UNIT-3INTEGRATION Definition of integration:
A function F (x) is called an anti derivative or integral of a function f (x) on an interval I if F ' ( x )=f ( x ) , For every value of x in I (i.e) If the derivative of a function F (x) w.r.to f (x), then we say that the integral of f (x) w.r.to x is F (x). (i.e) ∫ f ( x )dx=F (x )
Simple definite integrals Evaluate:∫
0
2
(x¿¿2+2x+2)dx ¿.Solution:∫0
2
(x¿¿2+2x+2)dx=∫0
2
x2dx+∫0
2
2x dx+∫0
2
2dx¿
¿ [ x33 ]0
2
+[ 2 x22 ]0
2
+[2 x ]02
¿ (2¿¿3−0)3
+(2¿¿2−0)+2 (2−0 )¿¿
¿ 83+4+4=8
3+8=8+24
3=323
Evaluate:∫0
π2
sin2 x dx . Solution: We know that,sin2 x=1−cos2 x
2
∴∫0
π2
sin2 x dx=∫0
π2
( 1−cos 2x2 )dx ¿ 12∫0
π2
(1−cos2 x )dx
¿ 12 [ x−( sin2 x2 )]0
π2
¿ 12 [( π2−0)− 12 (sin 2( π2 )−sin 0)]
¿ 12 [( π2 )−12 (sinπ−0 )]
¿ 12 ( π2 )= π
4¿
Evaluate:∫0
1 11+x2
dx.Solution: We know that, ∫ 11+x2
dx=tan−1 x+C
∴∫0
1 11+x2
dx=[ tan−1 x ]01
¿ tan−1 (1 )−tan−1 (0 )
¿ π4−0=π
4
Evaluate:∫0
π2cosx1+sinx
dx.Solution:Put u=1+sinx
dudx
=cosx⟹du=cosx dx When x=0 , u=1When x= π
2, u=1+1=2
∴∫0
π2cosx1+sinx
dx=∫1
2duu
= [ logu ]12
¿ log 2−log1=log 2¿
Evaluate .Solution:Using Integrationby parts ,∫u dv=uv−∫ vdu , Let u=log x ,dv=dx .∴du=1
xdx ,∫dv=∫ dx⟹ v=x
∴∫1
2
logx dx= [xlogx ]12−∫
1
2
x 1xdx
¿ [2 log2−1 log1 ]−[x ]12 ¿
¿2 log 2−0−(2−1 )¿2 log 2−1¿ log 22−1=log 4−1
Evaluate .Solution: We know that, ∫ tanx dx=−lo g (cosx )+C
∴∫0
π4
tanx dx= [−log (cosx ) ]0π4
¿−¿
¿−[ log( 1√2 )−log 1]¿−[ log 1−log√2−log1 ]¿ log √2
Evaluate: . Solution:We know that, ∫ exdx=ex+C∴∫
−a
a
ex dx=[ex ]−aa
¿ea−e−a
Evaluate: ∫cos 5x cos3 x dx .Solution: We know that, cosC cosD=1
2 [cos (C−D )+cos(C+D)]
∴∫cos5 x cos3 xdx=∫ 12 [cos (5x−3x )+cos (5 x+3 x )]dx
¿ 12∫ [cos2x+cos 8x ] dx
¿ 12 [∫cos2 x dx+∫cos 8x dx ]¿ 12 [ sin 2x2 + sin 8 x
8 ]+C
Evaluate: . Solution: We know that, ∫ 1
√a2−x2dx=sin−1( xa )+C
∫0
1 1√4−x2
dx=∫0
1 1√22−x2
dx
¿ [sin−1( x2 )]01
=[sin−1( 12 )−sin−1 (0 )]= π4
Integrate: ∫ 2 x1+x2
dx. Solution: Let u=1+x2du=2 xdx
∴∫ 2x1+x2
dx=∫ 1u du¿ log u+C=log (1+x2 )+C
Integrate: ∫ cosxsinx
dx . Solution: Let u=sinxdu=cosx dx
∴∫ cosxsinx
dx=∫ 1u du¿ log u+C=log (sin x )+C
Integrate:∫ sec 2 xtanx
dx. Solution: Let u=tanxdu=sec2x dx
∴∫ sec2 xtanx
dx=∫ 1u du¿ log u+C=log ( tan x )+C
Integrate: ∫ 4 x+32 x2+3 x+5
dx. Solution:Let u=2x2+3 x+5
du=(4 x+3)dx
∴∫ 4 x+32x2+3 x+5
dx=∫ 1u du¿ log u+C=log (2 x2+3 x+5 )+C
Integrate:∫cos3 x dx. Solution: We know that, cos3 x=4cos3 x−3cos x⟹4 cos3 x=cos3 x+¿3cos x¿
⟹cos3 x=14
¿
∴∫cos3 x dx=14∫¿¿¿
¿ 14 [ sin 3x3 +3 sinx ]+C
Integrate:∫sin 3 x cos2x dx. Solution: We know that, sinC sinD=1
2 [sin (C+D )+sin (C−D) ]
∴∫sin 3 x cos2x dx=∫ 12 [sin (3 x+2 x )+sin(3 x−2 x)] dx
¿∫ 12 [sin (3x+2 x )+sin(3 x−2 x)] dx
¿ 12∫ [sin5 x+sin x ]dx
¿ 12 [∫sin 5 xdx+∫sin x dx ]
¿ 12 [−cos5 x5
−cosx ]+C¿ −12 [ cos5 x5 +cosx ]+C
Evaluate:∫ cosx−sinxcosx+sinx
dx. Solution:Let u=cosx+sinxdu=(−sinx+cosx)dx
∴∫ cosx−sinxcosx+sinx
dx=∫ 1u du¿ log u+C=log (cosx+sinx )+C
Integrate:∫ dx(1+x2) tan−1 x . Solution:
Let u=tan−1 x
du= 11+x2
dx
∴∫ dx(1+x2) tan−1 x
=∫ duu
¿ log u+C=log ( tan−1 x )+C
Integrat:∫ 1+cosxx+sinxdx. . Solution:
Let u=x+sinxdu=(1+cosx)dx
∴∫ 1+cosxx+sinxdx=∫ 1u du
¿ log u+C=log ( x+sinx )+C
Evaluate:∫ sec 2(logx)x
dx . Solution:Let u=logx
du=1xdx∴∫ sec2(logx)
xdx=∫ sec2udu
¿ tanu+C=tan ¿¿
Evaluate:∫√1−cos2x dx . Solution:∫√1−cos2x dx=∫√2sin2 x [since2sin2θ=1−cos2θ ]¿√2∫ sinx dx¿√2 (−cosx )+C=−√2 cosx+C
Integrate: ..(L5) Solution: Let u=logxdu=1
xdx
∴∫ 1x (logx )n
dx=∫ 1undu
¿∫u−ndu
¿ u−n+1
−n+1+C=¿¿¿
Integrate:∫ sinxcos2 x
dx. .(L5) Solution:∫ sinxcos2x
dx=∫ sinx(cosx)(cosx)
dx
¿∫( sinxcosx )( 1cosx )dx
¿∫ tanx secx dx=secx+C
Evaluate ∫5 x4 . ex5dx. Solution: Let u=x5du=5 x4dx∴∫5 x4 ex
5
dx=∫ eudu¿eu+C=ex
5
+C
Evaluate ∫ e tanx
cos2 xdx. Solution:
Let t=tanx⟹dt=sec2 x dx
∴∫ e tanx
cos2xdx=∫ etanx . sec2 x dx
¿∫et dt=e t+C=etanx+C
Evaluate ∫ logxx dx . Solution:
Let t=logx⟹dt=1xdx
∴∫ logxx
=∫ t dt= t2
2+C=
(logx )2
2+C
Evaluate ∫ dxsin2 xcos2 x .(L6) Solution:
∫ dxsin2 xcos2 x
=¿∫ ¿¿¿¿
¿¿¿∫ sec2 xdx+∫ cosec2 xdx¿ tanx−cotx+C
Evaluate ∫sin 7 x .cos 5xdx. Solution:
We know that, sinCsinD=1
2 [sin (C+D )+sin (C−D)]
∴sin 7 x .cos 5xdx=12 [sin (7 x+5 x )+sin (7 x−5 x) ]
¿ 12 [sin (12 x )+sin (2x )]
∴∫sin 7 x .cos5 xdx=12∫ [sin (12 x )+sin (2 x)] dx
¿ 12 [−cos12 x12
− cos2 x2 ]+C¿−1
4 [cos2 x+ cos12 x6 ]+C Evaluate ∫ x . ex2dx. Solution:
Let u=x2⟹du=2x dx
⟹ du2
=xdx
∴∫ x . ex2
dx=∫ eu . du2 =12∫ e
udu=¿ 12eu+C=1
2ex
2
+C ¿
Evaluate ∫ sin(logx)xdx. Solution:
Let t=logx⟹dt=1xdx
∴∫ sin(logx)xdx=∫ sint dt=−cost+C=−cos ( logx )+C
Evaluate ∫ dxx . logx
. Solution:Let t=logx⟹dt=1
xdx
∴∫ dxx . logx
=∫ 1t dt=logt+C
Evaluate ∫0
π2
sec2 x dx . Solution:∫0
π2
sec2x dx=[ tanx ]0π2= tan π
2−tan0=∞
Evaluate ∫0
π2
sinx1+cosx
dx. Solution:Let u=1+cosx whenx=0 , u=1+cos0=1+1=2 du=−sinxdx x=π2 , u=1+cos π2=1+0=1
∴∫0
π2
sinx1+cosx
dx=∫2
1−duu
=¿∫1
2duu
¿
¿ [ logu ]12=log2−log 1=log2
Evaluate ∫cotx ¿log(sinx)
¿dx. Solution: Let t=log (sinx) and u=sinx ∴t=logu du=cosx dx⟹dt=1
udu
¿ 1sinx
cosx dx=cotx dx∴ ∫cotx ¿log(sinx)
¿dx=∫ dtt
=logt+C=log ( log ( sinx ) )+C
Evaluate ∫ 3−2 x(x2+x+1) dx. Solution:
Let 3−2x=A ddx
(x2+x+1 )+B ⟹3−2 x=A (2 x+1 )+B
Put x=−12;3−2(−12 )=A(2(−12 )+1)+B ⟹3+1=B ⟹B=4
Put x=0 ;3=A (0+1 )+B ⟹3=A+4 ⟹ A=−1
∴∫ 3−2 x(x2+x+1)
dx=∫−1 (2 x+1 )+4(x2+x+1)
dx¿−∫ (2 x+1 )(x2+x+1)
dx+4∫ 1(x2+x+1)
dx
¿−log (x2+x+1 )+4∫ 1
(x+12)2
+ 34
dx
¿−log (x2+x+1 )+4∫ 1
(x+12 )2
+(√32 )2 dx
¿−log (x2+x+1 )+4 [ 1√32 tan−1( x+12
√32
)]+C¿−log (x2+x+1 )+ 8
√3tan−1( 2x+1√3 )+C
. Evaluate ∫ x tan−1 x .dx. Solution: Use Integration by parts method ( i.e) ∫udv=uv−¿∫ vdu¿ Let u=tan−1x , dv=xdx du= 11+x2
dx , v= x22∴∫ x tan−1 x . dx=¿ x
2
2tan−1 x−∫( x
2
2) 11+x2
dx ¿
¿ x2
2tan−1 x−1
2∫( x2+1−11+x2
)dx
¿ x2
2tan−1 x−1
2 [∫( x2+11+x2
)dx−∫( 11+x2
)dx ]+C¿ x2
2tan−1 x−1
2 [∫dx−tan−1 x ]+C
¿ 12
[ x2 tan−1x+ tan−1 x− x ]+C
Evaluate ∫ x3ex2d x. Solution: Put t=x2⟹dt=2 xdx
⟹ dt2
=xdx ∫ x3ex2dx=¿ ∫ x2ex2 xdx¿∫ t e t dt2 =1
2∫ t et dt Use Integration by parts method Let u=t , dv=et dt du=dt , v=e t
∴ 12∫ t e
t dt=12 [ t et−∫ etdt ]
¿ 12
[ t e t−e t ]+C
¿ 12e t [t−1 ]+C
∴ ∫ x3 ex2
dx=12ex
2 [ x2−1 ]+C (since t=x2)
Evaluate∫0
π2sinxdx1+cos2x
. Solution: Let I=∫
0
π2
sinx1+cos2 x
dx Let t=cosx⟹dt=−sinx dx ⟹−dt=sinx dxWhen x=0⟹ t=cos0=1
When x= π2⟹ t=cos π
2=0
∴ I=∫1
0−dt1+t2
=−[ tan−1 t ]10
¿−[ tan−10−tan−11 ]¿−[0−π4 ]= π
4
Evaluate ∫0
π2
dxx2+5 x+6
.Solution:
Here∫ dxx2+5 x+6
=∫ dx( x+2 )(x+3)
Let 1( x+2 )(x+3)
= A( x+2 )
+ B(x+3)
¿A (x+3 )+B(x+2)
(x+2 )(x+3) ⟹1=A ( x+3 )+B (x+2) Put x=−2 ;1=A (−2+3 )+0 ⟹ A=1 Put x=−3 ;1=0+B (−3+2 ) ⟹B=−1
∫ dx( x+2 )(x+3)
=∫( 1(x+2 )
+(−1)(x+3 ) )dx
¿ log ( x+2 )−log ( x+3 )+C
¿ log( x+2x+3 )+C
Evaluate∫0
1 sin−1 x dx√1−x2 . Solution: Let t=sin−1 x ⟹dt= 1
√(1−x2)dx When x=0⟹ t=sin−1(0)
⟹ t=0 When x=1⟹ t=sin−1(1)
⟹ t=π2
∫0
1sin−1 x√1−x2
dx=∫0
π2
t dt=( t 22 )0
π2¿ 12 [ π2−0]= π
4
Evaluate ∫−1
1
log 3−x3+x dx. Solution:
Let f (x )=log [3−x3+x ]Then f (−x )=log [ 3+x3−x ]=log [ 3−x3+x ]
−1
¿−log [ 3−x3+x ]=−f (x ) ∴ f (x ) is an odd function of x. But we know that, ∫−a
a
f (x )dx=0 ,when f (x) is an odd function ofx∴∫
−1
1
log [3−x3+x ]dx=0 Integrate:∫ 7 x−6
x2−3 x+2dx.Solution:
Here 7 x−6x2−3 x+2
= 7 x−6( x−2 )(x−1)
Let 7 x−6( x−2 )(x−1)
= A( x−2 )
+ B( x−1 )¿
A (x−1 )+B( x−2)(x−2 )(x−1)
∴7 x−6=A ( x−1 )+B(x−2) Put x=1 ;7 (1 )−6=0+B(1−2) ⟹B=−1 Put x=2 ;7 (2 )−6=A (2−1 )+0 ⟹ A=8 ∴∫ 7 x−6
x2−3 x+2dx=∫ [ 8
( x−2 )+ (−1)
( x−1 ) ]dx¿8 log ( x−2 )−log ( x−1 )+C
Integrate:∫ 3 x+12 x2+x+1
dx.Solution:Let 3 x+1=A d
dx(2x2+x+1 )+B
Then ,3x+1=A (4 x+1 )+B
when x=−14;3(−14 )+1=0+B
⟹B=14
when x=0 ; A+B=1
⟹ A=1−14=34
∴∫ 3 x+12x2+x+1
dx=∫34
(4 x+1 )+ 14
2 x2+x+1dx
¿ 34∫
3x+12 x2+ x+1
dx+ 14∫
12x2+x+1
dx ¿ 34log (2 x2+x+1 )+ 1
4∫1
2 [(x+ 14 )2+(√74 )2]dx
¿ 34log (2 x2+x+1 )+( 14 )(12 )( 1√74 ) tan−1( x+
14
√74
)+C ¿ 34log (2 x2+x+1 )+( 1
2√7 ) tan−1( 4 x+1√7 )+C
Integrate:∫ 2x−1√ x2−5x−6
dx . Solution:Let 2 x−1=A d
dx(x2−5 x−6 )+B
⟹2x−1=A (2x−5 )+B when x=5
2;2( 52 )−1=0+B ⟹B=4 when x=0 ;−1=5 A+B
⟹−1=5 A+4⟹ A=1 ∴∫ 2x−1
√ x2−5x−6dx=∫ 1 (2x−5 )+4
√x2−5 x−6dx
¿∫ 2 x−5√x2−5 x−6
dx+4∫ 1√ x2−5x−6
dx
¿2√x2−5 x−6+4∫ 1
√(x−52 )2
−( 72 )2dx
¿2√x2−5 x−6+4 log [(x−52 )+√(x−52 )2
−(72 )2]+C
Integrate:∫ x+1√2x2+x−3
dx . (L5)Solution:Let x+1=A d
dx(2 x2+x−3 )+B
Then , x+1=A (4 x+1 )+B
when x=−14;−14+1=0+B
⟹B=34
when x=0 ;1=A+B⟹ A=14
∴∫ x+1√2x2+x−3
dx=∫14
(4 x+1 )+ 34
√2x2+x−3dx
¿ 14∫
(4 x+1 )
√2 x2+x−3dx+ 3
4∫1
√2 x2+x−3dx
¿ 14
(2 ) √2 x2+x−3+ 34∫1
√(x+ 14)2
−( 54 )2dx
¿14
(2 ) √2 x2+x−3+ 34 log [(x+ 14 )+√(x+ 14 )2
−( 54 )2]+C
Integrate:∫ x2 e−x dx.Solution: Use Integration by parts method Let u=x2 , dv=e−x dx du=2 xdx , v=−e− x ∴∫ x2e−x dx=x2 (−e−x )−∫(−e− x¿)2xdx=− x2e−x+2∫ xe− xdx ¿
Nowu=x ,dv=e− xdx du=dx , v=−e−x ∴∫ x2 e−x dx=−x2e− x+2 [− xe− x−∫ (−e− x)dx ]
¿−x2 e−x+2 [−xe− x+(−e−x ) ]+C
¿−e−x [ x2+2 x+2 ]+C
Evaluate ∫ x2 e3 xdx. Solution:Applying integration by parts, Let u=x2 , dv=e3 xdx du=2 xdx , v= e
3 x
3
∴∫ x2e3 xdx= x2e3x
3−∫ e3x
32xdx
¿ x2 e3x
3−23∫ xe3x dx−−−−−−−−−−(1)Again applying integration by parts Let u=x ,dv=e3 xdx
du=dx , v= e3x
3∴Equation (1) ⇒
¿ x2 e3x
3−23∫ xe3 x dx¿ x
2 e3x
3−23 {x . e3x3 −∫ e3 x
3dx }
¿ x2 e3x
3−2 x e
3 x
9+ 29∫e
3x dx ¿ x
2e3x
3−2 x e
3 x
9+ 29 ( e
3x
3 )+c ¿ x
2e3x
3−2 x e
3 x
9+2e
3x
27+c
Evaluate ∫0
∞
x e−x2
dx. Solution:Let t=x2
⟹dt=2 xdx⟹ xdx=dt2 when x=0 ; t=0 when x=∞;t=∞
∫0
∞
e−t dt2
=12∫0∞
e−t dt=12
[−e−t ]0∞
¿−12
[e−∞−e0 ]=−12
[0−1 ]=12
Evaluate:∫0
π2
cosx(1+sinx )+(2+sinx )
dx . Solution: Put t=sinx
⟹dt=cosxdx when x=0 ; t=sin 0=0 when x=π
2; t=sin π
2=1
∴∫0
π2
cosx(1+sinx )+(2+sinx )
dx=∫0
1dt3+2 t
=[ 12 log (3+2t )]0
1
¿ 12 [ log (3+2 )−log (3+0 ) ]
¿ 12 [ log (5 )−log (3 ) ]¿ 12 log( 53 )
properties of definite integrals Solution:(i )∫
a
b
f ( x )dx=−∫b
a
f ( x ) dx
( ii )∫0
a
f ( x )dx=∫0
a
f (a−x )dx
Evaluate ∫0
π2
√sinx cosx dx. Solution:
Let u=sinx when x=0;u=sin 0=0 ⟹du=cosxdxwhen x=π2;u=sin π
2=1
∴∫0
1
√udu=∫0
1
u12du=¿ [ u
12+1
12 +1 ]
0
1
=[u32
32 ]
0
1
=23 [1−0 ]=23 ¿
Evaluate ∫0
1
x ex dx. Solution:Let u=x ,dv=exdx du=dx, v=ex∫0
1
x ex dx=[ xe x]01−∫
0
1
ex dx=(1e1−0 )− [ex ]01
¿e−(e1−e0 ) ¿e−e+1=1. Evaluate:∫
0
π2
sin4xsin4x+cos4 x
dx. (L6)Solution:Let f (x )= sin 4 x
sin 4x+cos4 x−−−−−−−−−−−(I )
∴∫0
π2
f (x )dx=I (Let )−−−−−−−−−(1) Nowf ( π2−x )=
sin4 ( π2−x )sin4 ( π2−x )+cos4( π2−x )
¿( cosx )4
(cosx )4+( sinx )4−−−−−−−−−−−(II )
since sin( π2−x )=cosxcos (π2−x )=sinxBy the property of definite integral ∫0
π2
f (x )dx=∫0
π2
f ( π2−x)dx−−−−−−−−−−(2)From (1) and (2) we get
∫0
π2
f ( π2−x)dx=I−−−−−−−−−−(3)(1 )+(3 )⇒2 I=∫0
π2
f (x )dx+∫0
π2
f ( π2−x)dx
¿∫0
π2
sin4 xsin4 x+cos4x
dx+∫0
π2 (cosx )4
(cosx )4+ (sinx )4dx
[¿ I∧II ]¿∫0
π2 (sinx )4+ (cosx )4
(sinx )4+ (cosx )4dx
⟹2 I=∫0
π2
(1 )dx= [x ]0π2=π2
∴ I= π4
Evaluate:∫0
π2
log ( tanx )dx. Solution:Let I=∫
0
π2
log (tanx )dx−−−−−−−−−−−(1) ¿∫0
π2
log( tan( π2−x))dxBy the property ,∫
0
a
f (x)dx=∫0
a
f (a−x)dx
¿∫0
π2
log (cotx )dx−−−−−−−−−−−(2) (1)+(2) gives 2 I=∫
0
π2
[ log ( tanx )+log (cotx ) ]dx
¿∫0
π2
[ log ( tanx ) (cotx ) ]dx
¿∫0
π2
[ log (1 ) ] dx=0 [since l og1=0 ] ∴ I=0
Evaluate: ( sinx )32
(sinx )32+ (cosx )
32.
Solution:Let f (x )=
(sinx )32
( sinx )32+(cosx )
32
−−−−−−−−−−−(I )
∴∫0
π2
f (x )dx=I (Let )−−−−−−−−−−(1)Now f ( π2−x)=sin
32 ( π2−x )
sin32( π2−x)+cos
32 ( π2−x )
¿(cosx )
32
(cosx )32+(sinx )
32
−−−−−−−−(II )
since sin( π2−x )=cosxcos ( π2−x )=sinxBy the property of definite integral ∫0
π2
f (x )dx=∫0
π2
f ( π2−x)dx−−−−−−−−−−(2)From (1) and (2) we get∫0
π2
f ( π2−x)dx=I−−−−−−−−−−(3)(1 )+(3 )⇒2 I=∫0
π2
f (x )dx+∫0
π2
f ( π2−x)dx
¿∫0
π2 ( (sinx )
32
( sinx )32+(cosx )
32 )dx+∫0
π2 ( (cosx )
32
(cosx )32+ ( sinx )
32 )dx
[¿ I∧II ]¿∫0
π2 (sinx )
32+( cosx )
32
(sinx )32+( cosx )
32
dx
⟹2 I=∫0
π2
(1 )dx= [x ]0π2=π2
¿ π4 Typeequationhere .Area and Volume of Integration Formulae for area & volume in terms of integration.
Area=∫a
b
f ( x )dx
Volume=∫a
b
π [ f (x ) ]2dx
Evaluate ∫0
1
x 5x dx.(L6) Solution: Let u=x ,dv=5x dxdu=dx , v= 5x
log 5
∴∫0
1
x 5x dx=[x 5x
log5 ]0
1
−∫0
1 5x
log5 dx
¿( (1)51
log5 −0)− 1log 5 [ 5xlog5 ]
0
1
=5log5−
1( log 5 )2
[51−50 ]
¿ 5log 5
− 4( log5 )2
= 5 log 5−4( log 5 )2
Evaluate ∫−2
3
(4−x2 )dx . Solution:∫−2
3
(4−x2 )dx=[4 x− x3
3 ]−2
3
=(4×3−33
3 )−(4 (−2 )−(−2 )3
3 )¿ (12−9 )+(8−83 )=3+( 24−83 )
¿3+ 163
=253
Evaluate ∫ x2−5 x+1x
dx.Solution: ∫ x2−5 x+1
xdx=∫( x2x −5 x
x+ 1x )dx
¿∫( x−5+ 1x )dx¿ x
2
2−5 x+ logx+C
Using integration find the area of the region bounded by the lines 3 x−5 y−15=0 , x=1 , x=4 and the x-axis. (L3)Solution: The line 3 x−5 y−15=0 lies below the
Y
X
X=1 X=4
3x-5y-15=0
O
X
Y
X=4
x-axis in the interval x=1∧x=4
∴Required Area=∫1
4
(− y ) dx ¿∫1
4
(−15 ) (3 x−15 )dx¿ 35∫14
(5−x )dx
¿35 [5 x− x2
2 ]1
4
¿ 35 [5 (4−1 )−1
2(16−1 )]
¿ 35 [15−152 ]=92
Using integration, find the area of the region bounded by the parabolay2=16 x and the line x=4. Solution: The parabola y2=16 x is symmetrical about the x axis. ∴Required area ¿area AOCA+ area BOCB ¿2(area AOCA)
¿2∫0
4
ydx¿2∫0
4
√16 x dx¿8∫0
4
√ xdx
¿8× 23× [x 32 ]0
4
¿163 ×
(4 )32=163 ×8
¿ 1283