DEFINITE INTEGRATION

SECTION - I : STRAIGHT OBJECTIVE TYPE


sin x cos 
12.1  1  cos 2 x dx =  1  sin2 
0

(A) for no value of  (B) for exactly two values of  in (0, )

   
(C) for at least one  in  ,   (D) for exactly one  in  0, 
 2   2

1/ n
 n 3 3

1
dx
 (n  r ) 
 r 1
 
12.2 Let  =  , p = nlim
 
, then np is equal to
1  x 3
n 3n 
0  
 

(A) n 2 – 1 +  (B) n 2 – 3 + 3 (C) 2 n 2 –  (D) n 4 – 3 + 3

2
e|sin x| cos x
12.3 0 1 e tan x
dx =

(A) e (B) 1 (C) e – 1 (D) 0

12.4 Consider the integrals

1 1 1 x2 1 x2
2  
I1 =  x
e cos2 x dx, I 2 =  e  x cos2 x dx, I3 =  e 2 cos 2 x dx , I 4 =  e 2 dx
0 0 0 0
Then
(A) I2 > I4 > I1 > I3 (B) I2 < I4 < I1 < I3 (C) I1 < I2 < I3 < I4 (D) I1 > I2 > I3 > I4

2
12.5 0
x  x  x  ... dx is equal to ( x > 0)

19 17 13
(A) (B) (C) (D) Can’t determine
6 6 6

1
x
12.6 A function f(x) which satisfies the relation f(x) = ex + e f (t )dt , then f (x) is
0

ex ex
(A) (B) (e – 2) ex (C) 2ex (D)
2e 2
 0 2
| sin x | | sin x |
12.7 If m =  x 1
dx and n=  x 1
dx, where [.] represents greatest integer function, then
2
   2 0
  2
   

(A) m = n (B) m = –n (C) m = 2n (D) m = –2n

e
m
12.8 If Im =  ( n x )
1
dx , m  N, then I10 + 10I9 is equal to

e10
(A) e10 (B) (C) e (D) e – 1
10

1 1 1 1
12.9 If Sn =   + ........... + , n  N, then lim Sn is equal to
2n 2
4n  1 4n2  4 2
3n  2n  1 n

 
(A) (B) 2 (C) 1 (D)
2 6

1 1 2
et te t
12.10 Let A = 
0
dt
t  1 , then the value of t 2
1
dt
0

1 1 2
(A) A2 (B) A (C) 2A (D) A
2 2

x
1
12.11 The function f (x) =  t dt satisfies
1

x
(A) f (x + y) = f (x) + f (y) (B) f   = f (x) + f (y)
y
(C) f (xy) = f (x) + f (y) (D) None of these

a p

12.12 If f is a function with period P, then  f (x) dx is

a

(A) equal to f(a) (B) equal to f(p) (C) independent of a (D) None of these


2
12.13  (cos px  sin qx ) dx is equal to



(A) 0 (B) (C)  (D) 2
2

1
1 2x  1 
12.14  tan
0
 2
 1 x  x 
dx is equal to

(A) 0 (B) 1 (C) – 1 (D) None of these

 b
3 4 (b – a)m
12.15 The value of  (x – a) (b – x)
a
dx is
n
. Then (m, n) is

(A) (6, 260) (B) (8, 280) (C) (4, 240) (D) none of these

19
sin x
12.16 The absolute value of  1 x
10
8
dx , is

–7
(A) less than 10 (B) more than 10–7 (C) less than 10–6 (D) more than 10–6

a
 
12.17 Evaluate  n (cot a  tan x )dx ; a   0, 2 
0

(A) a n (sin a) (B) – a n (sin a) (C) n (cos a) (D) – n (cos a)

1 /4
12.18 S1 : In
 f (cot 1 x )dx putting cot –1 x = t may change the limits to  .......
1 3 / 4

S2 : If f(x) has removable discontinuities at finite number of points in (a, b) then if  f ( x )dx  F( x ) ,
b

 f ( x )dx = F(b) – F(a).

a

b
S3 : If f(x) has an infinite discontinuity in (a, b), then we can always write  f ( x )dx = F(b) – F(a)
a

where  f ( x )dx = F(x)

1
S4 : If f(x) : [0, 1]  R has single point continuity in (0, 1) then
 f ( x )dx can be evaluated.
0

(A) FTTF (B) TFFT (C) FFFF (D) TTFF


d  1   1
12.19 S1 :   x sin  dx = 2sin2   
dx  x 4 2
1

10 
2 5
S2 :  sin x cos 4 x dx =
8
0

x
S3 : If f(x) is even then f ( x )dx will always be odd.

0

S4 : If f(x) is even and  f ( x )dx is odd where f(x) is a continuous function then f(x) = 0 must have
0

exactly three roots in (– 1, 1).

(A) FTTF (B) TFFT (C) TTTF (D) FFFT

 
log(1  x 2 )
12.20 S1 : 
0
1 x2
dx =  log 2


sin2 x
S2 : e cos 3 ( 2n  1) x dx = 0
0

4
S3 :  | x  1 |dx = 5
0

2
2 4
S4 :  (1  x ) dx =
3
2

(A) FTTF (B) TFFT (C) TTTF (D) FFFT

SECTION - II : MULTIPLE CORRECT ANSWER TYPE

2
12.21 If f(x) is integrable over [1, 2], then
 f ( x ) dx is equal to
1

n 2n n 2n
1 r  1 r  r n r 
 f   f   (C) lim 1 1
(A) nlim
 n
r 1 n
(B) nlim
 n
r n1 n n  n 
r 1
f
 n 
 (D) nlim
  n
 f  n 
r 1

1
dx
12.22 If n =  n , n  N, then which of the following statements hold good?
0 1  x  2

 1
(A) 2n n + 1 = 2 n + (2n  1) n (B) 2 = 
8 4

 1  5
(C) 2 =  (D) 3 = 
8 4 16 48

2 50
12.23 If f(2 – x) = f(2 + x) and f(4 – x) = f(4 + x) and f(x) is a function for which f ( x )dx = 5, then   f ( x )dx is equal
0 0

to
46 51 52
(A) 125 (B)  f ( x )dx (C)  f ( x )dx (D)
4 1
 f ( x )dx
2

1 x
2
12.24 If F(x) =
x 2  ( 4t
4
 2F( t )) dt, then F(4) equals –

32 64 F(8) 11F( 8)
(A) (B) (C) (D)
9 9 28 28
 /2
12.25 If I =  e  sin x dx, where   (0, ), then
0

  –  
(A) I < (B) I > (e + 1) (C) I > e (D) I > 0
2 2 2

SECTION - III : ASSERTION AND REASON TYPE

/ 4 1  sin x
12.26 Statement-1 :  0
sec x
1  sin x
dx = 2 – 2

/4 sec x
1 
Statement-2 :  0 2
log( 2 – 1) +
1  2 sin 2 x
dx =
6 2
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True

1 2 4 2
12.27 Statement-1 :
 2 X x2 dx = – +
0  n2 (n2)2 (n2)3

1
Statement-2 :  e X (x – 1)ndx = 16 – 6e, then value of n is 3
0

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True

3
d e sin x 4 3e sin x
12.28 Statement-1 : If
dx
F(x) =
x
, x > 0 and 1 x
dx = F(k) – F(1) then one possible value of K is 64.

 1 cos ec
Statement-2 : If f(x) is a function satisfying f   + x2f(x) = 0  x  R0 then
x  sin 
f ( x ) dx = sin– cosec

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True

 2 x(1  sin 2 x )
12.29 Statement-1 :  1  cos 2 x
dx = 0

2
4
Statement-2 :  sin
2
x cos 2 xsin x  cos x  dx 
15
0

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
 n
n
12.30 Statement-1 :
 xdx = 2 , where {.} represents fractional part function and n  N.
0

n
n(n  1)
Statement-2 :  xdx =
0
2
, where [.] represents greatest integer function and n  N.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
 

12.31 Statement-1 :  x sin x cos 2 x dx   sin x cos
2
x dx
2
0 0

b b
ab
Statement-2 : 
a
x f ( x ) dx 
2  f ( x ) dx
a

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True


3
dx 
12.32 Statement-1 :  8 cos
0
2
x 1

18

2
3
dx 
Statement-2 :  8 cos
0
2
x 1

18

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True

f(x)
4t 3
12.33 Statement-1 : Let f be real valued function such that f(2) = 2 and f(2) = 1, then xlim
2  dt = 12
x2
2

v( x )
Statement-2 : Let f(x) =  g(t ) dt , then f(x) = g(v(x)) v (x) – g(u(x)) u(x)
u( x )

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
 /2
sin x 
12.34 Statement -1 : 1  
0
x  2

Statement-2 : If f(x) is continuous in [a, b] and m and  are greatest and least value of f(x) in [a, b], then
b
(b–a) 
 f ( x) dx  m(b – a)
a

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True

SECTION - IV : COMPREHENSION TYPE

Read the following comprehensions carefully and answer the questions.
Comprehension # 1

Definite integral of any discontinuous or non-differentiable function is normally solved by the property
b c b
 a
f ( x ) dx =  a
f ( x ) dx +  c
f ( x ) dx, where c  (a, b) is the point of discontinuity or non-differentiability..


12.35 The value of A =  [cos ec 1x ] dx, {where [ . ] denotes greatest integer function} ,is equal to
1

(A) cosec1 – 1 (B) 1 (C) 1 – sin 1 (D) none of these

100
12.36 The value of B =
 [sec 1 x ] dx, {where [ . ] denotes greatest integer function} ,is equal to
1

(A) sec 1 (B) 100 – sec 1 (C) 99 – sec 1 (D) none of these
B
12.37 The value of integral  [tan 1 x ] dx , {where [ . ] denotes greatest integer function} ,is equal to
A

(A) tan 1 (B) 100 – tan 1 – sec1 (C) 99 – sec1 (D) none of these

Comprehension # 2

/2 / 2 
Using integral  n(sin x ) dx = –  n(secx) dx = – n 2,
0 0 2
/2 /4 
n(tan x ) dx = 0 and n(1  tan x ) dx = n2 .
 0  0 8

/4  sin x  cos x 

12.38 Evaluate  n  dx =
 / 4  cos x  sin x 

n2
(A) n 2 (B) (C) 0 (D) –n 2
2
/4
12.39 Evalute  n (sinx + cosx)dx =
 / 4

n2 n2
(A) (B) (C) n 2 (D) 0
2 4
 /4
12.40 Evalute  n (sin2x) dx =
0

 n2 n2 n2

(A) (B) n 2 (C) (D) –
2 4 4

Comprehension # 3

b nb  c
b 1 r
Integral  f ( x ) dx can be represented as a limit of a sum of infinite series
 f ( x ) dx = nlim
  f 
a a
r  na  c
n n

where na +c  r  nb + c, r, n  N, c  R and any limit of sum of series of same form can be changed to
definite integral by replacing

1 r
(1) lim
n    (2)
n
 dx (3)
n
x

r  na  c 
(4) Lower limit = lim   = lim  =a
n   n 
min
n   n 

r  nb  c 
(5) Upper limit = lim   = nlim

 =b
n n
 max  n 

 n n n 1 
12.41 Find the value of lim     ..... 
n  (n  1) 2n  1 (n  2) 2(2n  2) (n  3 ) 3(2n  3) 2n 3 


  
(A) (B) (C) (D) none of these
3 2 4

1
12.42 The nth term of the corresponding series of  tan 1x dx is
0

 1 
(A) (B) tan–1(n – 1) (C) (D) tan–1n
4n n 2n

2n1
1 r
12.43 lim
n  n sec  n 
r 0
2
is

(A) sec 2 (B) tan 2 (C) sec2 (D) not defined

SECTION - V : MATRIX - MATCH TYPE

12.44 Match the following :

Column –  Column – 

1 6
cos x cos( x / 3)
(A) If  1 x
dx = k and  6  3  x
dx = mk , (p) 4
0 6  3

then m is

1
 1  3 
(B)   sin x    dx = , then k is (q) 1
1   4 k

{where [ . ] denotes greatest integer function}

3
3
(C) If f(x) = max (x – | x |, x + [x]) and  f ( x ) dx = – , (r) 0
k
3

then k is {where [ . ] denotes greatest integer function}

20

(D) If  1  cos x dx = 10k 2 , then k is (s) 2

0 

(t) 3

12.45 Match the following :

Column –  Column – 

10 [ x 2 ]dx 1
(A)  4 [ x 2  28 x  196 ]  [ x 2 ]
= (p)
101
{where [ . ] denotes greatest integer function}

2 x
(B)  1 x
dx= (q) 3

99 99 99
(C) lim 1  2  ...  n = (r)
1
n 
n100 3

1 1
(D) 5050  x 200 dx = , then  = (s) 1
1 

1
(t)
100
12.46 Column – I Column – II
1
3x 2
(A) 
1
1  4 tan x
dx = (p) 7

8
sin x 2 dx 1
(B) 
6
sin x 2  sin ( x  14 )2
= (q)
2
13
1
(C)
156  [x] dx =
1
(r) 1

0
1
(D)
  n2
/2
 n sin 2x dx = (s) 2

(t) 0

12.47 Column –  Column – 

n1 2 2
n r 
(A) Lim
n  r 1 n2
= (p)
4

4 
 
(B) (q)
tan x  cot x dx  2
0

1
3 
(C)  sin x cos 2 x dx  (r) –
4
1

2
sin3 x dx 
(D)  3
sin x  cos x 3
= (s)
2
0
(t) 0

SECTION - VI : INTEGER TYPE

1  2p  2p 2  2p 3  2p     2p  1 . 2p  3 . .......... . . 1 ,  
12.48 Limit cos 2n  cos 2n  cos 2n ......  cos 2  =  , p  N ,
n n     2p 2p  2 2 2
then find 

2
dx 
12.49 
0
a  bcosx  csinx
=
a b 2 c 2
2
2 2
where a > b  c > 0, then find 

12.50 A function f: R  R satisfies the equation f(x +y) = f(x) + f(y)  x, y R and is continuous throughout the
n

domain. If 1 + 2 + 3 + 4 + 5 = 450 where n = n  f ( x ) dx and f(x) = x, then find 

0

1
1 1 11  1
12.51  (5  2 x – 2 x
0
2
) (1  e( 2 – 4 x ) )
dx =

log
10
, then find 