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