Integral calculus

XII STANDARD

MATHEMATICS

PREPARED BY:R.RAJENDRAN. M.A., M. Sc., M. Ed.,K.C.SANKARALINGA NADAR HR. SEC. SCHOOL,CHENNAI-21

Evaluate: dxxx

x

3

0 3

)1.....(..........3

ILet 3

0

dxxx

x

dxxx

x

3

0 )3(33

3dxxafdxxf

aa

00

)()(property,By

)2....(..........3

3I

3

0

dxxx

x

dxxx

xx

3

0 3

32I

Adding (1) and (2)

3

0

dx 30)(x

2I = 3

I = 3/2

Evaluate:

3

6

tan1

x

dx

)1.....(..........sincos

cos

tan1 ILet

3/

6/

3/

6/

dxxx

x

x

dx

dxxbafdxxfb

a

b

a )()(property,By

Adding (1) and (2)

3/

6/

dx 3/6/)(

x

3/

6/ )6/3/sin()6/3/cos(

)6/3/cos(

dxxx

x

)2.(..........cossin

sin3/

6/

dxxx

x

3/

6/ cossin

sincos2

dxxx

xxI

63

62

I

12

I

Evaluate:

3

6

cot1

x

dx

)1.....(..........cossin

sin

cot1 ILet

3/

6/

3/

6/

dxxx

x

x

dx

dxxbafdxxfb

a

b

a )()(property,By

Adding (1) and (2)

3/

6/

dx 3/6/)(

x

3/

6/ )6/3/cos()6/3/sin(

)6/3/sin(

dxxx

x

)2.(..........sincos

cos3/

6/

dxxx

x

3/

6/ sincos

cossin2

dxxx

xxI

63

62

I

12

I

Evaluate: dxx

2

0

9

4sin

Let u = x/4, then dx = 4du

When x = 2, u = /2

When x = 0, u = 0

2

0

92

0

9 4 .sin4

sinILet

duudxx

2

0

9 sin4

duu

13

2

5

4

7

6

9

84

315

512

Find the area of the circle whose radius is a.

Equation of the circle whose center is origin and radius a units is x2 + y2 = a2.

Since it is symmetrical about both the axes,

The required area is 4times the area in the first quadrant.

x

y

The required area = dxya

40 dxxa

a

40

22 a

a

xaxa

x

0

12

22 sin22

4

a

aa

a

aaaa

a 0sin

20

2

0sin

224 1

221

222

0sin

21sin

24 1

21

2 aasq.units

224 2

2

aa

Find the area of the region bounded by the line y = 2x + 4, y = 1, y = 3 and y-axis

The required area lies to the left of y axis between y = 1 and y = 3 x

y

y =1 y =3

y =

2x+3

The required area = dyx3

1

)(

dyy

3

1 2

4dyy

3

1

)4(2

1

3

1

2

422

1

y

y

4

2

112

2

9

2

1

2

81249

2

1

2

8

2

1= 2sq.units

Find the area of the region bounded by x2 = 36y, y-axis, y = 2 and y = 4.

The required area lies to the right of y-axis between y = 2 and y = 4

x

y

y = 2

y = 4 x2 = 36y

The required area = dyx 4

2

dyy 64

2

4

2

2/3

2/36

y

422/3

3

2 6 y )24(4 2/32/3

)228(4 sq.units )24(8

Find the volume of the solid that results when the ellipse

(a > b > 0)is revolved about the minor axis.

The required volume is twice the volume obtained by revolving the area in the first quadrant about the minor axis (y-axis) between y = 0 and y = b

12

2

2

2

b

y

a

x

x

y

The required volume = dyxb

20

2

dyybb

ab )(2

0

222

2

dyybb

a b

)(20

222

2

by

ybb

a

0

32

2

2

32

32

33

2

2 bb

b

a

3

22

3

2

2 b

b

a

cu.units 3

4 2ba

Find the area between the curve y = x2 –x – 2, x-axis, and the lines x = – 2 , x = 4

Equation of the curve is y = x2 – x – 2

When y = 0,

x2 – x – 2 = 0

(x – 2)(x + 1) = 0

x = 2, – 1

The curve cuts x-axis at x = –1 and x = 2

The required area = A1 + A2 + A3

Where A1 is area above the x-axis between x = –2 and x = –1

A2 is area below the x-axis between x = –1 and x = 2

A3 is area above the x-axis between x = 2 and x = 4

The required area =

x

y

x=4x=-2

)(4

2

2

1

1

2

ydxdxyydx

)2()2()2(4

2

22

1

21

2

2

dxxxdxxxdxxx

)2()2()2(4

2

22

1

21

2

2

dxxxdxxxdxxx

4

2

232

1

231

2

23

223

223

223

xxx

xxx

xxx

42

4

3

82

2

1

3

1

2

2

1

3

14

2

4

3

8

4

2

4

3

88

2

16

3

64

6

20

6

32

6

7

6

20

6

4

6

7

6

203272047

sq.units 156

90

Find the area enclosed by the parabolas y2 = x and x2 = y

Equation of the parabolas are y2 = x………(1) = f(x) x2 = y………(2) = g(x)Sub (2) in (1) (x2)2 = x x4 – x = 0 x(x3 – 1) = 0 x = 0, 1If x = 1, y = 1The point of intersection is (1, 1) The required area = area between the

two curves from x = 0 to x = 1

Required area =

x

y

y2 = x

x2 = y

1

0

)()( dxxgxf

x=1

1

0

2 dxxx

1

0

32/3

32/3

xx

3

1

2/3

1

3

1

3

2

sq.units3

1