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KOCHI: Bldg.No.41/352, Mulloth Ambady Lane, Chittoor Road, Kochi - 11, Ph: 0484-2370094
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Definite - Integration
UNIT - 5
S O L U T I O N S
content
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w w w . m a t h i i t . i n
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1. ( )2
13
1 1 12 log log 2
0 0 00
1.
3 3
x x xc e dx e dx x dx
= = = =
2. ( ) ( ) / 4 / 4
2 2
0 0tan sec 1a xdx x dx
=
[ ] [ ] /4 /4 /4 /42
0 00 0sec 1 tan 1 .
4 xdx dx x x
= = =
3. ( )c / 2 / 2 / 2 / 2
2
0 0 0 02
sin sin 1sec tan .
1 cos 2 2 22cos
2
x x x x x xdx dx x dx dx
xx
+ += = +
+
/ 2
0
tan tan .2 2 4 2
xx
= = =
4. ( )/ 2
0sinxb Let I e x dx
= / 2/ 2
0 0cos cos
x xe x e x dx
= + / 2 / 2 / 2
0 0 0cos sin sin x x x
e x e x e x dx
= + ( ) ( )
/ 2/ 2
02 sin cos 1
x I e x x e
= = +
Hence ( )/ 2
/ 2
0
1sin 1 .
2
xe x dx e
= +
5. ( )2 2
2
211
1 1 1.
2
x x ec e dx e e x x x
= =
6. ( )a Put sin cos , x t x dx dt = =
so that reduced integral is
( ) ( )1 1
00
1 1log 1 log 2
1 2dt t t
t t
= + + + + 2 1 4
log log log .3 2 3
= =
7. ( )( )
/ 2
5 / 2/ 3
1 cos 1 cos
1 cos1 cos
x xb I dx
xx
+ =
( )/ 2
3/ 3
sin
1 cos
xdx
x
=
Now put 1 cos x t = Also, when1
,3 2
x t
= = and1
, 12
x t= =
Therefore,
121
31/ 21/ 2
3.
2 2
dt tI
t
= = =
Trivandrum: 0471-2438271 Kochi: 0484-2370094
w w w . m a t h i i t . i n
S O L U T I O N S
LEV EL - 1 (Fundamentals of Definite Integration)
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
8. ( )d Put2
1 1,t dt dx
x x= = then it reduces to
1/2 1/ 21/ 2 1
11
1.t t
ee dt e e e
e
= = = 9. (a) Put 2tan sec x dx d = =
As 1 0 04
x and x
= = = = , then
( ) / 4 / 4
/42
0
0 0
2 sec 2 tan 2 tan I d d
= =
10. ( ) ( )2
4 22
3
22
4 .4 2
ax bxc ax bx c dx cx c
+ + = + + =
Hence depends on c.
11. ( ) [ ]
/ 4 / 4
/ 6/ 6
1cos 2 tan2d ec x dx log x
=
1 1
tan tan log 3.2 4 6 2
log log = =
12. ( ) ( )1 1
log log log logb bb
aa ac Let I xdx x x xdx
x x= =
( ) ( ) ( )2 2 21
2 log log log2
b
a I x I b a = =
( )( ) ( )1 1log log log log log log2 2
bb a b a aba
= + = .
13. ( )a Put 2tan sec x dx d = = Also as 0, 0x = = and 1,4
x
= =
Therefore1 / 4
1 2
0 0tan sec xdx d
= 1
log 2 log 2.4 4 2
= =
14. ( )( )
1
20
dxd Let I
a b x b
= +
Put ( ) ( )t a b x b dt a b dx= + =
As 1 0 , x t a and x t b= = = = then
( )21 1 1 1 1 1
aa
bb
a b I dt
a b t a b t a b ab ab
= = = =
15. ( )/ 2
2/ 4
1cos
sina Let d
. Put sin cos ,t dt d = = then we have1
1
21/ 21/ 2
1 1
2 1.dtt t
= =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
16. ( )( )
11 2
3/ 20 2
sin
1
xb I dx
x
=
Put 1 21
sin1
x t dx dt x
= =
and sinx t=
Also 04
t to as
= 102
x to= / 4
2
0
1.sec log 2.
4 2 I t t dt
= =
17. ( )/2
0 2 cos
dxc I
x
=+
/2
0 2 2 2 22 sin 2 cos cos sin
2 2 2 2
dx
x x x x
=+ +
2
/ 2 / 2
0 02 2 2
sec2
sin 3cos 3 tan2 2 2
xdx
dx x x x
= =
+ +
Put 21tan sec ,2 2 2
x xt dt dx= = then
11
20
2 12 tan .
3 3 3
dtI
t
= = +
18. ( )d Put 12
1tan ,
1t x dt dx
x
= =+
then
/41 2 2
1 /4
20 0
tan.
1 2 32
x tdx tdt
x
= = = +
19. ( )d Put2
1 1t dt dx
x x= = as
2t
= and
[ ]2/
2 /21/ / 2
1sin
sin cos cos cos 1.2
xdx t dt t
x
= = = =
20. ( )2
/2
0sin
2 4
x xc Let I e dx
= +
1
00
12 sin 2 sin tan
4 4 11 1
tt e
I e t dt t
= + = + +
[ ]0
2 2sin 0 0.
2 2
te t
= = =
21. ( )b Put 1 ,x xe t e dx dt + = = then we have
( )( )1 11 12 2
1 11 1e e
t dtdt
t t
+ + = =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
[ ]1
1
2
1 1log log 1 1 log 2 2ee et t e
e e
+ = = + + +
1 1log 1.
2e
e
e e
+ = +
22. ( )a Let
/4
0
sin cos
9 16 sin 2
x x
I dxx
+
= + Put sin cos , x x t = then ( )sin cos x x dx dt + =
( )
0 0
221 1 25 169 16 1
dt dt I
tt = =
+ =0
1
1 1 1
10 5 4 5 4dt
t t + +
( ) ( )0
1
1 1log 5 4 log 5 4
10 4t t
+ ( )
1 1log 9 log1 log 3.
40 20= =
23. ( )c Let ( )/2
/4logsin cot
x I e x x dx
= + or
/ 2 / 2
/ 4 / 4log sin cotx x I e x dx e xdx
= + /2 /2
/4/4log sin log sinx xe xdx e x
= +
/2
/4logsinxe xdx
/ 2 / 4 / 41log sin log sin log 2.
2 4 2e e e
= =
24. ( )b Put 12
1sin ,
1t x dt dx
x
= =
then
[ ]1
1/ 2 /6 /6
020 0sin sin cos sin1
x x I dx t tdt t t t x
= = = +
3 1 1 3.
6 2 2 2 12
= + =
.
25. ( )a Put 2 cos 2 sin , x dx d = = then2 0
0 /2
2 1 cos2 sin
2 1 cos
xdx d
x
+ +=
( )
( )
/2
0
cos / 2
4 sin cossin / 2 2 2 d
= ( )/2
02 1 cos d
= +[ ]
/ 2
02 sin 2 1 2
2
= + = + = +
26. ( )c ( )220 0 01 sin
sec sec tan1 sin cos
dx xdx x x x dx
x x
= =
+
[ ] [ ] [ ]0
tan sec tan sec 1 0 1 1 2.x x
= = + = + + =
27.
22 2
00 0
( ) 1 sin sin cos 4 sin cos2 4 4 4 4
x x x x xc dx dx
+ = + = = 4(1-0-0+1) = 8
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
28. ( )b 11
1 1 2
0 0
cos cos 1 1. xdx x x x = =
29. ( )c /2
0
cos
1 cos sin
xdx
x x
+ +
( ) ( )
( ) ( ) ( )
2 2/2
20
cos / 2 sin / 2
2 cos / 2 2sin / 2 cos / 2
x xdx
x x x
=
+
( )
( )
2 / 2 / 2
0 0
1 tan / 21 11 tan
2 1 tan / 2 2 2
x xdx dx
x
= = +
=1 1
log log 2.4 4 22
+ =
30. ( )d . Let ( )/ 6
2
02 3 cos 3 I x x dx
= +
( ) ( )/ 6
/ 62 2
00
sin 3 sin 3 12 3 6 . 16
3 3 36
x x x x dx
= + = + .
31. ( )d Put 2sin 2sin cos x t dt x xdx= = .
Now / 2 1 1
1
4 2 00 0
sin cos 1 1 1tan .
1 sin 2 1 2 8
x xdx dt t
x t
= = = + + 32. ( )a Put 2tan sect x dt xdx= = .
Now / 4 1 1
6 2 6 7
00 0
1 1tan sec .
7 7 x xdx t dt t
= = =
33. ( )b Let / 6 / 6
2
30 0
sintan sec
cos
x I dx x xdx
x
= = .
Put 2tan sec ,t x dt xdx= = then we have
11 2 33
00
1.
2 6
t I t dt
= = =
34. ( )b Let/ 2
20
sin cos
cos 3cos 2
x xdxI
x x
=+ + . We put cos sin , x t xdx dt = = then
1 1
20 0
. 2 1
3 2 2 1
t dt I dt
t t t t
= = + + + + = [ ]1
02 log( 2) log( 1) [2 log 3 log 2 2 log 2]t t+ + =
[ ] [ ]9
2 log 3 3log 2 log 9 log8 log .8
= = =
35. ( )d( )
1 1
22 20 02 cos 1 cos 1 cos
dx dx
x x x =
+ + + +
( )
1
1 1
2 200
1 costansin sincos sin
dx xx
+ = = + +1 11 1tan cot tan cot . .
sin 2 2 sin
= =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
36. ( )b 2
31 1
21
1tan tan
1
x x I dx
x x
+ = + +
31 1
2 21tan cot
1 1
x xdx
x x
= + + + 3
12 .
2dx
= =
37. ( )a Put logx u= in 1,I so thatu
dx x du e du= = Also as x e= to2
, 1e u = to 2
Thus,2 2
11 1
.u x
e e I du dx
u x= = Hence 1 2.I I=
38 ( )a ( )
/ 2/ 4
/ 4/ 4
sin sin cos2
xx e
e xdx x x
=
( )/ 2
/ 4
1sin cos
2
xe x x
=
( )/ 2 / 2 / 41 1 11 0 .
2 22 2
ee e = =
39. ( )c ( )
( )
( )
( )
/ 2 / 2
2 20 0
1 2 cos 2 cos 2 3
2 cos 2 cos
x xdx dx
x x
+ + =
+ +
( )
/ 2 / 2
20 02 3
2 cos 2 cos
dx dx
x x
= + + ( )
21 1
220 0 2
14 6
3 3
dt tdt
t t
+=
+ +
2
x put tan t
=
( )1 1
220 0 22 12
3 3dt dt dt
t t= ++ +
1
1 1
2 2 20 00
1 12 123 6 3 6 3
dt t dt t t t
= + + + + + 1
2
0
12 .3 2
tt
= = +
40. ( )c ( )
02 2 2 2 21 cos sin 2 cos sin
2 2 2 2
dx
x x x xa a
+ +
( ) ( )0 2 22 2
1 cos 1 sin2 2
dx
x xa a
= + +
( ) ( ) ( )( )2 2 20
2;
1 1 / 1
dt
a a a t
=
+ + + { where tan
2
xt = }
( )
( )
( )1
2
0
12 1tan
1 11
a at
a aa
+ + = + ( )
1 1
2 2
2tan tan 0
11 aa
= = .
41. ( )b Required value ( )1
10
0
1 1.
10 10
x = =
42. ( )b Required value is/ 3
0
sin30.
3
x
=
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
43. ( )a / 4 / 4
0 0
1 tantan
1 tan 4
xdx x dx
x
+ = + /4
0
1log sec log 2.
4 2x
= + =
44. ( )b 1 1
20 0 1
x
x x xdx e dx
e e e =+ + Now put
x xe t e dx dt = =
Also as 0x = to 1, t = 1 to e , then reduced form is
1 1
2 11
1tan tan ,
1 1
e edt et
t e
= = + + 1 1 1
tan tan tan1
x yx y
xy
= + Q
45. ( )a 1 1 1
1 log 1 loge e ex x I dx dx dx
x x x
+= = +
[ ] ( )2
1
1
log 3log .2 2
e
e
exx + =
46. ( )c Integrate it by parts taking log 12
x +
as first function
22 2
1
00
1 1log 1
2 2 2 21
2
x x xdx
x
= + +
21
0
1 3 1log
2 2 2 2
xdx
x=
+ 1 3 1 1 3 3 2
log 2 4 log 3 4 log 2 log2 2 2 2 4 2 3
= + = +
On comparing with the given value3 3
,4 2
a b= = .
47. ( )b ( )
( )( )1 1
0 0
1
1 1 1
x x dxdxI
x x x x x x
+ += =
+ + + +
( )1 1 1
0 0 0
1 4 21 .
1 3
x xdx x dx xdx
x x+ += = + + =+
48. ( )c /4
/4
4 4 4 400
sin2 2sin cos4 4
sin cos sin cos
dd
=+ +
2/ 4
40
2tan sec4
tan 1
d
=
+
(Dividing numerator and denominator by 4cos }.
Now put 2 2tan 2 tan sec ,t d dt = = then the
reduced form is1 1
1
2
00
14 4 tan 4 0 .
1 4
dtt
t
= = =
+
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
49. ( )b ( )
( )
( )
1 1
3 30 0
1 2( 1)
1 1
xx e xe xdx dx
x x
+ =
+ +
or( ) ( ) ( )
1
1 1
2 3 20 0
0
2 1.41 1 1
x x xe e e edx dx
x x x
= =
+ + +
50. ( )a Here ( )( )
3
441 1
11xx
x xx x = =
++
( )3
2 2
41 1
1
1
x x dx dx
x x
= +
[ ] ( )2
2 4
11
1 1 32log log 1 log .
4 4 17x x
= + =
51. ( )d
1/ 2
1/ 2 1/ 21
2 2 21/ 4 1/ 4
1/ 4
2 12sin
1/ 21 1
2 2
xdx dx
x xx
= =
( )1/ 2
1
1/ 6sin 2 1 / 6.x = =
52. ( )a Put x t= or1
2dx dt x
= Also as 0x = to 2 so, 0t= to 2 . Therefore
22 2
0 00
3 32 3 2
log3
x ttdx dt
x
= =
( )2
23 1
log3=
53. ( )a ( ) [ ]2 2
00sin cos cos sin 0. x x dx x x
+ = + =
54 ( )a Let( )
/ 4
2 20
cos
cos 1 2 sin
x I dx
x x
=+
( )( )
/ 4
2 20
cos
1 sin 1 2sin
xdx
x x
= +
1/ 2
2 20
1 1 2
3 1 1 2dt
t t
= + + . By partial fractions, where sint x= 1/ 2
1
0
1 1 1 2log tan 2
3 2.1 1 2
tt
t
+ = +
( )( )
1
2 11 1log 2 tan 1
3 2 2 1
+ = +
( ) ( )21 1 1log 2 1 2 log 2 1
3 2 4 3 2 2 = + + = + +
55. ( )b / 2
20
sin
1 cos
x I dx
x
=+ Put cos sin x t xdx dt = = Then
0 1 11
2 2 01 0tan .
1 1 4
dt dt I t
t t
= = = = + +
56. ( )c [ ]2 2
11log log 2 log 2 2 1 xdx x x x= = +
4log 4 1 log 4 log log .ee
= = =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
57. ( )b 5
23
41 .
4 I dx
x
= + Solve further
58. ( )c We have2 2
sin cos1 1
0 0sin cos
x x
I tdt tdt = + Putting 2sint u= in the first integral and 2cost v= in the second integral, we have
0 /2sin 2 sin 2x x I u udu v vdv
=
/2
0 / 2 /2sin 2 sin2 sin 2
x x
u udu u udu v vdv
= +
I
/ 2 / 2 / 2
0 00
cos 2 1sin 2 cos 2
2 2
u uu udu udu
= = +
( )/ 2
/ 2
00
cos 2 1sin 2 .
2 4 4
u uu
= + =
59. ( )b ( ) ( ) ( )1 1 5 0 .....af x bf for each x ix x
+ =
. Replacing ( )1 , x by in ix
we get
( )1
5af bf x xx
+ =
..........(ii)
Eliminating1
fx
from ( )i and ( )ii , we get
( ) ( )2 2 5 5a
a b f x bx a bx
= + ( ) ( ) ( )2
22 2 2
11
log 52
ba b f x dx a x x a b x
=
( ) ( )log 2 2 10 log1 52
ba b a b a a b= + +
( )2
2 21
7 1 7log 2 5 log 2 5 .
2 2a a b f x dx a a b
a b
= + = +
60. ( )d ( )/ 4
2 2
0sec 1 tan
n
nI d
= / 4 / 2
2 2 2
0 0sec tan tan
n n
n I d d
= / 4
1
2 2
0
tan 11
1 1
n
n n n n I I I n n
= + =
. Hence 8 61 1
.8 1 7
I I+ = =
61. ( )b ( )
2 / 3 2 / 3
22 20 0
1
4 9 9 2 / 3
dx dx
x x=
+ +
2/ 3
1
0
1 1 1tan .
9 2 / 3 2 / 3 4 6 24
x = = =
62. ( )a 4 4
1 1 1
2 2 20 0 0
1 12
1 1 1
x x dx I dx dx
x x x
+ = = +
+ + +
( )1 1
2
20 01 2
1
dx I x dx
x = +
+
( )1
31
1
00
3 422 tan
3 3 2 6
x I x x
= + = + =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
63. ( )d 2 30
sina
I x x dx= . Put 3 2 3dt
x t x dx= =
[ ]3 3
3
00
1 1 1sin cos cos 1
3 3 3
a a I t dt t a = = = 3
11 cos .
3a =
64. ( )c / 4 / 4
0 0
sin costan cot
sin cos
x x x x dx dx
x x
+ + =
( )
/ 4
20
sin cos2
1 sin cos
x xdx
x x
+=
Put ( )sin cos , cos sin x x t x x dx dt = + = 0
212
1
dtI
t =
( )0
1
12 sin 2 0 / 2 .
2I t
= = =
65. ( )a 1 1
0 0
1 1 1.
1 1 1
x x x I dx dx
x x x
= =
+ + 1 1 1
2 2 20 0 0
1
1 1 1
x dx x
dx dx x x x
= = =
111 2
0 0sin 1 1.2 I x x
= + =
66. ( )c [ ]11
1log log log1 1.
e e
e I dx x ex
= = = =
67. ( )a 1
2logx x I dx
x= . Let log
dx x t dt
x= =
( )
log2
log 2
00
2 2 log .2
xx t
I t dt x
= = =
68.
/2
2 2 2 2
0
( )cos sin
dxd Ia x b x
= +Dividing the numerator and denominator by cos2x, we get
/ 2 / 22 2
2 2 2 2 2 2 2
0 0
(1/ cos ) sec
(sin / cos ) tan
x dx x dxI
a b x x a b x
= =+ +
Substituting btanx = t and b sec2x dx = dt and limit when x = 0, then t = 0 and when / 2 x then t = =
1
2 2
0 0
/ 1 1tan
dt b t I
a t b a a
= = +
1 11 1tan tan 0 02 2ab ab ab
= = =
69. ( )d ( ) ( ) / 4 5 / 4
0 / 4cos sin sin cos I x x dx x x dx
= + ( )
/ 4
2cos sin x x dx
+
[ ] [ ] [ ]5
44 40 2
4
sin cos sin cos sin cos x x x x x x
= + + + +
1 1 1 1 1 1 1 11 1
2 2 2 2 2 2 2 2I
= + + + +
2 1 2 2 2 1I = + or 2 1 2 2 2 1 4 2 2I = + + =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
70. ( )b 2
04 4
2
a a xdx a a + + 2 22 8 2 8 0a a a a +
( )( )4 2 0 2 4.a a a +
71. ( )b ( )
0 0
221 12 2 1 1
dx dxI
x x x = =
+ + + + ( )
01 1 1
1tan 1 tan 1 tan 0 .
4x
= + = =
72. ( )a 3 3
1
2 11
1tan .
1 3 4 12dx x
x
= = = +
73. ( )d ( )( )( )3
11 2 3 . I x x x dx= or
( )3
4 3 23
3 2
11
6 116 11 6 6 0.
4 3 2
x x x I x x x dx x
= + = + =
74. ( )c
( )
3 3 3
22 2 2
1 1
1 1
dx dx I dx
x x x x x x
= = =
( )( ) [ ]
3 3 3 3
222 2
1 1log 1 log
1dx dx x x
x x= =
= [ ] [ ]4
log 2 log1 log 3 log 2 2 log 2 log 3 log .3
= =
75. ( )a ( )
15
8 3 1
dxI
x x=
+Put 2 1tan tanx x = = , 22 tan secdx d =
( )
1
1
2tan 15
2 2tan 8
2tan sec
tan 3 tan 1I d
= + ( )1
1
2tan 15
2tan 8
2tan sec
sec 4 sec d
= ( )1
1
tan 15
2tan 8
2 tan sec
sec 4d
=
( )( )
1
1
tan 15
tan 8
2 tan sec
sec 2 sec 2d
=
+ ( )
( )
1
1
tan 15
tan 8
sec 21log
2 sec 2
= +
1 2 1 1 5
log log log .2 6 5 2 3
= =
76. ( )c 30
sinI d
= . Since sin is positive in integral ( )0,
( )3 20 0
sin sin 1 cos I d d
= =
( ) 20 0
sin sin cosd d
= + [ ]3
0
0
cos 4cos .
3 3
= + =
77. ( )b 1
1
0
1sin 2 tan
1
xdx
x
+ . Put cos ,x = then 1 1
1 cossin 2tan sin 2tan cot
1 cos 2
+ =
1sin 2 tan tan sin 22 2 2 2
= = ( ) 2 2sin sin 1 cos 1 x = = = =
Now, 1 11 20 0
1sin 2 tan 11
x dx x dxx
+ =
11
2 10
0
1 11 sin .2 2 4
x x x = + =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
78. ( )a 3 3 3
2 2 20 0 0
3 1 3 2
9 2 9 9
x x dxdx dx
x x x
+= +
+ + + ( )3
2 1
0
3 1log 9 tan
2 3 3
xx
= + +
( )3 1
log18 log92 3 4
= +
( )3
log 2 log 2 2 .
2 12 12
= + = +
79. ( )d ( )
2 2
41 15
4
111
dx dx
x xx
x
= + +
. Put 4 51 4
1 z dx dzx x
+ = =
17/1617/16
22
1 1 1 1 17log log 2 log
4 4 4 4 16
dzz
z
= = 1 32
log4 17
I =
80. ( )b ( )
3 3
2 22 2
1
1 1
x A B C I dx dx
x x x x x
+ = = + + or ( ) ( )( ) ( )21 1 1A x B x x C x x + + = +
Put 0,1, 1,x = we get 1, 2, 2 A B C = = =
3 3 3
22 2 22 2
1
dx dx dxI
x x x = +
[ ] ( )3
33
2 22
11 2 log 2 log 1x x
x
= +
1 1 3 16 12 log 2 log 2 log .
3 2 2 9 6I I = + =
81. ( )b 1
1log 1.log
e
e I x dx I x dx= =
[ ] ( ) ( )1
log log 0 1 1 1.e
I x x x e e e = = =
82. ( )c ( ) ( )
( )
2 2 / 2 / 2
20 0
sin cos sin cos
1 sin 2 sin cos
x x x x I dx dx
x x x
+ += =
+ +
( ) ( )/ 2 / 2
00sin cos cos sin I x x dx x x
= + = + or ( )1 1 2I = =
83. ( )d Let / 8 / 8
3 2
0 0cos 4 cos 4 .cos 4 I d d
= =
( )/8
2
01 sin 4 cos 4I d
=
Put sin 4 cos 44
dtt d = = when 0 ,
8to
= then t = 0 to 1
( )1
31
2
00
1 1 11 .
4 4 3 6
t I t dt t
= = =
84. ( )a We have8
3
2 31
1
xdx
x x
=
+ Put2
1 2 x t dx t dt + = =
when 3 8, x to= then 2 3t to= 2
3 3
2 2
2 2
5 3 22 2 3
1 1
t I dt or I dt
t t
= =
3
3
2
2 1 32 log 3 2 log .
2.1 1 2
t I t or I
t e
= = +
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
85. ( )a We have,1 11
2 2
00 02 x x x I x e dx I x e x e dx = =
( )1
02 2 0 1
x x I e xe e I e e e = = 2.I e =
86. ( )b We have, ( ) ( )2 2 21 , ; 1 , 1, 2 x x x x x x+ > + >
( )2 2
2 21 1
1 1, 1, 2
1 1
dx dxx
x xx x
<
87. ( )b On integrating both functions, we get
( ) ( )cot
tan2 2
//
1 1log 1 log log 1
2 2
xx
I eI e
t t t = + + +
2 2
2 2
1 1 1 1 1log sec log 1 log cot log log(cos ) log 1
2 2 x x ec x
e e e
= + + +
1log log 1.e
e
= = =
88. ( )a 3 / 4
/ 4 1 cos
dx
x
+3 / 4 3 / 4
2 2 / 4 / 4
1 cos 1 cos
1 cos sin
x xdx dx
x x
=
= ( )3 / 4
2
/ 4cos cot cosec x x ec x dx
( )3 / 4/ 4cot cos 2. x ec x
= + =
89. ( )a ( )
2
21 1
e dxI
x ln x=
+Let ( )
11 ln x t dt dx
x+ = =
Now, when2
1 , x to e= then 1 3t to=3
3
211
1 1 21 .
3 3
dtI
t t
= = = =
90. ( )b [ ]/2 /22
/4/4cos cot cot cot 1.
2 4ec xdx x
= = =
91. ( )c ( )
1/ 2log2 61
x
u
du
e
=
12
21
21
1 6
xe
ut dt as e t t
= =
+
( )
11 1
12 tan tan 16 4 12
xe
x
t e
= =
1 tan 1 3 4.3
x x xe e e
= = =
92. ( )c ( )( ) ( ) ( )2 1
1' ' I f g x f g x g x dx
=
Let ( )( ) ( )( ) ( )' ' f g x z f g x g x dx dz= =
When ( )1 , (1) x z f g= = when ( )2 , (2) x z f g= =
( (2)) ( (2))
( (1))( (1))
1
log
f g f g
f gf g I dz zz = =( ) ( )( )log ( (2)) log ( (1)) 0, 2 1 . I f g f g g g = = =Q
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
w w w . m a t h i i t . i n
S O L U T I O N S
LEVEL - 2 (Properties of Definite Integration)
1. ( )b ( ) ( )0 0
sin sin2
xf x dx f x dx
=
Since ( ) ( )0 0
1,
2
a a
xf x dx a f x dx= if ( ) ( ). f a x f x =
2. ( )c ( )/2
0
cot.....
cot tan
x I dx i
x x
=+
/2
0
cot2
cot tan2 2
x
dx
x x
= +
( )/2
0
tan.....
tan cot
xdx ii
x x
=+ . Now adding ( )i and ( )ii , we get
[ ]/2 /2
00
cot tan2 .
4tan cot
x x I dx x I
x x
+= = =+
3. ( )d/ 2 / 2 / 2
0 0 01 tan 1 cot1 tan
2
d d dI
= = =+ + +
. On adding,
/ 2
0
1 12
1 tan 1 cotI d
= + + + [ ]/ 2 / 2
00 .2 4d I
= = = =
4. ( )b ( )0
3 3 3
0
x xt t t
a a f x t e dt t e dt t e dt = = +
( ) ( ) ( )
03 3 3
.x
t t x
a a
df x d dt e dt t e dt x e
dx dx dx = + =
5. ( )b Let ( ) .f x x x= Then ( ) ( ) f x x x x x f x = = = .
Therefore1
10,x x dx
= ( By the property of definite integral ).
6. ( )d / 2 / 2
0 0
sinlog tan log
cos
x xdx dx
x
=
/2 / 2
0 0log sin log cos 0 x dx x dx
= = ( ) ( ){ }0 0 .a a
f x dx f a x dx= Q
7. ( )a / 2 / 2
0 0log sin log cos x dx x dx
= / 2 / 2 / 2
0 0 02 log sin cos log sin 2 log 2 I x x dx xdx dx
= =
0
1log sin log 2,
2 2tdt
= ( Putting 2x t= )
/ 2
0
1.2 log sin log 2
2 2
t dt
=
( ) ( ){ }2 log2 log2 .2 2
b b
a a I I I f x dx f t dt
= = = Q
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
8. ( )c ( )/ 2
0
cos sin1 .......
1 sin cos
x xdx i
x x
=
+
Now/ 2
0
cos sin2 2
1 sin cos2 2
x x
I dx
x x
=
+
( )/ 2
0sin cos ......
1 sin cosx x dx ii
x x = +
On adding, 2 0 0.I I= =
9. ( )d Let ( )2
log2
xf x
x
= +
( ) ( )1
2 2log log
2 2
x x f x f x
x x
= = = + +
1
1
2log 0.
2
xdx
x
= +
10. ( )c Let ( ) 17 4cos f x x x=
( ) ( ) ( )( ) ( )417
cos f x x x f x = =
Therefore,1
17 4
1cos 0. x xdx
=
11. ( )d Let ( )3/ 2
/ 2
3/ 2 3/ 20
sin.......
cos sin
xdxI i
x x
=+
3/ 2
/2
0
3/ 2 3/ 2
sin2
cos sin2 2
x
dx
x x
=
+
( )3/ 2
/2
3/ 2 3/ 20
cos.........
sin cos
xdxii
x x
=+
Adding ( )i and ( )ii , we get [ ]/2 /2
00
1 11 .
2 2 4 I dx x
= = =
12. ( )c ( ) / 4 / 4
0 0log 1 tan log 1 tan
4 I d I d
= + = +
/4
0
1 tanlog 1
1 tanI d
= + + ( ) / 4 / 4
0 0log 2 log 1 tan I d d
= + /4 /4
00
1 log 2log 2 log 2.
2 2 8I d
= = =
13. ( )d ( )
( )
2 2
0 0
sin 2 2sin 2
cos cos 2 I d d
a b a b
= =
2
0
sin 2
cosI d
a b
=
2
0
sin22 0 0.
cosI d
a b
= =
14. ( )a Put 1 x t dx dt = = . Also as 0x = to 1, 1t = to 0
Therefore, ( ) ( )( ) ( ) ( )1 0 1 1
0 1 0 01 . f x dx f t dt f t dt f x dx = = =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
15. ( )a ( )1/2
1/2
1cos log ..........( )
1
x I x dx i
x
= +
( )1/2
1/2
1cos log
1
x I x dx
x
+ =
1/2
1/2
1cos log .............( )
1
x I x dx ii
x
= + Adding ( )i and ( )ii , we get
1/ 2 1/2
1/2 1/2
1 12 cos log cos log
1 1
x x I x dx x dx
x x = + +
or 2 0I = or 0I = .
16. ( )d1 / 2
20 0
cos,
sin cos 41
dx d
x x
= =
++ ( Put sin , cos x dx d = = ).
17. ( )b
18. ( )c ( )1
21 1
1 11
1 1 2.2
x x dx x dx x
= = =
19. ( )b Let ( )30
sin ...... I x xd x i
= Also ( ) ( )3
0sin ...... I x xdx ii
=
Adding ( )i and ( )ii , we get ( )30 0
2 sin 3sin sin 34
I x dx x x dx
= =
0
cos 3 1 1 43cos 3 3
4 3 4 3 3 3
xx
= + = + =
. Hence,2
.3
I
=
20. ( )b 2 1 1 2
2 2 2 2
2 2 1 11 1 1 1 x dx x dx x dx x dx
= + +
( ) ( ) ( )1 1 2
2 2 2
2 1 11 1 1 x dx x dx x dx
= +
4 4 44.
3 3 3= + + =
21. ( )c Let ( )/ 2
0
cos......
sin cos
x I dx i
x x
=+
and/ 2
0
cos2
sin cos2 2
x
I dx
x x
=
+
( )/2
0
sin........
cos sin
x I dx ii
x x
=+
Adding ( )i and ( )ii , we get/2
02 1 .
2 4 I dx I
= = =
22. ( )d ( )/ 2
4 40
sin cos.......
cos sin
x x x I dx i
x x
=+
( )/ 2
4 40
cos sin2
..........sin cos
x x x
iix x
=
+
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
By adding ( )i and ( )ii , we get/ 2
4 40
cos sin2
2 cos sin
x x I dx
x x
=
+ 2
/ 2
40
tan sec
4 1 tan
x x I dx
x
=
+
Now, Put 2tan ,x t= we get2
1
2 00tan .
8 8 16
dtI t
I t
= = = +
23. ( )b
Let
/ 2
0 sin 4 I x dx
=
4 x is ve
when
4x
and ve+ when
4x
>
/ 4 / 2
0 / 4sin sin 2 2.
4 4 x dx x dx
= + =
24. ( )d
25. ( )b / 2
0sin cos x x dx
( ) ( ) ( ) / 4 / 2
0 / 4sin cos sin cos 2 2 1 . x x dx x x dx
= + =
26. ( )c [ ]/ 2 / 2
00 0cos 2 cos 2 sin 2. x dx x dx x
= = =
27. ( )b 4 4
/ 4 / 4 / 44 4
4 4 / 4 0 0
cos secsin 2 sec 2
sin tan
x xdx xdx xdx
x x
= =
Put tan ,x t= we get2
1
40
12
tdt
t
+
1 11 1
4 2
30 00 0
1 1 82 2 .
3 3t dt t dt
t t
= + = + =
28. ( )b 1.5 1 2 1.5
2 2 2 2
0 0 1 2 x dx x dx x dx x dx = + +
2 1.5
1 20 1 2 2 1 3 2 2 2 2.dx dx= + + = + =
29. ( )d ( ) ( )
( ) ( )0 0tantan
sec tan sec tan
x xx x I dx dx
x x x x
= =
+ +
0 0
tan sin2
2 sec tan 2 1 sin
x x I dx dx
x x x
= =
+ + 0 012 1 sindx
dxx
= +
On solving, we get2
1 .2 2
I
= =
30. ( )a Let( ) ( )
( ) ( )0 0tantan
sec cos sec cos
x xx x I dx dx
x x x x
= =
+ +
It gives20
sin
2 1 cos
x I dx
x
=
+
Now put cosx t= and solve, we get2
.2 2 4
I
= =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
31. ( )a 1
3 2
1sin cos 0. x x dx
= Since the function is an odd function.
32. ( )b Let ( ) ( )2sin 3
0cos 2 1x f x e n x dx
= +Since ( )( ) ( ) ( )cos 2 1 cos 2 1 2 1n x n n x + = + +
( ) ( )
2 2cos 2 1 sin sinn x and x x
= + = .Hence by the property of definite integral,
( )2sin 3
0cos 2 1 0xe n xdx
+ = ( ) ( )2 . f a x f x = 33. ( )b
1
1/ 1/ 1log log log
e e
e e x dx xdx xdx= +
[ ] [ ]1
1/ 1log log
e
e x x x x x x= +
( ) ( )1 1 2 1
1 0 1 1 2 2 1 .e ee e e e
= + + = =
34. ( )a [ ]( ) [ ] /2 /2 / 2
0 0 0sin sin x x dx xdx x dx
=
[ ]/ 2
2 2/ 2
00
, sin 0 .2 8
x x dx
= = =
Q
35. ( )c ( )1
01
n I x x dx=
( ) ( )( )1 1
0 01 1 1 1
n n I x x dx x x dx = =
( ) ( )1 11
0 0
1 1n n
x dx x dx+=
( )
( )
( )
( )
1 12 1
0 0
1 1 1 1
2 1 2 1
n nx x
n n n n
+ + = =
+ + + +
1 1.
1 2I
n n =
+ +
36. ( )c ( /6) ( / 2) ( /2) ( /3)2 2
( /6) ( /2) ( /2) ( /3)
[2 sin ] 1 2 2 1 x dx dx dx dx dx
+ + + +
+ + + +
= + + +
2 2 16 2 6 3 2 3
=
2 2 8 10 5
6 3 3 6 6 6 6 6 3
= = = =
37. ( )d ( )3
/2 /2
3 3 30 0
cos..........
1 tan sin cos
dx x I dx i
x x x
= =+ +
( )3
/ 2
3 30
sin.......................
cos sin
xdx ii
x x
=+
Adding ( )i and ( )ii , we get/ 2
02 .
4 I dx I
= =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
38. ( )a ( )
3 / 4 3 / 4
/ 4 / 41 sin 1 sin I d d
= =
+ + 3
4 4
+ = Q
3 / 4
/ 42
1 sinI d
=
+ .
On simplification, we get ( )2 1 tan .8I
= =
39. ( )b Since ( ) ( ) ( )b b
a a I xf x dx a b x f a b x dx= = + +
( ) ( ) ( )b b
a a I a b f x dx xf x dx = +
( ) ( ){ } f a b x f x given+ =Q
( ) ( ) ( ) ( )2 .2
b b b
a a a
a b I a b f x dx I xf x dx f x dx
+ = + = =
40. ( )a ( )0 0
sin sin I x xdx x x dx
= =
[ ]00
2 sin cos . I xdx x I
= = =41. ( )b
42. ( )a Adding/ 4
0.
4 4 I J dx I J
+ = = =
43. ( )d ( )5 5 5
1 1 13 1 3 1 x x dx x dx x dx + = +
( ) ( )3 5
1 33 3 12. x dx x dx= + =
44. ( )d 3
2............( )
5
x I dx i
x x=
+ Using the property
( ) ( )b b
a a I f x dx f a b x dx= = +
. .,i e change in ( )2 3 5 x x x= + = or dx dx=
( ) ( )2 3
3 2
5 5 .......5 5
x x I dx dx ii x x x x
= =+ +
Adding ( )i and ( )ii ,
3 3
2 2
52 1
5
x x I dx dx
x x
+ = =
+ [ ]3
2
13 2 1 .
2x I= = = =
45. ( )c Let ( )2 2cos 5 cos 5
0 0cos 3 cos 3 ,x x I e x dx e x dx
= =
( ) ( )0 0a a
f x dx f a x dx
= Q
2cos 5
0cos 3 2 0 0.x I e xdx I I I
= = = =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
46. ( )b / 2 / 2
0 0
1 cos.
41 tan cos sin
xdx dx
x x x
= =
+ +
47. ( )c 2 2
1 1 1
1 1 1
sin sin
3 3 3
x x x x I dx dx dx
x x x
= =
Here, ( )sin
3
xf x
x
=
is an odd function but
( )2
3
xf x
x=
is an even function
2 2 21 1 1
1 0 02 2 .
3 3 3
x x x I dx dx dx
x x x
= = =
48. ( )d Since ( ) 11sin f x x= is an odd function, therefore1
11
1sin 0.xdx
=
49. ( )d Given integral ( )2 2
2
2 2 I px s dx q xdx = + + or ( ) ( )
22
0
42 0 4 3
3 I px s dx p s= + + = +
Thus, to find the numerical value of I, it is necessary to know the value of p and s.
50. ( )c If ( )1
log ,1
xf x
x
+ = then ( ) ( )
1log
1
x f x f x
x
= = +
Therefore,1
1
1log 0.
1
xdx
x
+ =
51. ( )a ( ) / 2 0 / 2
/ 2 / 2 0
cos cos cos....
1 1 1 x x x x x x
I dx dx dx ie e e
= = +
+ + +
Putting x t= in0
/ 2
cos,
1 xx
dxe +
we get0 / 2
/ 2 0
cos cos
1 1
x
x x
x e x I dx dx
e e
= =
+ + or / 2 / 2
0 0
cos cos
1 1
x
x x
e x x I dx dx
e e
= ++ +
( )
( )
[ ] / 2 / 2 / 2
00 0
1 coscos sin 1.
1
x
x
e xdx xdx x
e
+= = = =+
52. ( )b [ ] ( ) ( )2
32 1 2
2 2 2
0 0 11
70 1 0 .
3 3
x x x dx x dx x dx
= + = + =
53. ( )b 2
3 32
0 0cos cos 0 xdx xdx
= = ( ){ }3 3cos cos .x x = Q
54.
2 / 2 / 2
0 0 0
( ) 2 log sin 2 log sin 4 log sina x dx x dx x dx
= =
14 log 2 2 log 2 2 log2 2
e e = = =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
Adding equations ( )i and ( )ii , we get
[ ]sin cos
/ 2 / 2 / 2
sin cos 00 0
2 22 1
2 2 2
x x
x x I dx dx x
+= = = = + Therefore, .
4I
=
64. ( )b ( ) ( )1 1/ 3 1
2 2 2
0 0 1/ 33 1 1 3 3 1 x dx x dx x dx = +
1 3 13 3
0 1/ 3 x x x x = +
1 1 1 1 4.
3 3 3 3 3 3 3 3
= + + =
65. ( )c 2/ 2 cos
2/ 2
sin
1 cos
xx I e dx
x
=
+ 2cos
2
sin
1 cos
xxe
x
+Q is an odd function
0.I =
66. ( )b ( ) ( ) ( )1.5 1.5
0.5 0.5
2 2 , I xf x dx x f x dx= =
( ) ( )
b b
a a f x dx f a b x dx = + Q
( ) ( ) ( ) ( )1.5 1.5 1.5
0.5 0.5 0.52 2 . x f x dx f x dx I I f x dx= = =
67. ( )a2
2
2
/ 2
0
2
x
xx
e dxI
e e
=
+ and
2
2
2
2/ 2
0
2
x
xx
e dxI
e e
=
+
0 0
( ) ( )
a a
f x dx f a x dx
=
Q
( )/ 2 / 2
002 1 .
4 I dx x I
= = =
68. ( )b ( ) ( )5 3 5
1 1 33 3 3 I x dx I x dx x dx= = +
( ) ( ) [ ] [ ]2 3 4 5
1 2 3 43 3 3 3 I x dx x dx x dx x dx = + + +
[ ] [ ]2 3 4 5 2 5
1 41 2 3 40 0 I dx dx dx dx x x = + + + = +
( ) ( )2 1 5 4 2.I = + =
69. ( )d 0 2
2 22 0 2
2 2 02 0
2 2
x x I x dx x dx x dx
= = + = +
( ) ( )2 2 4.= + =
70. ( )d Given ( ) ( ) f x f x = . We know that, ( ) ( ) ( )0
00
a a
a a f x dx f x dx f x dx
= = +
( ) ( ) ( )0 1 0
1 0 10 5 f x dx f x dx f x dx
+ = =
( )0
15. f t dt
=
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
71. ( )c ( ) ( ) ( )( )1 sin sin ,a a
a a I xf x dx x f x dx
= = ( ) ( )( )b b
a a f x dx f a b x dx= + Q
( ) ( ) ( ) 1sin sina a
a a x f x dx f x dx I
= =
1 2 2 1
22 . I I I I
= = =
72. ( )a 1/2
1/2
1cos1
x I x ln dxx
+ =
Since1
cos ln1
xx
x
+
is an odd function, ( )( )( ) f x f x = Q
0.I =
73. ( )b 2 2
1 1
1
1
log log loge ee e ee e
x x xdx dx dx
x x x
= +
2
1
21 0
1 10
log loge
e
x x
dx dx zdz zdzx x = + = + (Putting ( )log 1/ e x z x dx dz= = )
0 22 2
1 0
1 52 .
2 2 2 2
z z
= + = + =
74. ( )c ( ) ( ) ( )3 2 3
2 2 2 f x dx f x dx f x dx
= + cos sinxe xQ is an odd function
( ) ( )3 2 3
cos
2 2 2sin 2 0 2 3 2 2.
x f x dx e xdx dx
= + = + =
75. ( )a Let ( ) ( ) ( )( ) ( ) ( )( )x f x f x g x g x = + then,
( ) ( ) ( )( ) ( ) ( )( )x f x f x g x g x = +
( ) ( ) ( )( ) ( ) ( )( )0 0 x dx f x f x g x g x dx
= + = .
76. ( )c ( ) /3 /3
/6 /6
sin.......
1 cot sin cos
dx x I dx i
x x x
= =
+ +
( )/ 3
/ 6
cos.......
cos sin
x I dx ii
x x
=
+
Adding ( )i and ( )ii ,/3
/6
12 ; .
2 3 6 12 I dx I
= = =
77.2/ 3
/ 2
2 / 3 2 / 30
sin
sin cos
x I dx
x x
=+ or
2/ 3
/2
02 / 3 2/ 3
sin2
sin cos2 2
x
I dx
x x
=
+
or2/ 3
/ 2
2 /3 2/ 30
cos
cos sin
x I dx
x x
=+
( )
( )
2/3 2/3/2
2/3 2/30
sin cos2
sin cos
x x I dx
x x
+ =
+[ ]
/2 /2
00
12 .
2 4 I dx I x
= = =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
78. ( )a Let ( ) ( )2log 1 f x x x= + +
Now, ( ) ( ) ( )( )( )
2
2 2
2
1log 1 log 1
1
x x f x x x x x
x x
+ + = + = +
+ +
2 22 2
2
((1 ) )log log1 log( 1 ) log( 1 ) ( )
( 1 )
x x x x x x f x
x x
+ = = + + = + + = + +
( ) ( ) ( )0, .a
a f x if f x f x
= = Q
Hence, ( )1
2
1
log 1 0x x
+ + =
79. ( )d ( )2
cos sin I ax bx dx
= or ( )2 2cos sin 2 cos sin I ax bx ax bx dx
= +
( )2 2cos sin 2 cos sin I ax bx dx ax bx dx
= + or ( )2 202 cos sin 0 I ax bx dx
= +
0
1 cos 2 1 cos 22
2 2
ax bx I dx
+ = + or ( )0 2 cos 2 cos 2 2 . I ax bx dx
= + =
80. ( )b 0 0
1 cos 2cos
2
x I dx x dx
+= =
[ ] [ ]/2 /2
0 /20 /2cos cos sin sin I xdx xdx x x
= =
or sin sin 0 sin sin 1 1 2.2 2
I
= = + =
81. ( )c
82. ( )b 2sin 3
0cosx I e xdx
= ( ) ( ) ( )2sin 3
0cos ........
x I e x dx i
=
( )2sin 3
0cos ..........x I e xdx ii
= Adding ( )i and ( )ii , we get 2 0 0.I I= =
83. ( )a
9 1 4 9
0 0 1 42 2 3 4 x dx dx dx dx + = + +
( ) ( )2 12 3 36 16 2 9 20 31.= + + = + + =84. ( )c
2 1 22 2 2
0 0 1 I x dx x dx x dx = = +
[ ]1 2 2
10 10 2 1.dx dx x= + = =
85. ( )c [ ]x xe is a periodic function with period 1.
[ ] [ ]1000 1
0 01000 ,
x x x xe dx e dx
= [ ] ( )1
00, 0 1 1000 1000 1 .
x x if x e e = < < = = Q
86. ( )c ( )1 1
1/ 1/ ......
ana
n n
n n
x x I dx dx i
a x x a x x
= = + +
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
1
1
1 1
1 1 1 1
an
n
a x dxn n
a a x a xn n n n
+
= + + +
( ) ( )
b b
a a f x dx f a b x dx = + Q
( )
1
1 ......
an
n
a x
I dx ii x a x
= + Adding ( )i and ( )ii , we get [ ]
( ) 11/
1/1/2 1
a n an
nn I dx x
= =1 1 2 2
2 .2
na na I a I
n n n n
= = =
87.( )c / 2
0sin 2 log tan , I x x dx
= / 2
0 sin 2 log tan ,2 2 I x x dx
= ( ) ( )
0 0
a a
f x dx f a x dx = Q / 2 / 2
0 0sin 2 log cot sin 2 log tan x x dx x x dx
= = 2 0 0. I I I I = = =
88. ( )a [ ]1/ 2 1/ 2
1/ 2 1/ 2
1log
1
x I x dx dx
x
+ = + . If ( )1
log ,1
xf x
x
+ = then
( ) ( )1 1
log log1 1
x x f x f x
x x
+ = = = +
[ ]1/ 2
1/ 20 I x dx
= + (being integral of odd function)
( )0 1/ 2 0
1/ 21/ 2 0
11 0 .
2dx dx x
= + = =
89. ( )b [ ]2
002sin 0. 2 cos 0 xdx dx x
+ = +
( ) ( )2 cos cos 0 2 1 1 4.= = =
90. ( )c ( )/ 2
3
/ 23sin sin 0, I x x dx
= + = ( Q Function ( )33sin sinx x+ is an odd function).
91. ( )c 1 1/ 2 1
0 0 1/ 2
1 1 1
2 2 2 I x x dx x x x x dx
= = +
1/ 2 12 2
0 1/ 2
1 1
2 2 x x dx x x dx
= +
1/ 2 1
2 3 3 2
0 1/ 24 3 3 4
x x x x = +
1 1 1 1 1 1 1 .
16 24 3 4 16 24 8 = + + =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
92 . ( )a ( ) ( )8 5 8
0 0 55 5 5 17 I x dx x dx x dx= = + = .
93. ( )d ( ) ( )2 1 2
0 0 11 1 1 I x dx x dx x dx= = + +
1 22 2
0 1
1.2 2
x xx x
= + + =
94. ( )d [ ] [ ] [ ] [ ] [ ]2 1 0 1 2
2 2 1 0 1 x dx x dx x dx x dx x dx
= + + +
1 0 1 2
2 1 0 12 1 0 1dx dx dx dx
= + + +
[ ] [ ] [ ]1 0 2
2 1 12 0 x x x
= + + +
( ) ( ) ( )2 1 2 0 1 2 1 2 1 1 4.= + + + + = + + =
95. ( )d ( )1 1 1
1 1 1
20 0 0
1tan tan tan 1
1dx x dx x dx
x x
= +
( )1
11 1 2
00
12 tan 2 tan log 1 log 2.
2 2 xdx x x
= = + =
96. ( )b ( ), 0b b
a a
xdx dx a b
x= <
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
1 1 22 2 2
2 1 13 3 3
x x x x x x
= + +
( )2 2 2 1 10 28
2 9 3 1 6 .3 3 3 3 3 3
= + + + = + =
100. ( )a Given ( ) 1 f x x=
( ) ( ) ( )2 2 1 2
0 0 0 11 1 1 . f x dx x dx x dx x dx = = +
( )
1 22 2
0 1
1 11 2 2 1 1.
2 2 2 2
x xx x
= + = + =
101 . ( )b Let ( ) ( )/2
0 0sin sin I xf x dx A f x dx
= =
Now, ( ) ( ) ( )( )0 0
2 sin sin I xf x dx x f x dx
= +
( ) ( )0 0
sin sin f x dx f x dx
= =
( )/ 20
2 2 sin I f x dx
= ( ) ( )/ 2
0 0sin sin . I f x dx A f x dx
= = Hence .A =
102. ( )c Put ( )sin cos sin cos x x t x x dx dt + = = Also as 0x = to2
, t = 1 to 1.
Since here limit is 1 to 1, therefore the value of integral will be zero, ( ){ }0a
a f x dx =Q
103. ( )c Given function ( ) [ ]11
1log log log1
x x L x dt t x
t= = =
( ) log , L x x = Hence ( ) ( ) ( ).L xy L x L y= +
104. ( )c For 0 1,x< < we have2
1 ,x
e e< < so that2 21 1 1 1
0 0 0 01 1 .
x xdx e dx e dx e dx e< < <
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
108. ( )a ( ) 4 4 40 0
cos cos cosx x
g x t dt t dt t dt
+ ++ = = + ( ) ( )g f x= +
( ) ( ) ( )40
cos ,x
f x u du g x t u= = = + Q
( ) ( ) ( ).g x g x g + = +
109. ( )d 2 21 1 1
0 0 01 ,x xdx e dx e dx + = + but 21
0
xe dx
is not integrable.
110. ( )a ( )
( ) ( )( )
2
0.........
2
a f x I dx i
f x f a x=
+
( )
( ) ( )( )
2
0
2...........
2
a f a x I dx ii
f a x f x
=
+
Adding ( )i and ( )ii , we get2
02 2 .
a
I dx a I a= = =
111. ( )b 0 0
sin sin sinn n n
n
x dx x dx x dx
+ += +
2 1 cos .n = + .
112. ( )a ( ) / 4 / 4
2 2
0 0tan sec 1 tan
n n
nu xdx x x dx
= = /4
1 / 4 4
2 2 2
20 0
0
tansec tan tan
1
nn n
n
x x x dx x dx u
n
= = 2
1
1n n
u un
+ =
113. ( )a Put2
;2
x dx d
= = As 0 1, 02
x to to
= = . Then it reduces to
/ 2
0
2 2log sin log 2 log 2.
2d
= =
114. ( )c Put sin ,x = we get1 / 2
20 0
log log sin .cos
cos1
xdx d
x
=
/ 2
0log sin log 2.
2d
= =
115. ( )b / 2
0cot I x x dx
= Integrating by parts, we get
( )/ 2/ 2
0 0log sin log sin x x x dx
log 2 log 2.2 2I
= =
116. ( )b Let 1x t+ = when 2 0, x to= then 1 1t to=
( )1
3
12 cos I t t t dt
= + +
Since 3t and cost t are odd functions
[ ]1 1
112 2 4. I dt t
= = =
117. ( )a On differentiating both sides, we get
( )
2
sin sin cos cos xf x x x =
( ) ( )2 2
1 1 1sin 3.
sin 3 f x f x f
x x
= = =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
118. ( )b 2 2
0 0
1 1sin sin sin
2 2
n n
x x dx x dx =
[ ]/ 2 / 2
00
22 sin 2 cos 2 .
2
n x dx n x n
= = =
119. ( )a Since ( ) 0,a
a f x dx
= if ( ) ( ) f x f x = Therefore, 3 0.
a
a
dx
x x=
+
120. ( )a ( ) / 3 / 3
/ 6 / 6
cos........
1 tan cos sin
dx x I dx i
x x x
= =
+ +
( )/3
/6
sin.................
cos sin
x I ii
x x
=
+
(Since ( ) ( )b b
a a f x dx f a b x dx= + )
Adding ( )i and ( )ii we get,/3
/6
12 .
2 3 6 12 I dx I
= = =
121.4
4 4
sin
sin cos
x I dx
x x
=
+
( )4
/ 2
4 40
sin2 2 .......
sin cos
x I dx i
x x
= +
4
/ 2
04 4
sin2
4
sin cos2 2
x
I dx
x x
= +
or
( )4
/ 2
4 40
cos4 ............
cos sin
x I dx ii
x x
=+
Adding ( )i and ( )ii we get,/ 2
02 4 4 2
2 I dx
= = =
122. ( )d Since, f is continuos function. Let 1x t= .dx dt =
When 3 5, x to= then 2 6t to=
Therefore, ( ) ( ) ( )
5 6 6
3 2 21 1 . f x dx f t dt f x dx = =
123. ( )b We have, 2 21
lim ......1 4 2n
n n
n n n
+ + + + +
2 2 21 1 2
2
lim lim
1
n n
n nr r
n n
r n rn
n
= =
= =+
+
1
22 01
2
1lim ,
11
n
nr
dx
xrn
n
=
= =+
+
( )1 1
00
1, lim .
n
nr
r Applying formula f f x dx
n n
=
=
=1
1 1 1
0tan tan 1 tan 0 .
4x
= =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
124. ( )b Let3 3 3 3
1 4 1lim ......
1 2 2nS
n n n= + + +
+ +2
3 3 3 3 3 3
1 4lim ......
1 2n
n
n n n n= + + +
+ + +
2 2
3 3 31 1 3
3
lim lim1
n n
n nr r
r rS r n r
nn
= = = =+ +
2
31
1lim .
1
n
nr
r
nn r
n
=
= +
Applying the formula, we get ( )( )2 2
11 13
3 30 0 0
3 1 1 1log 1 log 2.
1 3 1 3 3e e
x x A dx dx x
x x= = = + =
+ +
125. ( )b 99 99 99 99
100 1001
1 2 .......lim lim
n
n nr
n r
n n =
+ + +=
199 1001
99
01 0
1 1
lim .100 100
n
nr
r x x dx
n n =
= = = = 126. ( )b Let
1/ 1/ ! 1 2 3 4
lim lim . . . ....
n n
nn n
n nP
n n n n n n = =
1 1 2log lim log log ... log
n
nP
n n n n = + + +
1
1log lim log
n
nr
rP
n n == or
( ) ( )1 1
00
1log log log 1 .P x dx x x x P
e= = = =
127. ( )b ( )
2 2
2 201
1 /lim 5 1.
11 /
n
nr
r n x L dx
n xr n
=
= = = ++
128. ( )d 1 1 1 1
lim .....1 2 2n n n n n
+ + + + + +
1 1 1 1lim .....
1 2n n n n n n = + + + + + + +
1 1 1 1lim 1 .....
1 21 1 1
n nn
n n n
= + + + + + + +
1
00
1 1 1lim
11
n
nr
dxrn x
n
=
= = + +
( )1
0log 1 log 2 log 1 log 2.e e e ex= + = =
129. ( )a Let22 2
1 1
1lim lim
1
n n
n nk k
k
k nI
n k n k
n
= =
= =
+ +
( ) [ ]1
12
2 00
1 1log 1 log 2 .
1 2 2
x I dx x
x = = + = +
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
130. ( )b( )2 2
1 1 1lim .....
1ny
n n n n n n
= + + + + +
( )
1 1 1lim ....
1 11 1
ny
n nn n
n n
= + + + + +
( )
1 1 1lim 1 ....
1 11 1
ny
n n
n n
= + + + + +
or
( )1
1 1lim ,
11
n
nk
yn
kn
=
=
+
Put1k
xn
= and
1dx
n=
1
1
00
lim lim 2 11
n
nn
n
n n
dxy x
x
= = + +
2 1 2 12 lim 1 2 lim 2
n n
n ny
n n
= =
12 lim 2 2 2 2 2.
ny
n = =
131. ( )a 1 11
1 2 3 .....lim lim
p p p p pn
p p
n n r
n r
n n+ +
=
+ + + +=
1
11
01 0
1 1lim
1 1
p pnp
nr
r x x dx
n n p p
+
=
= = = = + +
132. ( )b ( )1
1/
01 0
1lim . 1
nr n x x
nr
e e dx e en =
= = = =
133.( )d / 2
4 4
0 0sin 2 sin x dx xdx
= Applying gamma function,( ) ( )
( )
/ 24
0
5 / 2 . 1/ 2 32 sin 2 .
2 6 / 2 8xdx
= =
134. ( )a Let ( ) ( )1 12
2 5x
F x y t dt = = and ( )2 2 0 2x
F x y t dt = = .Now point of intersection means those point at which 2
1 2 15 6 y y y y x x= = = + and 2
2.y x=
On solving, we get 2 2 65 65
x x x x= + = and2 36 .
25y x= = Thus point of intersection is
6 36, .
5 25
135. ( )b ( ) ( ) ( ) ( )00
' ' .b c b c
f x a dx f x a f b c a f a
+ = + = +
136. ( )c ( ) 1 1' 0 ,2 2F x x x = >
Hence the function is increasing on 1 1,2 2
and therefore ( )F x has
maxima at the right end point of1 1
,2 2
.
( )1 / 2
1
1 3.
2 8 M a x F x F t dt
= = =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
137. ( )b ( )/ 2
2 2
/ 2sin cos sin cos x x x x dx
+
/ 2 / 23 2 2 3
/ 2 / 2sin cos sin cos x xdx x xdx
= +
/ 22 3
0
2 40 2 sin cos 0 2
15 15 x xdx
= + = + =
138.
( )3
20
( ) tan ,
1
dxa Putting x we get
x x
=+ +
( ) ( )
/ 2 / 22
3 3
0 0
sec cos
tan sec 1 sin
d d
=
+ +
/2
2
0
1 1 1 3
2(1 sin ) 8 2 8
= = + = +
139. ( )d ( )( ) ( ) ( ) ( )
2 22 2 4 221 1 2 2 1
' .2 .2 2 1 x x x x x
f x e x e x xe e + + + += =
( ) ( )' 0, , 0 . f x x >
140. ( )c ( ) ( ) ( ) ( )1
logx x f x dx xe f x f x dx f x
x= + = + .
On differentiating both sides, we get
( ) ( )0 ' f x f x= + . We know ( ) ( ), . x x xd
e e f x ce
dx
= =
141.( )c ( )3 / 2 / 2
3 / 2 3
0 0sin cos sin cosd d
= Applying gamma function,
/23 / 2 3
0
31
3 122 2
sin cos3
3 222
2
d
+ +
= + +
( )
( )
5
5 / 4 2 84.
9 5 52 13 / 4 452. .
4 4 4
= = =
142.20
1 1log
1 I x dx
x x
= + + . Put2
tan sec x dx d = =
( )2
/ 2
20
seclog tan cot
secI d
= + ( )
/ 2
0log tan cotI d
= +( )2/ 2
0
1 tanlog
tanI d
+ =
/ 2 / 2
0 02 log sec log tan I d d
=
{ } / 2 / 2
0 02 log sec log tan 0I d
= = Q / 2
02 log cosI d
=
/2
02 log 2 log cos log 2
2 2I
= = Q log 2.I =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
143. ( )a ( )
20 2
log
1
x x I dx
x
=
+ . Put 2tan sec x dx d = =
( )/ 2 240
tan log tansec
secI d
= ( )
/ 2
0sin cos log tan d
=
( )/2
0
1sin 2 log tan 0
2d
= = { }/2
0sin 2 log tan 0 .d
=Q
144. ( )d Given ( ) 1 12 tan 2 tan1
t t
tt
dx f t x t
x
= = = +
Differentiating with respect to t, ( ) ( )2
2 2' ' 1 1.
1 2 f t f
t= = =
+
145. ( )a ( )3
2log
x
xF x t dt = Applying Leibnitzs theorem,
( ) 3 3 2 2' log logd d
F x x x x xdx dx
= ( )2 23log .3 2 log 2 9 4 log . x x x x x x x= =
146. ( )c ( ) ( )1
1 1 1
2
0
2sin sin 1 sin 0 .1 2
xIx
= = = +
147. ( )a ( ) 2 2 22 2 0 f x x x x = = or4 2 2 0x x+ =
or ( )( )2 22 1 0x x+ = 2 1 0, 1.x x = =
148. ( )c ( )( ) ( ) 220 0 0
1 11
2 22
1 11 1
xdxxdx
dxx xx x
+ = +
+ ++ +
( ) ( )2 1 000
1 1 1 1log 1 log 1 tan2 2 2 2
x x x
= + + + +
10 0 0 .
2 2 4
= + + =
149. ( )d ( )3 3
sin 24 4
sin sin
31 1
3 3x x xd e xF x e dx e dxdx x x x
= = Put 3 23 x t x dx dt = =
( ) ( ) ( ) ( )sin
64 64
1 164 1 ,
te
F t dt F t dt F F
t
= = =
On comparing, 64.k =150. ( )b Since , ( )
0sin .
x
f x t tdt = Now, according to Leibnitzs rule, ( ) ( )' sin . 1 0 sin . f x x x x x= =
151. ( )b 2 2 2 22 2 2 2 2 21 1 2 4 3 9 1
lim sec sec sec .... sec 1n n n n n n n n
+ + + +
=2 2
2 2
2 2 21 1
1lim sec lim sec
n n
n nr r
r r r r
n n n n n = == .
Given limit is equal to the value of integral1
2 2
0sec x x dx
1 12 2 2 2
0 0
1 12 sec sec ,
2 2 x x dx t dt Put x t = = = [ ]10
1 1tan tan1.
2 2t= =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
152. ( )c Given curve logy x= and 1x = , 2x = .
Hence required area ( )2 2
11log log xdx x x x= =
( )2 log 2 1 log 4 1 .sq unit= =
153. ( )c
Required area
4 4
1 1 2
y
x dy dy= = 4
3/ 2
1
1 2 7
.2 3 3y= = sq. unit.
154. ( )b Let the ordinate at x a= divide the area into two equal parts
Area of
44
222
8 81 4 AMNB dx x
x x
= + = =
Area of22
81 2
a
ACDM dxx
= + =
On solving , we get 2 2 ,a = since 0 2 2.a a> =
155. ( )d Required area is2
1 20
4 . . A A y dx y dx sq unit
+ = + =
156. Required Area =
/4/4
0 0
cos 2 sin 2(sin 2 cos 2 )
2 2
x x x x dx
+ = +
1 cos sin cos 0 sin 0 1 .2 2 2
sq unit = + + =
157. ( )d Area ( )
( )
43/2
4
0
0
3 4 2 1123 4 56 .
3. 3/ 2 9 9
x x dx sq unit
+= + = = =
158. ( )a 2y x= and 22 2 0, 2 y x y y y= = =
Required area ( )2
32
2 2
00
42
3 3
y y y dy y
= = =
sq. unit.
159. ( )b Obviously, triangle ACB is right angled at C.
Required area 12
AC BC = 1
2 2 2 2 42
= = sq. unit
Y
C
O AX
B3 4y x= +
Y
XO1A
2A
2
34,
2
( )2,3A
C B
Y
O XM D N
( ), 0a
( )2, 0 ( )4 , 0
Y
C
A
B
X
2y x= +
( )2,4
( )2,0
2y x=
( )0,22
x
=
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
160.
/3 /3
1 2
0 0
3 3( ) cos , cos 2
2 4d A x dx A x dx
= = = =
1 2: 2 :1A A =
161. ( )b Shaded area
2 2 23 4
11 1
1 1 15
(16 1) .4 4 4 ydx x dx x A sq unit = = = = =
162. ( )b The curve is symmetric about x axis ,
Shaded area [ ]/ 2 / 2
002 sin 2 cos 2. xdx x
= = =
163. ( )a The curves y x= and sin y x x= + intersect at ( )0,0 and ( ), .
Hence area bounded by the two curves ( )0 0 0
sin sin x x dx x dx x dx
= + =
[ ] ( ) ( )0
cos cos cos 0 1 1 2.x
= = + = + =
164. ( )c Required area [ ] ( )/3 /3
002 tan 2 log sec 2 log 2 . x dx x
= = =
165. ( )b 2
2
0
32 2 1 3 .
log2
kx kdx k= =
Now check from options, only ( )b satisfies the above condition.
166. ( )d ( )2 2 2
1 1
1 2 1 1 1 1bb
f x dx b b x = + = + + = +
( ) 22 2
21
2 1 1
d x x f x x
dx x x = + = =
+ +
167. ( )a Area bounded by the curve ( ), y f x x axis= and the ordinates 1x = and x b= is ( )1
b
f x dx
From the question ( ) ( ) ( )
1
1 sin 3 4b
f x dx b b= +
. Differentiate with respect to b , we get
( ) ( ) ( ) ( ).1 3 1 cos 3 4 sin 3 4 f b b b b= + + +
( ) ( ) ( ) ( )3 1 cos 3 4 sin 3 4 . f x x x x= + + +
168. ( )b Given 2 4x y=
Shaded area2
2 23
00
1 8 2.
4 12 12 3
xdx x = = = =
169. ( )a ( )2
3 222
00
4 164 .
3 2 3
x x x x dx sq unit = =
Y
OX
siny x=2
/ 2
2x =Y
OX
1x =
3y x=
X
Y
2x =
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
170. ( )c Given curve is ( )3 10
2 3 102
x y x x y
x
+ = + =
Required area is4 4
3 3
3 10
2
x ydx dx
x
+=
( )4
33 16 log 2 3 16 log 2 . x x sq unit = + = +
171. ( )b Required area = area of OABC - area of OBC
163/ 2
16
00
6416 4 64 .
3 / 2 3
xx dx
= = =
172. ( )d Required area ( )2
2
02 2
x x x dx =
23
2
0
2
log 2 3
xx
x
= +
4 8 1 3 44 .
log 2 3 log 2 log 2 3= + =
173. ( )b Required area / 2 / 2
2
0 0
1 cos 2sin .
2
x A x dx dx
= =
[ ] [ ] / 2 / 2
0 0
1 1sin 2 .
2 4 4x x
= =
174. ( )c Required area 3 2 20 3
43
x dx x dx= + 3 22
2 1
30
1 44 sin
2 2 2 23
x x xx
= + +
3 3 2.
2 2 3 3
= + =
175. ( )c We have 24 y x x= and 0, 0, 4y x= =
Required area ( )42 3
42
00
44
2 3
x x x x dx
= =
64 3232 . .
3 3sq unit= =
176. ( )d Shaded area [ ]/ 2 / 4
00tan log sec x dx x
= =( )log sec / 4 log sec 0 log 2 log1 log 2.= = =
A
O
B
C
Y
( )16,4
2y x=
4y =
x
X
Y3x y=
( )2,0
/4x =
( )/4,0
( )/2,0( )/2,0
Y
X
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
177. ( )a Solving 0y = and 24 3 , y x x= + we get 1, 4.x =
Curve does not intersect x axis between 1x = and 4x = .
Area ( )4
2
1
1254 3 .
6 x x dx
= + =
178. 2( ) 4 2c We have y ax y ax= =
We know the equations of lines x = a and x = 4aTherefore, the area inside the parabola between the lines
44 4 4 3/21/2
2 2 23 / 2
aa a a
a a a a
x A y dx ax dx a x dx a
= = = =
1/ 2 3/ 2 3/ 2 1/ 2 3/ 2 24 4 28(4 ) (8 1)
3 3 3a a a a a a = = =
179. ( )b 1,y x= if 1x > and ( )1 ,y x= if 1x .Differentiate both sides . . ' 'w r t t , we get ( ) ( )2 2 4 22 2t f t t t f t t = =
Put2
,5
t = we get 4 2 .25 5
f =
15. ( )b The equation of curve is2
2 2 1 1
2 4y x x x x y x y
= = =
This is a parabola whose vertex is1 1
,2 4
. Hence point of intersection of the curve and the line
( )2 1 0 . ., 0 x x mx x x m i e x = = = or 1x m= 11 2 3 2 2 3 3
2
0 0
9 (1 ) (1 ) (1 )( ) (1 )
2 2 3 2 2 3 6
mm x x mx m m m
x x mx dx m
= = = =
( )3 1/36 9
1 27 1 27 3 22
m m m
= = = = =
Also, ( )3 31 3 0m = ( ) ( ) ( )( )21 3 1 9 1 3 0m m m + + =
( ) ( )2
1 3 1 9 0m m + + = 2 2
2 1 3 3 9 0 5 13 0m m m m m + + + = + =
5 25 52. .,
2m i e m
= is imaginary Hence,m = -2.
16. ( )b Curve ( )2 32 y a x x = is symmetrical about x axis and passes through origin. Also3
02
x
a x and 0x < . So curve does not lie in 2x a> and 0x < , curve lies
wholly on 0 2 .x a Area3/ 2
2 / 22 4
0 08 sin ,
2
a xdx a d
a x
= =
( Put 22 sinx a = )2
2 3 1 38 . .4 2 2 2
aa
= = ( Applying Gamma function )
Y
X
O
2x a=
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Trivandrum: 0471-2438271 Kochi: 0484-2370094
Y
C
O BAX
17. ( )b Solving the equations,
( )2 2 9................ x y i+ = 2 8 .................( ) y x ii=
Put ( )2 8 y x in i= , 2 8 9 0x x+ = ( )( )9 1 0x x + =
. . 9 1i e x or =
9x = gives imaginary value of y for equation ( )ii hence neglected.
( )1, 0A and ( )3,0B
Required area ( ) ( )1 3
0 12 y dx for ii y dx for i = +
( )1 31/ 2 2 2
0 12 2 2 3 x dx x dx = +
313/ 2 2
1
0 1
9 92 2 2 sin
3 / 2 2 2 3
x x x x = + +
=14 2 9 8 9
2 sin3 2 2 2 2 3
x
+
12 2 9 19 sin .3 2 3
= +
18. ( )b The lines are 1, 0 y x x=
1, 0, 1, 0, 1, 0y x x y x x y x x= < = + = +