Integral Calculus and its applications for XII Std (State

Integral Calculus and its applications for XII Std (State Board) by Dr. D. Senthilkumar

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

Integral Calculus and Its Applications Exercise 7.1 1. Sin2 2.

.

=

– 3cos + 3cos

=4 4 (

3. 4.

dx =

+

sin-1

+c

= 2 sin cos Put, sin x = t = sin-1

5. Put

+c

=t

6.

=

tan-1

7.

=

log

+c +c

8. Sin-1x = t 9. = 2 sin cos Put, t = 10. 11. 12.

Exercise 7.2 1. f(x) = - f(x) f(x) is an odd function. 2. f(x) = - f(x) f(x) is an odd function. 3. 4. f(x) = - f(x) f(x) is an even function.

( 5. f(-x) = f(x) f(x) is an even function. 6. f(-x) = - f(x) f(x) is an odd function. Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

7. 8. 9. 10.

Exercise 7.4 1. Find the area of the region bounded by the line x-y = 1 and (i) x - axis, x= 2 and x = 4. (ii) x - axis, x= -2 and x = 4. Proof: Consider x-y = 1 When x = 0, -y = 1

(0, -1)

y = -1 When y = 0, x = 1

(1,0)

When x = 2, 2-y = 1

(2,1)

-y = 1 - 2 y=1 When x = 4, 4-y = 1

(4,3)

-y = 1 - 4 y=3 When x = -2, -2-y = 1

(-2,-3)

-y = 1 + 2 y = -3 The area lies above the x – axis. A= = = =

-

= (8 – 4) – (2 – 2) = 4 sq. units. The area below the x – axis. Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

A= = = = =

-

=2+2 = 4 sq. units. 2. Find the area of the region bounded by the line x-2y -12 = 0 and (i) y - axis, y= 2 and x = 5. (ii) y - axis, y= -1 and x = -3. Proof: Consider x-2y -12 = 0 When x = 0, -2y = 12

(0,-6)

x = -6 When y = 0, x = 12

(12, 0)

When y = 2, x-4 -12 = 0

(16, 2)

x = 16 When y = 5, x-10 -12 = 0

(22, 5)

x = 22 When y = -1, x+2 = 12

(10, -1)

x = 10 When y = -3, x+6 = 12

(6, -3)

y=6 The area lies right of x – axis. A= = = =

-

= Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= 85 – 28 = 57 sq. units. The area lies right of y – axis. A= = = =

-

= = 37 – 21 = 16 sq. units. 3. Find the area of the region bounded by the line y = x - 5 and the x – axis between the ordinate x = 3 and x = 7. Proof: Consider y = x - 5 When x = 0, y = -5

(0,-5)

When y = 0, x = 5

(5, 0)

When x = 3, y = 3 - 5 (3, -2) x = -2 When x = 7, y = 7 - 5

(7, 2)

y=2 The area lies right of x – axis. A = A1 + A2 = = =

+ +

= =

+ +

=

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= 50 – 25 – 50 +29 = 4 sq. units. 4. Find the area of the region bounded by the curve y = 3x2 - x and the x – axis between x = -1 and x = 1. Proof: Consider y = 3x2 – x y = x(3x – 1) When y = 0, x = 0, x = When x = -1, y = 4 (-1, 4) When x = 1, y = 2

(1, 2)

The area lies right of x – axis. A = A1 + A2+ A3 = =

–

–

+

=

+

= (0) +

+ (0) +

=

+

= =2+ =

=

sq. units.

5. Find the area of the region bounded by x2 = 36y y - axis, y= 2 and y = 4. Proof: The equation x2 = 36y is of the form x2 = 4ay, it is a parabola open upwards

x2 = 4(9)y. When y = 2,

=4

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

x2 = 72 When y = 4,

=4 = 144

The area lies right of x – axis. A= = = = = =4 =8

sq. units.

6. Find the area included between the parabola y2 = 4ax and its latus return. Proof: The required area is twice the area bounded by the curve y = 2

, x=0, x= a and

x-axis.. A= = = = = = = =

sq. units.

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

7. Find the area of the region bounded by the ellipse

between the two latus

returns. Proof: ,

a > b.

=

5= = = = e = ae = 3

=2

Since the curve is symmetrical about both axes, the required area is 4 times the area in the first quadrant. A= = dx =

+

sin-1

+c

= = = =

sq. units.

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

8. Find the area of the region bounded by the parabola

and the line 2x – y =4.

Proof: Consider 2x – y = 4. When x = 0, y = -4 (0, -4) When y = 0, x = 4

(2, 0)

= 4x

(1)

= 4x

(2)

Solving (1) and (2) 2x = 4 + y x=4+y Substitute x value in (1) =4 =2 = 8 + 2y

y(y (y + 2) (y – 4) = 0 y = -2, y = 4 When y = -2, x =

=1

y = 4, x =

= 4

(1, -2) (4, 4)

The parabola and the straight line interests at (1, -2) and (4,4). Required area = = =

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= = = = = 15 - 6 = 9 sq. units. Note: Required area A = A1 - A2 A1 = area between the straight line 2x – y = 4, y-axis and y = -2 and y = 4. A2 = area bounded by the parabola

y-axis and the linear y = -2 and y = 4.

9. Find the common area enclosed by the parabolas 4

= 9x and 3

= 16y

Proof: 4

= 9x =

3

x it is of the for

. (Open rightwards)

= 16y =

Solving

y it is of the for

. (Open upwards)

= 9x

(1)

= 16y

(2)

=x Substitute the value of x in equation (2)

= 16y = 27y

y(y Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

y = 0, y = 3 When y = 0, x = 0 y = 3, x =

=4

(0,0) (4, 3)

The two parabolas intersects at (0,0) and (4,3) Required area A = A1 - A2 A1 = area bounded by the parabola 4

= 9x, the x-axis x = 0 and x = 4.

A2 = area bounded by the parabola

x-axis x = 0 and y = 4.

A= = =

= =4 =4

2-4

=8 = 4 sq. units.

10. Find the area of the circle whose radius is a. Proof: The equation of the circle is

.

Since it is symmetrical about both axes, the required area is 4 times the area in the first quadrant. The first quadrant area is bounded by the curve y =

, x = 0, x = a and

x – axis. A= = =4

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

=4 = =

sq. units.

Example 7.18 : Find the area of the region bounded by the line 3x − 2y + 6 = 0, x = 1, x = 3 and x-axis. Solution: Consider 3x − 2y + 6 = 0 When x = 0, -2y + 6 = 0

(0,3)

-2y = -6 y=3 When y = 0, 3x + 6 = 0

(-2,0)

3y = -6 y = -2 When x = 1, 3 - 2y + 6 = 0

(1,4.5)

-2y + 9 = 0 -2y = -9 y = 4.5 When x = 3, 9 - 2y + 6 = 0

(3,7.5)

-2y + 15 = 0 -2y = -15 y = 7.5 A= = = = = =

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= =

8

= 12 sq. units. Example 7.19: Find the area of the region bounded by the line 3x − 5y − 15 = 0, x = 1, x = 4 and x-axis. Solution: Consider 3x − 5y− 15= 0 When x = 0, -5y - 15 = 0

(0,-3)

-5y = 15 y = -3 When y = 0, 3x -15 = 0

(5, 0)

3y = 15 y=5 When x = 1, 3 - 5y - 15 = 0 -5y - 12 = 0 -5y = 12 y= When x = 4, 12- 5y - 15 = 0 -5y - 3 = 0 -5y = 3 y= A= = = = =

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= = = = = = sq. units. Example 7.20: Find the area of the region bounded y = x2 − 5x + 4, x = 2, x = 3 and the x-axis. Solution: Consider y = x2 − 5x + 4 When x = 0, y = 4 When y = 0, x2 − 5x + 4 = 0

(0, 4) (1, 0) (4, 0)

x2 − 4x− 1x + 4 = 15 x(x – 4) – 1 (x - 4) = 0 x = 1, 4 When x = 2, y = 4 – 10 + 4 = 8 - 10 = -2 When x = 3, y = 9 – 15 + 4 = -15 + 13 = -2 Required area is below the x – axis. A= = = = =

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= = = = =

sq. units.

Example 7.21: Find the area of the region bounded by y = 2x + 1, y = 3, y = 5 and y – axis. Solution: Consider y = 2x + 1 When x = 0, y = 1 When y = 0, x =

(0, 1) (

, 0)

When y = 3, 3 = 2x + 1 2 = 2x x=1 When y = 5, 5 = 2x + 1 4 = 2x x=2 Required area is below the x – axis. A= = = = = = = =

= 3 sq. units.

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

Example 7.22: Find the area of the region bounded y = 2x + 4, y = 1 and y = 3 and y-axis. Solution: Consider y = 2x + 4 When x = 0, y = 4

(0, 4)

When y = 0, 2x + 4 = 0

(-2, 0)

2x = -4 x = -2 When y = 1, 1 = 2x + 4 2x = -3 x= When y = 3, 3 = 2x + 4 2x = -1 x= Required area lies to the left of y – axis. A= = = = = = = = = 2 sq. units. Example 7.23: (i) Evaluate the integral

.

(ii) Find the area of the region bounded by the line y + 3 = x, x = 1 and x = 5. Solution: Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= = = = 12 – 12 =0 Consider the line y = When x = 0, y = -3 y = 0, x = 3

(0,-3) (3, 0)

x = 1, y = -2 x = 5, y = 2 From the diagram A1 lies below x – axis. A1 = = = = = = =6-4 = 2 sq. units. As A2 lies about the x – axis. A2 = = = = =

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

=

-6

=8-6 = 2 sq. units. Required area A = A1 + A2 =2+2 = 4 sq. units. Example 7.24: Find the area bounded by the curve y = sin 2x between the ordinates x = 0, x = π and x-axis. Solution: To find the points where the curve y = sin 2x meets the x-axis, put y = 0. Sin 2x = 0 Sin 2x = sin n ,

n

2x = n x= i.e., x = 0,

,

,....

n ,

The vales of x between x = 0 and x =

are x = 0, , .

A1 lies about the x – axis. A2 lies below the x – axis. A= = =

+

=

+

= =

+ +

=1+1 = 2 sq. units. Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

Example 7.25: Find the area between the curves y = x2 − x − 2, x-axis and the lines x = − 2 and x = 4 Solution: Consider y = x2 − x − 2 When y = 0, x2 − x− 2 = 0

(-1, 0) (2, 0)

x2 − 2x + 1x - 2 = 0 x(x – 2) + 1 (x - 2) = 0 (x + 1) (x - 2) = 0 x = -1, 2 When x = 0, y = -2

(0,-2)

When x = -2, y = 4 + 2 – 2 y=4 When x = 4, y = 16 – 4 - 2 = 16 - 5 y = 10 A = A1 + A2+ A3 = A1 =

–

= = =

+

= = = =

sq. units.

A2 =

–

= Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= = = = = = =

sq. units.

A1 = =

–

= = = = = = =

sq. units.

A = A1 + A2+ A3 = = = Required area = 15 sq. units. Example 7.26: Find the area between the line y = x + 1 and the curve y = x2 − 1. Solution: =x+1

(1)

y = x2 − 1

(2)

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

Solving (1) and (2) = x2 − 1 x2 − 1 x2 − x2 – 2x x(x (x + 1) (x – 2) = 0 y = -1, y = 2 When x = -1, y = 0

(-1, 0)

x = 2, y = 3

(2, 3)

=x+1 x

0

-1 2

-2 1

y

1

0

-1 2

3

y = x2 − 1 x

0

1

y

-1 0

-1 2

-2

0

3

3

A = A1 - A2 = =

–

= = = = = = = = sq. units. Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

Example 7.27: Find the area bounded by the curve y = x3 and the line y = x. Solution: = x3 y=x

(1) (2)

Solving (1) and (2) x3 = x3 x(x2 x (x - 1) (x + 1) = 0 x = 0, x = 1, x = -1 When x = 0, y = 0 x = 1, y = 1 A = A1 + A2 =

+

= = = = = sq. units. Example 7.28: Find the area of the region enclosed by y2 = x and y = x – 2. Solution: y2= x is a parabola. y = x – 2 is a straight line passing through (0, -2) (2, 0). y2= x

(1)

y=x-2

(2)

To find the point of intersection of the parabola and the straight line solving (1) and (2) y2 = Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

y2 y2 – 2y y(y (y + 1) (y – 2) = 0 y = -1, y = 2 When y = -1, -1 = x - 2

(1,-1)

-1 + 2 = x x = 1, When y = 2, 2 = x - 2

(4, 2)

4=x A= = = = = = = =

+

= = sq. units. Example 7.29: Find the area of the region common to the circle x2 + y2 = 16 and the parabola y2 = 6x. Solution: x2 + y2 = 16 is a circle with centre at origin and radius 4. y2 = 6x is a parabola open right wards. To find the point of intersection of the circle and parabola solution these two equations. x2 + y2 = 16

(1)

y2 = 6x

(2)

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

Solving x2 x2 x (x + 8)-2 (x + 8) = 0 (x - 2) (x + 8) = 0 x = 2, x = -8 When x = 2, y2 = 12 y=

2

(2, 2

)

(2, 2

)

When x = -8, y2 = - ve y = imaginary Omit x = -8 Required area = 2(A1 + A2) A1 = area bounded by the parabola

= 6x, x = 0, x = 2 and x-axis.

A1 = =

=2 =2 = A2 = area bounded by

x = 2, y = 4 and x-axis.

A2 = 2 =2 =2 =2 =2 = Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= = = A = A1 + A2 = = = = = =

sq. units.

Example 7.30: Compute the area between the curve y = sin x and y = cos x and the lines x = 0 and x = π. Solution: Sin x = cos x = Sin x = cos x =

⇒x= ⇒x=

= 225 Sin 225 = sin (180 +45) = - sin 45 = Cos 225 = cos (180 +45) = - cos 45 = -

.

From the figure we see that cos x > sin x for 0 ≤ x < and sin x > cos x for

< x < π.

A= =

+

= =

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= = =2

sq. units.

Example 7.31: Find the area of the region bounded by the ellipse Solution: , =1 = y= Area of the ellipse = 4

Area of the ellipse in the first quadrant.

I= = = = = = = ab sq. units. Example 7.32: Find the area of the curve y2 = (x − 5)2 (x − 6) (i) between x = 5 and x = 6 (ii) between x = 6 and x = 7. Solution: y2 = (x − 5)2 (x − 6) y = (x − 5) This curve cuts the x-axis at x = 5 and at x = 6. When x takes any value between 5 and 6, y2 is negative. ∴ The curve does not exist in the interval 5 < x < 6. Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

Hence the area between the curve at x = 5 and x = 6 is zero. (ii) y = (x − 5) y2 = (x − 5)2 (x − 6) A=2 Put t = x – 6 =1 dt = dx x

6

7

t

0

1

=2 =2 =

=2 =2 =2 =

sq. units.

Example 7.33: Find the area of the loop of the curve 3ay2 = x(x − a)2. Solution: 3ay2 = x(x − a)2 Put y = 0 x(x − a)2 = 0 x = 0, x = a,a. (0,0), (a,0) ,(a,0) Here a loop is formed between the points (0, 0) and (a, 0) about x-axis. Since the curve is symmetrical about x-axis, the area of the loop is twice the area of the portion above the x-axis. Required area = =

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= = = = = = = Required area =

sq. units.

Example 7.34: Find the area bounded by x-axis and an arch of the cycloid x = a (2t − sin 2t), y = a (1 − cos 2t). A cycloid is the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slippage. Solution : The curves crosses x-axis when y = 0. ∴ a(1 − cos 2t) = 0 ∴ cos 2t = 1 ∴ cos 2t = cos 2nπ 2t = 2nπ, n є z ∴ t = 0, π, 2π, … ∴ One arch of the curve lies between 0 and π x = a(2t − sin 2t) = a(2 − cos 2t.2) dx = 2a(1 − cos 2t) dt A= –

=

–

= = = = = = =

odd (or) π

even

sq. units

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

Hint: VOLUME V= V= Example 7.35: Find the volume of the solid that results when the ellipse that results when the ellipse

is revolved about the minor axis.

Solution: Volume of the solid is obtained by resolving the right side of the curve the y-axis. When x = 0,

about

y2=b2

Volume V = =2 =2 =2 = =

cubic units.

Example 7.36: Find the volume of the solid generalid when the region enclosed by and x=0 is resolved about the y-axis. Solution:

V= = = =

cubic units.

Problem 11: Find the volume of the solid x-axis. Solution: X 0 1 Y 0 2 Required Volume

-1 2

2 5

is resolved about the

-2 5

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

V= = = = = = = = =

cubic units.

Problem 12: Solution: Put

is revolued about x-axis,

.

∴ The loop is formed between V= = = = = = = = = =

cubic units.

Problem 13: Solution: Required Volume V=

is revolued about the y-axis.

= = = =

cubic units.

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

Problem 14:

is revolved about major axis

.

Solution: V= The required volume is twice the volume obtained by revolving the area in the first quadrant about the x-axis.

Volume V = 2 =2 =2 = =

cubic units.

Problem 15: Derive the formula for the volume of a right circular cone with radius and height . Solution: To find the volume of a cone with base radius and height , the revolving the area of a triangle whose vertices are (0,0), ( ,0) and ( ) about the x-axis. The equation of the straight line joining (0,0) and ( is

The volume of the cone is obtained by revolving the area bounded by and x-axis V=

,

= = = =

cube units

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

Problem 16: The area of the region bounded by the curve x-axis, x=1 and find the volume of the solid generated by revolving the area mentioned about x-axis. Solution: The required volume is obtained by revolving the area bounded by the curve x=1, and x-axis. About x-axis V =

and

= = = = = = = =

cubic units.

EXERCISE 7.5 Problem 1: Find the perimeter of the circle with radius . Solution: The perimeter of the circle in 4 lines the length of the arc of the circle in the first quadrant between x=0 and x= . Perimeter = 4

,

Perimeter = 4 = = = Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= =2 Problem 2: Find the length of the curve and . Solution The curve crosses the x- axis when y = 0

between

One arch of the curve lies between 0 and Therefore and gives only the length of the half arc. Length =

(1)

= = = = = = Therefore The length = = = = =4 Problem 3: Find the surface area of the solid generated by revolving the arc of the parabola bounded by its latus rectum. Solution: The required surface area of the solid is obtained by revolving the area bounded by and x-axis about x-axis. S= , Differentiating we get

1+ Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= =2 S=

= = = =

= = =

square units

Problem 4: Prove that the curved surface area of a sphere of radius intercepted between two parallel planes at a distance and from the centre of the sphere is and hence deduct the surface area of the sphere . Solution The required surface area of the solid is obtained by revolving the area bounded by -axis about -axis. S= , differentiate this we get

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

S= = = To find the total surface area of the sphere take S= = = Example 7.37: Find the length of the curve Solution: when

square units. and

square units between

and

The curve is symmetrical about x-axis. Length = Differentiating

1+

= = = =

Therefore

L= =

=

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= = = = = = =

=

Example 7.38: Find the length of the curve Solution: x = a cos3t, y = a sin3t is the parametric form of the given astroid, where 0 ≤ t ≤ 2π x = a cos3t = =

= = = Since the curve is symmetrical about both axes, the total length of the curve is 4 times the length in the first quadrant.

Length = = π

= π

= = = = =

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

Example 7.39: Show that the surface area of the solid obtained by revolving the arc of the curve y = sin x from x = 0 to x = π about x-axis is 2π [ + log (1 + )] Solution: dx =

+

log

+c

y = sin x = cos x

= Surface area = = Put cos x = t - sin x = sin x dx = - dt x

0

t

1

-1

= = = = = = sq. units. Example 7.40: Find the surface area of the solid generated by revolving the cycloid x = a(t + sin t), y = a(1 + cos t) about its base (x-axis). Solution: When y = 0, a(1 + cos t) = 0 cos t = -1 t=. x = a(t + sin t) , y = a(1 + cos t) = a(1 + cos t), y = a(- sin t) = - a sin t = = Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.

= = = = = Surface area = = = = = = Put = x = dt = 2dx x

0

t

0

= = = 32 = 32 =

sq. units.

Dr.D.Senthilkumar, Assistant Professor, Department of Mathematics, Govt. Arts College.