE b
F(a) is called the definite integral of f(x) between the limits a and b. ! f (x) dx = "F(x)# = F(b) – F(a) b
Definition:
a
a
d
$F( x )% & f ( x ) dx differential calculus. Note : The word limit here is quite different as used in differential For a cont. bounded function f (x) in (a, b) where
a = nh Lim h[f (a )(f (a ( h ) ( f (a ( 2h ) ( ....... ( f (a ( n ' 1 h )] where b – a h *0 n *)
Lim h h *0 n *)
n '1
b
r 0
a
F(a) +& f (a ( rh) & ! f (x) dx = F(b) – F(a)
computing the limit of an infinite sum is called evaluating the definite integral by first principle. Very Important: Important: b
!
If f (x) dx =0, then the equation equation f (x) = 0 has atleast one root in (a, b) provided f is continuous in (a, b).
1.
a
Note that the that the converse is not true. Consider the example: example: If 2a + 3b + 6c = 0 then the QE ax2 + bx + c = 0 must have a root in (0, 1) b 1 b / , Lim ! f n ( x )dx & ! 1 Lim f n ( x ) .,dx ; / , 0 n *) n *) / a 0 a
2.
3
e.g.
1 t 3 . Lim ! /1 ' , n *) 3 / n -, ' a 0 a
n
&
t 2 dt
2 2
b
g '1 ( b )
a
g '1 ( a )
3
(n 2 N) N) find find 'a'. 'a'. [Ans [Ans.. ln
$
%
$
%
2 ( 3 wrong ans : ln 2 3 ' 3 ;
1 2
ln 2 ]
! f (x) · d$g(x)% = ! f (x ) · g' (x) dx .
3.
b
d
! dx $f(x)% & "f(x)#
4.
b
if f(x) is continuous continuo us in (a, b) however if f(x) is discontinuous in (a, b) at x = c 2 (a, b)
a
a
b
'
d
! dx
c b $f (x)% & "f (x)#a ( "f (x)#c(
a
'
0 1 1 d 1 '1 1 . . '1 1 5 '1 1 5 !'10 // dx 0 / cot x -, -,, &cot x 34 '1 ( cot x 34 0( =6 – 1 0 / 346 -,. ( 64 & 62 1
e.g.
5.
Remember the values of the following def. integral (a)
! sin x dx = ! cos x dx = 1
6 / 2
!
sin x dx &
! sin
2
x dx &
0
6 / 2
!
3
0
6.
(b)
0
0
(c)
6 / 2
6 / 2
6 / 2
cos x dx & 3
0
2 3
!
! cos
2
x dx &
0
6 / 2
(d)
6 / 2
sin x dx & 4
0
6 / 2
!
cos 4 x dx &
0
6 4 36 16
If g (x) is the inverse of f (x) and f (x) has domain x 2 [a, b] where f (a) = c and f (b) = d then the value of b
d
a
c
ac) ! f (x) dx + ! g( y) dy = (bd – ac) Explanation: e.g.
f : [0, 1] * [e, e 1
I=
!e
e
x
f (x) = e
]
e e
!
dx ( 2 ln (ln x ) dx
0
if
e
e ex
then f – 1(x) = 2 ln(ln x);
Bansal Classes
hence I = e
e
Ans. ]
Page # 13
Illustrations: (Evaluating definite integrals by finding antiderivatives) sin x ( 1
8
!
1.
[Ans. 2 [cos2 – cos3] ]
dx ;
x (1
3
6 / 4
! cos 2x
2. (a)
4 ' sin 2x dx =
sin '1 x
!
3.
dx =
1 ' x2
0
8
62 8
7
1
!x
6.
!n(1
( 2x) dx =
dx
! 4 ( 5 sin x =
8.
!n 2
8
0
ex
! 1( e
1 3
!n2
x
(B)
2008
9.
(8 > 7)= 6; 5.
!
The value of the integral
0
(A) (2008)2
dx
! (1 ' 2x ) 2
0
dx = ln
0
(A*)
[Ans: – 2; wrong ans 4, 2]
1 / 2
(x ' 7 )(8 ' x) 7.
' t) dt 1' t
!n (1
0
dx
3 ln 3
0
6 / 2
!
4.
!
; (b)
3
0
1
1' e '2
8'3 3
4 3
3
;
2
ln 2
1 3 ( 1 . 1 // ,, ; l ( n ( 2 3 ) l n = or 1 ' x2 2 0 2 -
(C)
2 3
ln 2
(D) 2 ln
3 2
1 3x 2 ' 8028x ( (2007) 2 ( 1 . dx / , equals 2008 0
(B) (2009)2
(C*) 2009
(D) 1
e
! (x ( 1)e lnx dx x
10
[Ans. e + (e – 1)ee ]
1
11. (a)
29
3 ( x ' 2) 2
3
! 3( 3
3
[Ans. 11(a) 2 – x2 ' 4
4
12.
!
x
2
6 16
14.
4
2
ln 2 –
3
3
dx =
! 3(
dx ; (b)
( x ' 2) 3
2
3
26 3 3
( x ' 2) 3
( x ' 2) 2
Wrong ans.
[Wrong Ans.
32
2
3 32
,
9
,
256
8 3
ln
3 8
1 sin x ( sin 2x ( sin 3x ( ...... ( sin 7 x .
! /0 cos x ( cos 2x ( cos 3x ( ...... ( cos 7x -, dx
9 8
;
15.
! '
3
cot
'1
36
x dx =
4
1 b
17
26 3 3
eb
a
)
18.
!x
2 n (1
(
1 2
1 2 , 0 3
3
ln 2 ; 1 –
2 3
ln 2 ; 2 –
2 3
ln 2 (
26 3 3
]
2 e 2 ' e '2 x d ( ! nx ) 13. = (wrong ans : 0) 2 '1
!
]
[Ans. 1
ln /
2
16.
!
1 8
ln 2 ]
xe ' dx x
(Ans. 1 –
0
2 e
)
g
x n '1 n ' 2 x 2
z
Prove that
'
3
'
1
0
'1
1
( ( n ' 1)( a ( b) x ( nabj b n '1 ' a n '1 dx & 2( a ( b ) ( x ( a ) 2 ( x ( b) 2
e ' x dx , n 2 N [Ans. 2
0
1 n! . / , 0 2 -
] 1
1
!
19. Assume that f '' is continuous and that f (1) = 3, f '(1) = 2 and f ( x ) dx = 5. Find the value of
! x f ' ' (x) dx . 2
0
0
[Ans. 6] 20.
! 3
x 2 dx
(A) 186
( x ' 3)(5 ' x )
(B)
117 6
(C)
8
15 6 4
(D*)
336 2
26
GENERAL ASKING:
! [cos x(1 ( x) ( (1 ' x) sin x] dx 0
6
21.
Let I =
2
cos x
! a cos x ( b sin x
6
dx and J =
0
2
sin x
! a cos x ( b sin x dx , where a > 0 and b > 0. 0
Compute the values of I and J.
Bansal Classes
Page # 14
1
[Ans. I =
a
2
(b
2
1 a6 b . . // ( b ln1 / , ,, ; 0 a - 0 2
1 b6 b . . // ' a ln1 / , ,, 0 a - 0 2
1
J=
a
2
(b
2
]
b
Note-5 If f(x) is always +ve in (a, b) then
! f (x) dx is always > 0 (When a < b) a
Example : 26
dx
! 5 ' 2 cos x the substitution tan x2 = t is obviously wrong.
1.
0
6
6
dx
! 1 ( cos
2
0
x
2
sec x dx
&!
tan x = t is wrong.
2 ( tan x 2
0
ELEMENTARY DEFINITE INTEGRAL AFTER THE FIRST LECTURE Evaluate the following definite integrals. 1
!
Q.1 Let
16 ( 9x
0
Q.4 Given f ' (x) = 1
5' 4x
'1
6 / 4
2
0
cos x x
Q.6
$sin
0
3
x
!n2
dx 9 ( 4x
! 2
1 1 ' / 0 !n x
2
( cos3 x%
. , dx Q.7 2 !n x sin '1 3
!
dx Q.10
2
13
6 / 2
!
!
a. Q.2
Q.3
0
1 !1 //0
1 x ln x
x cos x cos 3x dx Q.14
0
! 0
ln x .
(
x
, dx , -
sin 4 x
( cos4 x
1 ( x 2 dx Q.11 x 3
Q.15
5 ( 4 sin x
0
sin 2 x
dx
Q.8
!
cos x dx (1 ( sin x ) (2 ( sin x )
0
x
dx
!
6 / 2
6 / 4
1
6 / 4
Q.13
& ln a . Find
e
x e'x dx
36 2 6 . 1 36 . 1 , f / , = a, f / , = b. Find the value of the definite integral ! f ( x ) dx . 0 2 0 2 62
sin 2 x . cos2 x
!
Q.9
(!
e
x dx
!
Q.5
2
dx
! 2
3
!
6 / 2
dx ( x ' 1)
Q.12
( x ' 1) (5 ' x)
2
x2
' 2x
12
2
dx
1 x ' 1 . !3 2 /0 3 ' x -,
dx
dx
! 1 ( cos 9·cos x 92(0, 6)
Q.16
0
ln 3
ex (1
2
!
Q.17
e2x
0 2
Q.21
(1
6 / 4
!
dx Q.18
dx
! x $x (1% 4
cos 2x
1 ' sin 2 x dxQ.19
0
6 / 2
Q.22
1
!
! 0
sin : cos :
$a
2
sin
2
x 3' x
1 / 2
dx
Q.20
! 0
dx
$1' 2x % 2
1' x 2
: ( b 2 cos2 :% d: a;b (a > 0, b > 0)
0
36 4
! $(1 ( x ) sin x ( (1 ' x) cos x % dx , (b)
Q.23(a)
3
0
6
! 6
x
(1 ( x cos x · ln x ( sin x ) dx Q.24
sin x
1
! x (tan'1 x)2 dx 0
2
Q.25 Suppose that f, f ' and f '' are continuous on [0, l n 2] and that f (0) = 0, f ' (0) = 3, ln 2
f (ln 2) = 6, f ' (ln 2) = 4 and
!e
'2x
ln 2
· f ( x ) dx = 3. Find the value of
0 1
Q.26
!x 0
1
Q.28
!
dx 2
( 2 x cos 7 ( 1
( 0
Q.30
b
Q.27
dx
!
1( x 2
a
( 1)(ln x ( 1) dx x 4x ( 1
1
Q.29
2x
· f ' ' ( x ) dx .
0
where '6<7<6
x x (x 2x
! e'
!
x5
0
1( x2 1' x 2
where a=
e ' e '1 2
& b=
e2 ' e
'2
2
dx
Suppose that the function f, g, f ' and g ' are continuous over [0, 1], g (x) ; 0 for x 2 [0, 1], f (0) = 0, g(0) = 1
!
6,
2009
f (1) =
=
2
>( f ' (x ) · g(x )=g
f ( x ) · g ' ( x ) g ( x ) ' 1 2
0
Bansal Classes
and g (1) = 1. Find the value of the definite integral,
2
g (x )
2
>dx .
(x) ( 1
Page # 15
6 / 4
!
Q.31
sin 9 ( cos 9 9 ( 16 sin 29
0
6 / 2
6
d9 Q.32
!
9 sin 9 cos9 d9 2
!
Q.33
0
(2 ( cos x )
0
6 / 2
1 ( 2 cos x 2
!
e
1 d 1 1 . . !'1//0 dx /0 1 ( e1 / x -, -,, dx 1
Q.37
! ln(x e
Q.38
x x
1
1
0
A
1 36
x ( sin x 1 ( cos x
dx
' x2 dx 2 (1( x) 1 ' x 2
x .
1 116 ( x .5 dx ,3 4 -4
! ?@cos /0 8 ' 4 -, ' cos /0 8
Q.39
)
!
Q.36 6
dx
!
Q.34
0
4 / 3
2 x 2 ( x ( 1 dx Q.35 Let A = then find the value of e A. 3 2 ( ( ( x x x 1 3 / 4
dx
2
2
0
6
Q.40 If f(6) = 2 &
! (f(x)+ f BB(x)) sin x dx = 5, then find f(0) 0
b
Q.41
|x|
!
x
a
6 / 2
!
Q.43
0
!n 3
dx
!
Q.42
f (x)dx, where f(x) = e 'x + 2e'2x + 3e'3x + .. )
!n 2
sec x ' tan x
cosec x
sec x ( tan x
1 ( 2 cosec x
1
dx
! x f ' ' (x) dx , where f (x) =cos(tan
– 1x)
Q.44
0
Q.45(a) If g (x) is the inverse of f (x) and f (x) has domain x 10
1
2
5], where f (1) = 2 and
! f (x) dx ( ! g( y) dy .
f (5) = 10 then find the value of (b)
5
2 [1,
1
Suppose f is continuous, f (0) = 0, f (1) = 1, f ' (x) > 0 and
1
! f (x) dx = 3 . Find the value of the definite 0
1
!
'1 integral f ( y) dy . 0
ANSWER KEY Q.1 21/3 · 31/2 Q.2 Q.8 ln
Q.15
Q.21
4
Q.9
3
Q.16
3
4
2 1 6
!n
6 1 e . Q.3 2 (a ' 3b ) Q.5 ' Q.4 /0 , 2 e 2 2
6 ln 3
Q.10
2
9
6 1
1
ln
32
sin 9
Q.22
17
Q.17
Q.11
6
Q.12
1 1 6
. / ( ln 3 ' ln 2 , 2 0 6 -
' b3 a 2 ' b2
1 a3 3
6
Q.23 (a) 2
3 2
– 1 +
Q.18
1 3
2 (1 ;
6
1
Q.6 e '
6
6' 3
Q.13
6
Q.19
16
36 2
1 6 2 . / , (b) / 6 ' 4 , 0 -
7
Q.25 13
Q.26
Q.30 2009
Q.31
Q.37
2 1( e
Q.43 6 /3
1 if 7 ;0; if 7 &0 2sin7 2 1
ln 3 Q.32
20
Q.38 ln 2 Q.44 1 '
Bansal Classes
Q.39 3
2 2
'
4 9
Q.27 1 Q.33
1 2
Q.28 0 Q.34
2
Q.40 3
Q.45
(a) 48 (b) 2/3
6 2
Q.35
Q.29 16 9
Q.41 | b | – | a |
2 !n 2
Q.14
Q.20
Q.24
Q.7 2 3
tan'1
1 2
!n
6 4
1 3
$2( 3 %
6 1 6 . 1 / '1,( !n2 4 0 4 - 2
36(8 24
Q.36 Q.42
6 2 1 2
Page # 16
PROPERTIES
OF DEFINITE WITH ILLUSTRATIONS
INTEGRAL
(A) PROPERTIES: ________________________________________________________________________________ b
b
P
!
1
!
f(x) dx =
a
P
f(x) =
a
a
!
2
f(x) dx = –
a
!
f(x)dx
b
b
c
!
3
;
f(t)dt
a
b
P
b
!
!
f(x) dx +
a
f(x) dx
provided f has a piece wise continuity
c
when f is not uniformly defined in (a, b) Illustrations 3 / 2
!
(i)
6
2
x[ x ] dx
1 ( cos 2 x
!
(ii)
2
0
0
'1 e
!
(iv)
ln x dx
(A*)
2 –
'e
e
!
(vi)
!
| 1 + 2cosx | dx =
3
f(x) dx –
+ 2 3
(B)
26 3
2
!
[x2 – x +1]dx (A) 1, (B)
! '
+ 3 + 3 3
2
!
f(x) dx +
(C*)
4 6 / 3
(C*)
5' 5
e
e
2
f(x) dx =
a
f (x) dx 26 3
(D)
+ 4 3
9' 5 2
0
a
a
4
!
2 6 / 3
3' 5
0
P
2
(D)
26
4 6 / 3
0
0
26
2
1 ' sinx dx (Ans. 4 2 )
2 6 / 3
(A)
(C) –
0
26
(viii)
(B) 2e
26
15 A (v) ! ? x ( 3 dx 24 '1 @
!
! C5x ' 9Cdx 0
2
3
(vii)
3
(iii)
dx = 2
if f ( x )isodd
! $f (x) ( f (' x ) % dx = [ 2 f (x)dx ! a
if f ( x )iseven
0
0
Proof :
Explain 1' x
1 / 2
!
Illustrations : (i)
sec x ln
1( x
' 1 / 2 2
! '
(iii)
(iv)
!
f(x) dx where f(x) =
x
7
| x sin 6x | dx [Ans.
36 ( 1
1
62
' 3x5 ( 3x3 ' x ( 1 2
cos x
' 6 / 4
32
! '
1 1 1 ( x . .,dx ' 1 [ x ] l n / ( / , = !'1 / 2 /0 2 0 1 ' x - -,
dx = 0 ; (ii) 6 / 4
| 1 ' x 2 | dx
2
(v)
1 / 2
= 2
]
2 1 '1 x '1 x ( 1 ., / tan tan ( (vi) ! / 2 ,dx x x 1 ( '10 3
b
b
P
5
!
f(x) dx =
a
f(a + b ' x) dx or
a
6
Illustrations : 1.
!
(A*) 6 (B) 26 (C) 36 (D) 56 2
2
sin 3 x
! sin x ( cos x
dx [Ans.
6 / 3
! 6
6 '1 4
100
sin2x ln (tan x) dx (Ans : zero); 4.
/ 6
Bansal Classes
!
f(x) dx =
0
0
3.
a
a
!
50
f(a ' x) dx
0
6
] 2.
1 . 1 sin 2008 x / , · !'6 2 /0 (2007) x ( 1 -, sin 2008 x ( cos2008 x dx 2
!n x !n x
!
( !n (150 ' x)
3 6 / 8
dx (25)
5.
! 6
/ 8
!n
[Ans.
6 4
]
1 4 ( 3 sin x . / , dx (zero) 0 4 ( 3 cos x Page # 17
6 / 4
!
6.
ln (1 + tan x) dx =
0
6 / 4
6 8
1
9.
!
6
cot – 1(1 – x + x2) dx (Ans. x sin x
! 6
11.
1 ( sin x
/ 4
6 / 2
x sin x cos x
!
14.
sin x ( cos x 4
0
1
! '
17.
dx (Ans. 6[
3
4
x4
1 / 3
1' x
4
dx
&
10.
6
6 / 4
62
15.
16
!
!x
! ax
ln x dx
( bx ( a
2
2
Home Work after P-5 : 1. sin x ' cos x
! 1 ( sin x cos x
!
x( a
2
2
6
dx (Ans. zero) 6
! sin x dx (Ans. 0)
7.
0
11.
16
ln 2
1 1 .
! ln/0 x ' 1 -,dx = 0
16.
0
dx 1 ( tan (nx ) n
6 ln 2 3 3
]
! f (cos 2x) cos x dx 6
2
(a > 0) (Ans.
2
sin 2 x
! 1 ( sin x cos x
4.
12
6
(Ans.
6 4n
2
! '6
sin 2 x
0
dx (Ans.
6 3 3
1
) 5.
! 0
ln (1 ( x )
1( x
2
1 2
ln ( 2 ( 1) )
dx (Ans.
6 8
· ln 2)
3 cos x ' cos x dx (Ans. 4 ) 3
2
1
)
2
! sin x ( cos x dx (Ans.
) 2.
0
sin 8x
6
1
6
64
'x
0
0
&
( 1%
6 ( 1' x )
=0
dx
a 2
!
2
0 3a 2
8.
2
( 2x ( 4
0
2n
= 6 /2
1
[Ans.
! f (sin 2x) sin x dx =
Prove that:
6
$17 ( 8x ' 4 x % $e
1 ( cos 2 x ( sin 2x
62
6.
tan x
dx
ln x dx
0
6
0
1( 2
1 2 x . dx , 0 1 ( x 2 -
Evaluate:
3.
!
=
' 10x ( 25
x dx
0
0
2
2x
2
dx
! 0
x 2 dx
2
– ( 2 – 1)]) 12. 4
6
8.
cos '1 /
)
6
! 2
GENERAL: For a > 0, b > 0, I =
(2)
1 6 . , 0 4 -
dx = /1 '
3
)
(1)
tan 2 x
1 ( ex / 4
– ln 2)
2
0
3 6 / 4
! '6
7.
ln 2
! tan
9.
0
'1 1 2 x ' 1
. dx / , 2 0 1 ( x ' x 6
! '6
(cos px ' sin qx ) 2 dx , p, q 2 I (Ans. 26)
2
!
12.
6
(Ans. 0) 10.
!
sin 29 sin 9 d9 (Ans.
0
sin 8x · ln (cot x ) cos 2 x
0
2
dx
6 4
)
(Ans. 0 )
2
13.
Evaluate :
!' $x f ( x ) ( x · f " ( x) ( 2%dx , where f (x) is an even differentiable function (Ans. 8) 3
2
2
36 2
14.
I=
! 6
[2 sin x ] dx
15.
2
dx
If for al real number y, [y] is the greatest integer less than or equal to y, then the value of the integral 3 6 / 2
! 6
[2 sin x] dx is
(A)
'6
(B) 0
/ 2
6
17.
[Ans. 0]
12
2
16.
ln x
! 1( x
!e 0
e cos x cos x
( e ' cos x
Bansal Classes
dx
[Ans.
6 2
(C*)
'
6 2
(D)
6 2
]
Page # 18
2a
P
!
6
a
a
!
f(x) dx =
0
!
f (x) d x +
0
f (2a ' x) d x
D
0
0
[2
if f (2 a
' x) & ' f ( x )
if f (2 a
' x) & f (x )
a
!
f ( x) dx
0 26
!
ILLUSTRATIONS : (i) I =
sin4 xdx
0 26
!
(ii) I =
cox x dx = 0 (iii) I =
1 ( 2 sin 2 x
0
6 / 2
!
dx
!
5
0
(vi)
6
6
6
ln (sin x)dx = '
2
0
! 0
6
1
! 0
!0 //0
(vii)
'1
x .
, , -
x 6
(viii)
x
0
tan
! x$sin
(a)
2
%
(sin x ) ( cos (cos x ) dx [Ans.
2x
0
26
( (cos x )
x (sin x )
! (sin x)
(c)
2n
0
2n
!
4
6
x sin x cos x d x =
0
na
!
P 7
2n
62 2
dx , n 2 N
dx
( (cos x )
26
(ix)
6 2
2n
, n 2 N
!
!
(iv)
x
6
sin x sin 4 x
0
= 0 (v)
!
x l n (sin x) d x (Ans. –
0
6
sin3 x cos3 x dx =0
0
6
· ln 2 );
62 2
· ln 2 );
2
!
( 2 cos 2 x )ln (sin 2x ) dx [Ans. –
0
2
ln 2 '
63 16
! x $sin(cos
6 2
ln 2 ]
]
6
]; (b)
!
2
%
2
x ) cos(sin x ) dx [Ans.
0
62 4
(sin 1) ]
[Ans. 62]
[Ans.
62]
36 2 128
a
f(x) dx = n
2
36
[Ans. 2
! (sin x)
(c)
dx (Ans.
dx
x (sin x ) 2 n
6
0
3
0 26
! 1 ( 2 sin
=2
sin '1 x
!
ln(1 – cos x)dx = – 6 ln 2; ) 1
dx
1 6 x . dx = ' ln 2 ; ln sin / 0 2 -,
1
ln 2 ;
2
f(x)dx where f(a + x) = f(x) n
2I
0
0
Illustrations: 2n
(i)
!
6
0
1 sin x ' A /0 ?@
!
5. dx [ . ] denotes greatest integer function. Note that: A 34 -, ?@ 200 6
1000
(ii)
sin x 2
e
x'[x]
dx = 1000(e – 1) (iii)
n6 ( v
(v)
!
2000 6
!
1 ( cosx dx = 400 2
0
0
| cos x | dx where
0
6 2
sin x 2
(iv)
! 0
5 = 0, n2N 34
dx 1 ( e sin x
[Ans. 4n]
(Ans. 10006)
< v < 6 & n 2 N = (2n + 2 – sinv) 2 n6
(vi)
! max . $sin x, sin ' (sin x )%dx equals (where n 2 I) 1
The value of the definite integral
0
(A)
n (6 2
' 4)
2
Bansal Classes
(B)
n (6 2 4
' 4)
(C*)
n (6 2 ' 8) 4
(D)
n (6 2
' 2)
4
Page # 19
(B)
DERIVATIVES OF ANTIDERIVATIVES (LEIBNITZ RULE) If f is continuous then h (x )
d
! f (t) dt = f ( h(x) )·hB(x) – f (g(x) ). gB(x) (integral of a continuous function is always differentiable)
dx
Theory :
g (x ) x2
d
!
dx
f(t) dt = f(x2).2x – f(x) . 1
(note that integral must not contain any function of x.)
x
! f (t ) dt = F(t) + C
Let
D
FB(t) = f(t)
Note : If f is continuous on [a, b] then the x
!
function defined by g(x) & f ( t )dt (a F x F b)
x2
!
E
2
f(t) dt = F(x ) – F(x)
a
is continuous on [a, b] and differenti able in (a, b) and g ' (x) & f(x).
x
x2
d
E
!
dx
f(t) dt = F B(x2) . 2x – FB(x) = f(x2) . 2x – f(x) ]
x
x2
e.g.
dt
! 1(
Let G(x) =
t
2
1.
If
x
(x > 0). Find G ' (9).
[Ans.
!
f(x) =
sin '1
!
t dt +
0
tan x
t dt
!
2.
1( t
1/ e
e 3x
e
2
y
, If
d y d x2
x2
d dx
!
6 4
sint (t ' x)dt; (b)
d
x2
dx ! ( x
dt
( t2)
2
x
x
(A) 2
2
!
!
z sin z d z & y =
1
z2 cos z2 d z , find
t
t2
! z lnz dz
(b) If x =
3. (a)
= k y, find k.
t2
6. (a) If x =
t dt =
0
= 1
)
cos '1
(x > 0)
[Ans. 48]
2
1 ( 4 t2
0
2
]
dt x > 0. Find derivative of f (x) w.r.t. ln x when x = ln 2
t
dt
!
dt t (1 ( t
1/ e
!n
2x
!
+
t
!
4. f (x) =
5. x =
cot x
5
cos2 x
sin 2 x
2 [0, 6 /2]
9
1
and y =
1
!z t
2
lnz dz find
2
(C) – 8
(B*) 4 dy dx
dy
.
(A*) – t2
dx
(D) – 4
(B) – 2t2
(C) 1 (D) – 1/t2
x2
7. Limit x * 0
! cos t
dt
0
x sin x x
! 9. Limit x *0
2
(Ans : 1)
8. f(x) =
! 0
t
sin 2
2
dt
t
then find Limit x *0
f ' ( x) x
&
1 4
2
x e t dt
0
1' e z2
10. If y =
x
(A*) – 1
x2
dx
! 1( x
3
0
find
d2y dz 2
(B) –
at z =1
1 2
(C) 0
(A*) – 2 (B) – 4
(D) – 2
(C) –
1 2
(D) –
1 4
x
11.
!
t Let f (x) is a derivable function satisfying f (x) = e sin( x ' t ) dt and g (x) = f '' (x) – f (x). Find the range 0
of g (x).
Bansal Classes
[Ans.
"'
2,
#
2 ]
Page # 20
ASKING PROBLEM: Lim
(1)
x
1
x *0
x
!t
3
2
t
1 dt = (1 3
4
0
x
Evaluate Lim x
(2)
x *)
t 2 ' x 2 . 1 e / !0 0 -,dt
[Ans. 1/2 ]
x
!
(3)
f ( t )dt = x cos (6x), for x > 0, f (4) is equal to
(A) 2
1
(B*) 1 (C)
2
0
Lim
(4)
x *0
1
x
(1 ' tan 2 t ) x!
1t
dt = Lim (1 ' tan 2x )
1x
=e
x *0
0
Lim ' x *0
1
(D)
4
tan 2 x x
= e – 2 Ans. (using L'hospital's Rule) x
Let f (x) be a continuous function such that f (x) > 0 for all x G 0 and $ f ( x ) %
101
(5)
100
of $ f (101) %
0
is
(A) 100
(C)
!
= 1 + f ( t ) dt . The value
(B*) 101
(C)
1
101
(D)
100
(101)100
DEFINITE INTEGRAL AS A LIMIT OF SUM Fundamental theorem of integral calculus
b
!
f (x) d x = Limit h [ f (a) + f (a + h) + f (a + 2 h) + ...... + f (a + n ' 1 h) ] * h 0 n*)
a
b
or
!
n
'1
r
0
+&
f (x) d x = Limit h h*0 n*)
a
where b ' a = nh
f (a + r h)
Note: Evaluating a definite integral by evaluating the limit of a sum is called evaluating definite integral by first principle or by a b initio method. D n h = 1 , we have Put a = 0 & b = 1 1
!
1
f (x) d x =
n
0
n
'1
+&
r
1 r . /0 , ; n
f
0
4
!
Examples : For C.B.S.E.
(x
2
replace
!
' x) dx ;
!' e
x2d x ;
a
(x
2
* dx; H *
( 1) dx
0
1
!
n
3
1 b
1
x
b
!
!
cos x d x ;
a
a
1
;
!
(2x
2
;
r
*
n
( 5) dx ;
0
b
dx ;
4
!
2
!
( x ( e 2 x ) dx ;
0
2
! e' dx x
1
b
dx x
4
x
;
! sin x dx a
EXAMPLES n2
1.
Limit
2.
Limit 1 n * ) n
3.
4.
n * )
Limit
3 / 2
( 1
1 n
(1
(
2
Limit n * )
Lim n *)
(n ( 2 ) 2
1
(
(
n * )
2 3/2
( ....... (
( ...... (
n
(2
3
( ......... ( 2
n2
"
n2
( (n ' 1) 2 #
A n(1 ? n 2 ( 12 ( @ 1 2
n n
2
n ( n ( 1) ( 2n ( 1)
Bansal Classes
n
4
n
n
2
(3 ( ...... ( ( 32
(B) tan '1 2 +
(
n (n ( 2) 2( 2n ( 2)
1 2
]
[ Ans. : l n 4 ]
4n
=
!
x dx
[ Ans. :
0
(2 ( ( 22
ln5
[ Ans. :
3/ 2
1
n n
(A*) tan '1 2 + 5.
(
(n ( 1) 2
n2
16 3
]
5 3 5n4 3
1 2
(
ln2
(C) tan '1 2 +
1 2
ln3
(D) tan '1 2 +
1 2
ln4
n (n ( 3) 3 (2n ( 3)
......... up to n terms
Page # 21
n . 1 1 '1 1 . 1 / , has the value equal to Lim / tan ,/+ n *) 0 n - 0 k &1 1 ( tan $k n % -,
6.
1 ( ln (cos 1)
(A)
(B)
2
1 ( ln (sin 1) 2 1 / n
"(n ( 1) (n ( 2) ....... (n ( n)#
1 ' ln (sin 1 ( cos 1)
(C)
(D*) 1+ ln (sin 1 + cos 1)
2
4
7.(a)
Limit
(b)
Limit
(c)
Lim
(D)
EVALUATING INTEGRALS DEPENDENT ON A PARAMETER (NOT IN IIT JEE SYLLABUS)
n * )
n
A ?$1 ( @
n * )
n
$ *)
2n
Cn
2 n2
1n
22 n2
4 n2
.
[wrong answer 4/e ; 4/e 2]
e
% . $1 ( % . $1 ( %
1 n2
%
= 32 n2
(A*) 4
6 n2
$
....... 1 (
n2 n2
2n n2
%
5 3 = 4
4 e
(C) 4/e2
(B) 4/e
(D) 2/e
Differentiate I w.r.t. the parameter within the sign of integrals taking variable of the integrand as constant. Now evaluate the integral so obtained as usual as a function of the parameter and then integrate the result to get I. Constant of integration to be computed by giving some arbitrary values to the parameter so that I is zero. 6 / 2
! sin(x ( c) dx
Explain by taking I =
0
EXAMPLES: tan '1 a x
1
!
1(a). I =
1 ' x2
x
0
)
1.(b)
!
Evaluate
d x
tan '1 ax ' tan '1 x
1
!
2.
0
!n
x
2
)
(1 ' a 2 x 2 ) (1 ' x ) 2
d x (a2 < 1)
! x(1 ( x ) dx
3.
[Ans:
'
tan 1 (ax)
6
[Ans:
2
2
0
1
!
4.
!n
1' x
0
6 / 2
!
6.
0
(1 ' a 2 x 2 )
ln
2
6
$
1' a
2
!
-I
5.
!
!n
(1 ( a cos x ) cos x
0
sin x
2
ln a ]
%]
6
dx
6
ln (a ( 1) ]
d x (a2 < 1)
1 1 ( a sin x . / , 0 1 ' a sin x -
[Ans.
'1
l n (a2 cos2 x + b2 sin2 x) dx
d x (| a | < 1) = 6sin – 1a
[Ans: 6sin – 1a]
(| a | < 1)
6 / 2
7.
a 2 ( 1 ., K J]
dx where a is a parameter..
x
0
6 N 1 M!n/ a ( 2 L 0
[Ans:
[Ans. 6 l n
0
a
(b 2
]
(E)
ESTIMATION OF DEFINITE INTEGRAL AND GENERAL INEQUALITIES IN INTEGRATION: Not all integers can be evaluated using the technique discussed so far.
(a)
For a monotonic increasing function in (a, b) b
(b – a) f(a) <
! f (x ) dx < (b – a) f(b) a
(b)
Foa a monotonic decreasing function in (a, b) b
f(b). (b – a) <
! f (x) dx < (b – a) f(a) a
(c)
For a non monotonic function in (a, b)
Bansal Classes
Page # 22
b
f(c) (b – a) <
! f (x) dx < (b – a) f(b) a
(d)
In addition to this note that b
b
! f (x) dx
! | f (x) | dx
<
a
equality holds when f (x) lies completely above the x-axis
a
Illustrations : (1) (a) 2
(3)
! 1 / 2
(4)
6 / 2
6 128
<
! (sin x)
10
6 / 4
x
4 dx
dy
(
x
dx <
6 4
& 0 when x =
dx
6
, (b) 1 <
2
!
sin x x
0
1 ln 4
O
2 3
6
dx <
;
2
(2)
e '1 3
! 2 ( lnx < e 2' 1 ; e
<
dx
1
)
Given that f satisfies | f(u) – f(v) | < |u – v | for u and v in [a, b] then prove that b
(i)
(5)
(F) (a)
f is continuous in [a, b] and
(a)
General
1 2
1
P! 0
4'x
(x
!
(ii)
dx 2
2 f ( x ) dx ' (b ' a ) f (a ) < (b ' a ) 2 a
P
5
6
6
;
6
1F
(b)
2
!
1 ' sin 3 x dx
0
F
1 2
$
%
2 ( ln (1 ( 2 ) .
DETERMINATION OF FUNCTION Find a function f, continuous for all x (and not zero every where), such that x
f 2(x) =
f ( t ) sin t
! 2 ( cos t dt
[ Ans :
0
1 2
ln
3 2 ( cos x
]
6 / 2
(b)
If f(x) = x +
! sin(x ( y) .f (y) dy
where x and y are independent variables, find f(x).
0
(G)
WALLI'S THEORM & REDUCTION FORMULA 6 / 2
! sin
n
x cos m x dx =
[(n ' 1)(n ' 3) ....1or 2 ] [ (m ' 1)(m ' 3)...1 or 2]
0
6
where K =
K (m, n are non-negative integer)
(m ( n ) ( m ( n ' 2) ...1or 2
if m , n both are even
A2 ? ?@ 1
otherwise
Examples: 36 2
26
(a)
! x sin
6
4
x cos xdx
(b)
! cos
4
6 / 2
! x (sin x )
If un =
0
6 / 4
(c)
! (tan 9)
If In =
0
n
n
dx , n > 0, then prove that u = n
n
u n '2 (
!
'1 n If un = x tan x dx then prove that (n +1) u n + (n – 1) un – 2 = 0
(e)
n '1
sin nx
! sin x dx ,
1 n
2
]
; n2un – (n – 1)un – 2 = 1
d9 then prove that n (I + I ) = 1 n+1 n – 1
1
(d)
16
0
0
(b)
56
3x · sin 2 6x dx [Ans.
6 2
'
1 n
n 2 N
Bansal Classes
Page # 23
(H)
DIFFERENTIATING AND INTEGRATING SERIES
(1)
Find the sum of the series
(a)
x
2
1. 2
'
x
3
x
(
2.3
4
' ..... ( ('1)
3.4
n 1 x (
n (1
n (n ( 1)
(d)
If CxC < 1 then find the sum of the series
(c)
Let f (x) =
1 2
tan
x 2
A
(
1 2
2
x
tan
2
x
x
! f (x) dx = – ?@ln cos 2 . cos 2 [Note : Limit cos
x
n *)
2
. cos
x 2
2
. cos
x 2
3
(
2
2
1 2
3
. cos
......cos
tan x 23 x
2
n
x 2
3
( ...... 1 1( x
&
(
( .....)
......cos sin x x
[ Ans: (x + 1) ln (1+ x) – x ]
|x|<1 2x 1( x
( 2
4x 3
8x 7
1( x
1( x
( 4
( ......) . 8 1
then prove that f (x) =
x
[Ans.
1 1'x
]
– cot x
1 sin x . 5 / , = l n – 3 0 x 2n 4 x
]
)
(d)
Prove that
+& 1 /0 3n1( 1 ' 3n1( 2 . -, = 3 63
n 0
(I)
SOME INTEGRALS WHICH CANNOT BE FOUND IN TERMS OF KNOWN ELEMENTRY FUNCTIONS. sin x
(1)
!
(6)
!
x tan x dx
(10)
!
(1 ( x )
x
dx
2 1 / 3
Bansal Classes
(2) (7) dx
!
cos x
!e
x 'x 2
(11)
dx
dx dx
! ln x
(3)
!
(8)
!e
x
(12)
!
1 ( k 2 sin 2 x dx k 2 R
sin x dx 2
dx
(4) (9)
! sin x
2
x3
! 1( x
5
dx
(5)
! cos x
2
dx
dx
Page # 24
