Class 12 - Maths - Application of Integrals

                                                                        Exercise 8.1

Question 1:

Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.

Answer:

The area of the region bounded by the curve, y2 = x, the lines, x = 1 and x = 4, and the x-axis is

the area ABCD.

Class_12_Application_Of_Integrals_Figure1

So, area of ABCD = ʃ14 y dx

                               = ʃ14 √x dx

                               = [x3/2/(3/2)]14 

                               = (2/3)[43/2 - 13/2]

                               = (2/3)[8 – 1]

                               = (2/3) * 7

                               = 14/3 units

Question 2:

Find the area of the region bounded by the curve y2 = 9x and the lines x = 2, x = 4 and the x-axis in the first quadrant.

Answer:

The area of the region bounded by the curve, y2 = 9x, the lines, x = 2 and x = 4, and the x-axis is

the area ABCD.

Class_12_Application_Of_Integrals_Figure2

So, area of ABCD = ʃ24 y dx

                               = ʃ24 3√x dx

                               = 3[x3/2/(3/2)]24 

                               = 3[43/2 - 23/2]

                               = 2[8 – 2√2]

                               = (16 – 4√2) units

Question 3:

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

Answer:

The area of the region bounded by the curve, x2 = 4y, y = 2, and y = 4, and the y-axis is the area

ABCD.

Class_12_Application_Of_Integrals_Figure3

So, area of ABCD = ʃ24 x dy

                               = ʃ24 2√y dy

                               = 2[y3/2/(3/2)]24 

                               = (4/3)[43/2 - 23/2]

                               = (4/3)[8 – 2√2]

                               = (32 – 8√2)/2 units

Question 4:

Find the area of the region bounded by the ellipse x2/16 + y2/9 = 1

Answer:

Given, equation if ellipse is x2/16 + y2/9 = 1

It can be observed that the ellipse is symmetrical about x-axis and y-axis.

      Class_12_Application_Of_Integrals_Ellipse                                           

Area bounded by ellipse = 4 * Area of OAB

So, area of OAB = ʃ04 y dx

                             = ʃ04 3√(1 – x2/16) dx

                             = (3/4)ʃ04 √(16 – x2) dx

                             = (3/4)[x√(16 – x2)/2 + (16/2) * sin-1 x/4]04 

                             = (3/4) [4√(16 – 42)/2 + (16/2) * sin-1 4/4] – (3/4) [0 + (16/2) * sin-1 0/4]

                               = (3/4)[8 * π/2]

                               = 3π units

Therefore, area bounded by the ellipse = 4 * 3π = 12π units

Question 5:

Find the area of the region bounded by the ellipse x2/4 + y2/9 = 1

Answer:

Given, equation if ellipse is x2/4 + y2/9 = 1

It can be observed that the ellipse is symmetrical about x-axis and y-axis.

       Class_12_Application_Of_Integrals_Ellipse1                                 

Area bounded by ellipse = 4 * Area of OAB

So, area of OAB = ʃ12 y dx

                             = ʃ12 3√(1 – x2/4) dx

                             = (3/2)ʃ12 √(4 – x2) dx

                             = (3/2)[x√(4 – x2)/2 + (4/2) * sin-1 x/2]12 

                             = (3/2) [2√(4 – 22)/2 + (4/2) * sin-1 2/2] – (3/2)[1√(4 – 12)/2 + (4/2) * sin-1 1/2]

                             = (3/2)[2π/2]

                             = 3π/2 units

Therefore, area bounded by the ellipse = 4 * 3π/2 = 6π units.

Question 6:

Find the area of the region in the first quadrant enclosed by x-axis, line x = √3y and the circle    x2 + y2 = 4

Answer:

The area of the region bounded by the circle x2 + y2 = 4, x = √3y, and the x-axis is the

 area OAB.

Class_12_Application_Of_Integrals_Figure4

The point of intersection of the line and the circle in the first quadrant is (√3, 1).

Area OAB = Area ∆OCA + Area ACB

Area ∆OCA = (1/2) * OC * AC = (1/2) * √3 * 1 = √3/2

Area ACB = ʃ32 y dx

                  = ʃ32 √(4 – x2) dx

                  = [x√(4 – x2)/2 + (4/2) * sin-1 x/2]√32 

                  = [2 * π/2 - (√3/2) * √(4 - 3) - 2sin-1 √3/2]

                  = [π - √3/2 – 2(π/3)]

                  = [π/3 - √3/2]

Therefore, area enclosed by x-axis, x = √3y and the circle x2 + y2 = 4 in the first quadrant

= √3/2 + [π/3 - √3/2] = π/3 units

Question 7:

Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line x = a/√2

Answer:

The area of the smaller part of the circle, x2 + y2 = a2 cut off by the line x = a/√2 is the area

ABCDA.

Class_12_Application_Of_Integrals_Figure5

It can be observed that the area ABCD is symmetrical about x-axis.

                                                        

So, Area ABCD = 2 * Area ABC

Now, Area of ABC = ʃa/√2a y dx

                                 = ʃa/√2a √(a2 – x2) dx

                                 = [x√( a2 – x2)/2 + (a2/2) * sin-1 x/a]a/√2a 

                                 = [a2/2 * π/2 - (a/2√2) * √( a2 - a2/2) - a2/2 * sin-1 1/√2]

                                 = a2/2 * π/2 - (a/2√2) * (a/√2) - a2/2 * (π/4)

                                  = a2π/4 - a2/4 - a2π/8

                                  = (a2/4)(π - 1 - π/2)

                                  = (a2/4)(π/2 - 1)          

So, area of ABCD = 2[(a2/4)(π/2 - 1)] = (a2/2)(π/2 - 1)

Therefore, The area of the smaller part of the circle, x2 + y2 = a2 cut off by the line x = a/√2 is

(a2/2)(π/2 - 1) units.

Question 8:

The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value of a.

Answer:

The line, x = a, divides the area bounded by the parabola and x = 4 into two equal parts.

 

Class_12_Application_Of_Integrals_Figure6So, Area OAD = Area ABCD

It can be observed that the given area is symmetrical about x-axis.

⇒ Area OED = Area EFCD  

Now, Area of OED = ʃ0a y dx

                               = ʃ0a √x dx

                               = [x3/2/(3/2)]0a 

                               = (2/3)[a3/2]  ………………1

Again Area of EFCD = ʃa4 √x dx

                                   = [x3/2/(3/2)]a4 

                                   = (2/3)[43/2 - a3/2]

                                   = (2/3)[8 - a3/2]  ………………2

From equation 1 and 2, we get

(2/3)[a3/2] = (2/3)[8 - a3/2]

=> [a3/2] = [8 - a3/2]

=> 2a3/2 = 8

=> a3/2 = 4

=> a = 42/3

Therefore, the value of a is 42/3

Question 9:

Find the area of the region bounded by the parabola y = x2 and y = |x|.

Answer:

The area bounded by the parabola, x2 = y and the line y = |x| can be represented as:

The given area is symmetrical about y-axis.

Class_12_Application_Of_Integrals_Parabola

So, Area OACO = Area ODBO

The point of intersection of parabola x2 = y, and line, y = x, is A (1, 1).

     Area of OACO = Area ∆OAB – Area OBACO

⇒ Area of OACO = Area of ∆OAB – Area of OBACO

Now, area of ∆OAB = (1/2) * OB * AB = (1/2) * 1 * 1 = 1/2

and area of OBACO = ʃ01 y dx

                                    = ʃ01 x2 dx

                                    = [x3/3]01 

                                    = 1/3

Hence, Area of OACO = Area of ∆OAB – Area of OBACO

                                       = 1/2 - 1/3       

                                       = 1/6

Therefore, the required are = 2 * 1/6 = 1/3 units.

Question 10:

Find the area bounded by the curve x2 = 4y and the line x = 4y – 2.

Answer:

The area bounded by the curve, x2 = 4y, and line, x = 4y – 2, is represented by the shaded area

OBAO.

Class_12_Application_Of_Integrals_Figure7

Let A and B be the points of intersection of the line and parabola.

Coordinates of point A is (-1, 1/4).

Coordinates of point B is (2, 1).

We draw AL and BM perpendicular to x-axis.

It can be observed that,

Area OBAO = Area OBCO + Area OACO   ……... (1)

Then, Area OBCO = Area OMBC – Area OMBO

                                = ʃ02 (x + 2)/4 dx - ʃ02 x2/4 dx

                                = (1/4)[x2/2 + 2x]02 – (1/4)[x3/3]02

                                = (1/4)[2 + 4] – (1/4)[8/3]

                                = 3/2 - 2/3

                                = 5/6    

Similarly, Area OACO = Area OLAC – Area OLAO

                                       = ʃ-10 (x + 2)/4 dx - ʃ-11 x2/4 dx

                                       = (1/4)[x2/2 + 2x]-10 – (1/4)[x3/3]-10

                                       = -(1/4)[(-1)2/2 + 2(-1)] – (-1/4)[(-1)3/3]

                                       = -(1/4)[1/2 - 2] – 1/12

                                       = 1/2 - 1/8 – 1/12

                                       = 7/24

Therefore, required area = 5/6 + 7/24 = 9/8 units.

Question 11:

Find the area of the region bounded by the curve y2 = 4x and the line x = 3.

 

Answer:

The region bounded by the parabola, y2 = 4x, and the line, x = 3, is the area OACO.

The area OACO is symmetrical about x-axis.

Class_12_Application_Of_Integrals_Parabola_1

So, Area of OACO = 2 (Area of OAB)

Now, Area of OACO = 2ʃ03 y dx 

                                    = 2ʃ02√x dx

                                   = 4[x3/2/(3/2)]03 

                                   = (8/3)[33/2]

                                   = 8√3

Therefore, the required area is 8√3 units.

Question 12:

Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2 is                                                                                                                                                           

  A. π                            B. π/2                                 C. π/3                                     D. π/4

Answer:

The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is

represented as:

Class_12_Application_Of_Integrals_Circle

So, area of OAB = ʃ02 y dx

                            = ʃ02 √(4 – x2) dx

                            = [x√(4 – x2)/2 + (4/2) * sin-1 x/2]02              

                            = 2(π/2)

                            = π units

Hence, the correct answer is option A.

 

Question 13:

Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is

  1. 2 B. 9/4 C. 9/3  D. 9/2

Answer:

The area bounded by the curve, y2 = 4x, y-axis, and y = 3 is represented as:

Class_12_Application_Of_Integrals_Figure8

Now, Area of OAB = ʃ03 x dy   

                                  = ʃ0(y2/4) dx

                                  = (1/4)[y3/3]03 

                                  = (1/12)[33]

                                  = 27/12

                                  = 9/4 units  

Thus, the correct answer is B.

                                                 Exercise 8.2

Question 1:

Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y.

Answer:

The required area is represented by the shaded area OBCDO.

        Class_12_Application_Of_Integrals_Circle_&_Parabola                                          

Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we get the point of

intersection as B(√2, 1/2) and D(-√2, 1/2).

It can be observed that the required area is symmetrical about y-axis.

So, Area OBCDO = 2 * Area OBCO

Now, draw BM perpendicular to OA.

Therefore, the coordinates of M is (√2, 0).

Therefore, Area OBCO = Area OMBCO – Area OMBO

                                         = ʃ0√2  {√(9 – 4x2)/4} dx - ʃ0√2 x2/4 dx

                                         = (1/2)ʃ0√2 √(9 – 4x2) dx – (1/4)ʃ0√2 x2 dx

                                        = (1/4)[x√(9 – 4x2) + (9/2) * sin-1 2x/3]0√2  - (1/4)[x3/3]0√2            

                                        = (1/4)[√2 * √(9 – 8) + (9/2) * sin-1 2√2/3] - (1/12)(√2)3

                                        = (√2/4) + (9/8) * sin-1 2√2/3 - √2/6

                                        = (√2/12) + (9/8) * sin-1 2√2/3

                                        = (1/2)[√2/6 + (9/4) * sin-1 2√2/3]

Therefore, the required area OBCDO = 2 * (1/2)[√2/6 + (9/4) * sin-1 2√2/3]

                                                                   = [√2/6 + (9/4) * sin-1 2√2/3] units

Question 2:

Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1

Answer:

The area bounded by the curves (x – 1)2 + y2 = 1 and x2 + y2 = 1, is represented by the shaded

area as:

    Class_12_Application_Of_Integrals_Figure9                                                

On solving the equations (x – 1)2 + y2 = 1 and x2 + y2 = 1, we get the point of intersection as

A(1/2, √3/2) and B(1/2, -√3/2).

It can be observed that the required area is symmetrical about x-axis.

So, Area of OBCAO = 2 * Area of OCAO

Now, join AB which intersects OC at M, such that AM is perpendicular to OC.

The coordinates of M is (1/2, 0)

Now, area of OCAO = Area of OMAO + Area of MCAM

                                    = ʃ01/2  {√{1 - (x – 1)2} dx - ʃ1/21 √(1 – x2) dx

                                    = [(x - 1)/2 * √{1 - (x – 1)2} + (1/2) * sin-1 (x - 1)]0√2

                                    + [x/2 * √(1 - x2) + (1/2) * sin-1 x]1/21  

                                    = [(-1)/4 * √{1 - (-1/2)2} + (1/2) * sin-1 (1/2 - 1) - (1/2) * sin-1 (-1)]

                                    + [(1/2) * sin-1 1 - 1/4 * √{1 – (1/2)2} - (1/2) * sin-1 1/2]               

                                    = [-√3/8 + (1/2)(-π/6) - (1/2)(-π/2)] + [(1/2)(π/2) - √3/8 -(1/2)(π/6)]   

                                    = [-√3/4 - π/6 + π/2]

                                    = [2π/6 - √3/4]  

Therefore, required area OBCAO = 2 * [2π/6 - √3/4] = [2π/3 - √3/2] units.

Question 3:

Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3

Answer:

The area bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3 is represented by the shaded

area OCBAO as:

Class_12_Application_Of_Integrals_Figure10

Then, Area OCBAO = Area ODBAO – Area ODCO

                                   = ʃ0(x2 + 2) dx + ʃ0x dx

                                   = [x3/3 + 2x]03  + [x2/2]03

                                   = (9 + 6) – 9/2

                                   = 15 – 9/2

                                   = 21/2 units

 

 

Question 4:

Using integration finds the area of the region bounded by the triangle whose vertices are    (–1, 0), (1, 3) and (3, 2).

Answer:

BL and CM are drawn perpendicular to x-axis.

It can be observed in the following figure that,

Class_12_Application_Of_Derivatives_Triangle

Area (∆ACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA)    ............. (1)

Equation of line segment AB is

       y - 0 = {(3 - 0)/(1 + 1)}(x + 1)

=> y = (3/2)(x + 1)

So, area of ALBA = ʃ-13(x + 1)/2 dx

                               = (3/2)[x2/2 + x]-11 

                               = (3/2)[1/2 + 1 – 1/2 + 1]

                               = 3 units

Equation of line segment BC is

      y - 3 = {(2 - 3)/(3 - 1)}(x - 1)

=> y = (1/2)(-x + 7)

So, area of BLMCB = ʃ1(1/2)(-x + 7) dx

                                  = (1/2)[-x2/2 +7 x]13 

                                  = (1/2)[-9/2 + 21 + 1/2 - 7]

                                  = 5 units

Equation of line segment AC is

       y - 0 = {(2 - 0)/(3 + 1)}(x + 1)

=> y = (1/2)(x + 1)

So, area of ALBA = ʃ-1(1/2)(x + 1) dx

                               = (1/2)[x2/2 + x]-13 

                               = (1/2)[9/2 + 3 – 1/2 + 1]

                               = 4 units

Therefore, from equation (1), we get

Area (∆ABC) = 3 + 5 – 4 = 4 units

Question 5:

Using integration find the area of the triangular region whose sides have the equations   y= 2x +1, y = 3x + 1 and x = 4.

Answer:

Class_12_Application_Of_Integrals_Triangle_1

The equations of sides of the triangle are y = 2x +1, y = 3x + 1, and x = 4.

On solving these equations, we obtain the vertices of triangle as A(0, 1), B(4, 13), and C(4, 9).

It can be observed that,

Area (∆ACB) = Area (OLBAO) –Area (OLCAO)

                        = ʃ0(3x + 1) dx - ʃ0(2x + 1) dx

                        = [3x2/2 + x]04 - [2x2/2 + x]04

                        = (24 + 4) – (16 + 4)

                        = 28 – 20

                        = 8 units

Question 6:

Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is                                                

A. 2 (π – 2)                        B. π – 2                         C. 2π – 1                    D. 2 (π + 2)

Answer:

The smaller area enclosed by the circle, x2 + y2 = 4, and the line, x + y = 2, is represented by the

shaded area ACBA as

Class_12_Application_Of_Integrals_Circle_1

It can be observed that,

Area ACBA = Area OACBO – Area (∆OAB)

                    = ʃ0{√(4 – x2)} dx - ʃ02 (2 - x) dx

                    = [x√(4 – x2)/2 + (4/2) * sin-1 x/2]02  - [2x – x2/2]02            

                    = 2 * π/2 – (4 - 2)

                    = π – 2 units

Thus, the correct answer is B.

Question 7:

Area lying between the curve y2 = 4x and y = 2x is                                                                                  

A. 2/3                            B. 1/3                          C. 1/4                                D. 3/4

Answer:

The area lying between the curve, y2 = 4x and y = 2x, is represented by the shaded area

OBAO as:

Class_12_Application_Of_Integrals_Figure11

The points of intersection of these curves are O (0, 0) and A (1, 2).

We draw AC perpendicular to x-axis such that the coordinates of C are (1, 0).

Area OBAO = Area (∆OCA) – Area (OCABO)

                    = ʃ02x dx - ʃ02 2√x  dx

                    = 2[x2/2]01  - [x3/2/(3/2)]01            

                    = |1 – 4/3|

                    = |-1/3|

                    = 1/3 units

Thus, the correct answer is B.

 

 

 

                                                Miscellaneous Exercise on Chapter 8

Question 1:

Find the area under the given curves and given lines:

(i) y = x2, x = 1, x = 2 and x-axis   (ii) y = x4, x = 1, x = 5 and x –axis

Answer:

(i). The required area is represented by the shaded area ADCBA as:

Class_12_Application_Of_Integrals_Area_Under_Curve_1

Area ADCBA = ʃ1y dx

                       = ʃ12 x2 dx

                       = [x3/3]12               

                       = 8/3 – 1/3

                       = 7/3 units

(ii). The required area is represented by the shaded area ADCBA as

Class_12_Application_Of_Integrals_Area_Under_Curve_2

Area ADCBA = ʃ1x4 dx

                       = [x5/5]15               

                       = 55/5 – 1/5

                       = 54 – 1/5

                       = 625 – 1/5

                       = 625 – 0.2      

                       = 624.8 units

Question 2:

Find the area between the curves y = x and y = x2

Answer:

The required area is represented by the shaded area OBAO as

Class_12_Application_Of_Integrals_Area_Under_Curve_3

The points of intersection of the curves, y = x and y = x2 is A (1, 1).

Now, draw AC perpendicular to x-axis.

So, Area (OBAO) = Area (∆OCA) – Area (OCABO)

                              = ʃ01 x dx - ʃ01 x2 dx

                              = [x2/2]01 – [x3/3]01

                              = 1/2 - 1/3

                              = 1/6 units

Question 3:

Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y= 1 and     y = 4.

Answer:

The area in the first quadrant bounded by y = 4x2, x = 0, y = 1, and y = 4 is represented by the

shaded area ABCDA as:

Class_12_Application_Of_Integrals_Area_Under_Curve_4

So, area of ABCD = ʃ1x dy

                               = (1/2)ʃ14 √y dy

                               = (1/2)[y3/2/(3/2)]14               

                               = (1/3)[ 43/2 - 1]

                               = (1/3)[8 - 1]

                               = 7/3 units

Question 4:

Sketch the graph of y = |x + 3|and evaluate ʃ-6|x + 3| dx

Answer:

The given equation is y = |x + 3|

The corresponding values of x and y are given in the following table.

x

-6

-5

-4

-3

-2

-1

0

y

3

2

1

0

1

2

3

 

On plotting these points, we obtain the graph of y = |x + 3|as follows:

 

 Class_12_Application_Of_Integrals_Area_Under_Curve_5

 

It is known that (x + 3) ≤ 0 for -6 ≤ x ≤ -3 and (x + 3) ≥ 0 for -3 ≤ x ≤ 0

Now, ʃ-6|x + 3| dx = -ʃ-6-3  (x + 3) dx + ʃ-3(x + 3) dx

                                    = -[x2/2 + 3x]-6-3 - [x2/2 + 3x]-30

                                    = -[{9/4 - 6} – {36/2 - 18}] – [0 – 9/2 + 9]

                                    = -[-9/2] – [-9/2]

                                    = 9/2 + 9/2

                                    = 9 units

Question 5:

Find the area bounded by the curve y = sin x between x = 0 and x = 2π.

Answer:

The graph of y = sin x can be drawn as

Class_12_Application_Of_Integrals_Area_Under_Curve_6

Now, required area = Area OABO + Area BCDB         

                                    = ʃ0π  sin x dx + |ʃπ2π  sin x dx|

                                    = [-cos x]0π + [-cos x]π

                                    = [-cos π + cos 0] + |-cos 2π + cos π|

                                    = 1 + 1 + |-1 - 1|

                                    = 2 + |-2|

                                    = 2 + 2

                                    = 4 units

Question 6:

Find the area enclosed between the parabola y2 = 4ax and the line y = mx.

Answer:

The area enclosed between the parabola, y2 = 4ax and the line y = mx is represented

by the shaded area OABO as:

Class_12_Application_Of_Integrals_Area_Under_Curve_7

The points of intersection of both the curves are (0, 0) and (4a/m2, 4a/m).

Now, draw AC perpendicular to x-axis.

So, Area OABO = Area OCABO – Area (∆OCA)    

                            = ʃ04a/m2 2√(ax) dx + ʃ04a/m2 mx dx

                            = 2√a[x3/2/(3/2)]04a/m2 – m[x2/2]04a/m2

                            = (4√a/3) * (4a/m2)3/2 – (m/2) * (4a/m2)2

                            = 32a2/3m3 - (m/2) * (16a2/m4)

                            = 32a2/3m3 - 8a2/m3   

                            = 8a2/3m3  

Question 7:

Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12.

Answer:

The area enclosed between the parabola, 4y = 3x2 and the line, 2y = 3x + 12 is represented by

the shaded area OBAO as:

Class_12_Application_Of_Integrals_Area_Under_Curve_8

The points of intersection of the given curves are A (–2, 3) and (4, 12).

Now, draw AC and BD perpendicular to x-axis.

So, Area OBAO = Area CDBA – (Area ODBO + Area OACO)   

                            = ʃ-2(3x + 12)/2 dx + ʃ-2(3x2/4) dx

                            = (1/2)[3x2/2 + 12x]-24 + (3/4)[x3/3]-24                                               

                            = (1/2)[24 + 48 – 6 + 24] - (1/4)[64 + 8]

                            = 90/2 – 72/4

                            = 45 - 18         

                            = 27 units

Question 8:

Find the area of the smaller region bounded by the ellipse x2/9 + y2/4 = 1and the line  x/3 + y/2 = 1.

Answer:

The area of the smaller region bounded by the ellipse x2/9 + y2/4 = 1and the line x/3 + y/2 = 1

is represented by the shaded region BCAB.

         Class_12_Application_Of_Integrals_Area_Under_Curve_9                                                       

Now, Area BCAB = Area (OBCAO) – Area (OBAO)  

                               = ʃ02 √(1 - x2/9) dx - ʃ02(1 – x/3) dx

                               = (2/3)[ ʃ0√(9 - x2) dx] - (2/3)ʃ0(3 – x) dx

                               = (2/3)[(x/2)√(9 - x2) + (9/2)sin-1 x/3]03 - (2/3)[(3x – x2/2)]03

                               = (2/3)[9/2 * π/2] – (2/3)[9  - 9/2]

                               = (2/3)[9π/4 – 9/2]

                               = (2/3)[(9/4)(π – 2)]

                               = 3(π – 2)/2 units

Question 9:

Find the area of the smaller region bounded by the ellipse x2/a2 + y2/b2 = 1 and the line    x/a + y/b = 1.

Answer:

The area of the smaller region bounded by the ellipse x2/a2 + y2/b2 = 1 and the line                         

x/a + y/b = 1 is represented by the shaded region BCAB.

        Class_12_Application_Of_Integrals_Area_Under_Curve_10                                                

Now, Area BCAB = Area (OBCAO) – Area (OBAO)  

                               = ʃ0b√(1 - x2/a2) dx - ʃ0b(1 – x/a) dx

                               = (b/a)[ ʃ0√(a2 - x2) dx] - (b/a)ʃ0(a – x) dx

                               = (b/a)[(x/2)√( a2 - x2) + (a2/2) * sin-1 x/a]0a - (b/a)[(ax – x2/2)]0a

                               = (b/a)[a2/2 * π/2 – (a2  - a2/2)]

                               = (b/a)[a2π/4 – a2/2]

                               = (ba2/2a)(π/2 – 1)

                               = (ab/2)(π/2 – 1)

                               = (ab/4)(π – 2) units

Question 10:

Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-axis.

Answer:

The area of the region enclosed by the parabola x2 = y, the line, y = x + 2, and x-axis is

represented by the shaded region OABCO as:

         Class_12_Application_Of_Integrals_Area_Under_Curve_11                                             

The point of intersection of the parabola x2 = y, and the line, y = x + 2, is A (–1, 1).

Now, Area OABCO = Area (BCA) + Area COAC

                                  = ʃ-2-1  (x + 2) dx + ʃ-1x2 dx

                                  = [x2/2 + 2x]-2-1 + [x3/3]-10                                              

                                  = [1/2 - 2 – 4/2 + 4] + 1/3

                                  = 5/6 units

Question 11:

Using the method of integration find the area bounded by the curve |x| + |y| = 1                   

[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x– y = 1]

Answer:

The area bounded by the curve |x| + |y| = 1 is represented by the shaded region ADCB as

The curve intersects the axes at points A (0, 1), B (1, 0), C (0, –1), and D (–1, 0).

It can be observed that the given curve is symmetrical about x-axis and y-axis.

Class_12_Application_Of_Integrals_Area_Under_Curve_12

Now, Area ADCB = 4 * Area OBAO

                               = 4ʃ0(1 - x) dx

                               = 4[x - x2/2]01                                              

                               = 4[1 – 1/2]

                               = 4 * 1/2

                               = 2 units

Question 12:

Find the area bounded by curves {(x, y): y ≥ x2 and y = |x|}.

Answer:

The area bounded by the curves {(x, y): y ≥ x2 and y = |x|} is represented by the shaded region

as shown below:

 Class_12_Application_Of_Integrals_Area_Under_Curve_13

 

 It can be observed that the required area is symmetrical about y-axis.

Now, Required area = 2[Area(OCAO) - Area(OCADO)]

                                    = 2[Area(OCAO) - Area(OCADO)]

                                    = 2[ʃ0x dx + ʃ0x2 dx]

                                    = 2[x2/2 – x3/3]01                                              

                                   = 2[1/2 – 1/3]

                                   = 2 * 1/6

                                   = 1/3 units

Question 13:

Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A (2, 0), B (4, 5) and C (6, 3).

Answer:

The vertices of ∆ABC are A (2, 0), B (4, 5), and C (6, 3).

Class_12_Application_Of_Integrals_Area_Under_Curve_14

Equation of line segment AB is

     (y - 0) = {(5 - 0)/(4 - 2)}(x - 2)

=> y = 5(x - 2)/2   ………….1

Equation of line segment BC is

     (y - 5) = {(3 - 5)/(6 - 4)}(x - 4)

=> (y - 5) = -(x - 4)

=> y – 5 = -x + 4

=> y = -x + 9    ………….2

Equation of line segment CA is

      (y - 3) = {(0 - 3)/(2 - 6)}(x - 6)

=> (y - 3) = 3(x - 4)/4

=> y – 3 = 3x/4 - 3

=> y = 3x/4 - 6   

=> y = 3(x - 2)/4     ………….3

So, Area (∆ABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)        

                             = ʃ25(x - 2)/2 dx + ʃ4(-x + 9) dx - ʃ23(x - 2)/4 dx

                             = (5/2)[x2/2 – 2x)]24 + [-x2/2 + 9x]4- (3/4)[x2/2 – 2x)]26

                             = (5/2)[8 – 8 – 2 + 4] + [-18 + 54 + 8 - 36]  - (3/4)[18 – 12 – 2 + 4]

                             = 5 + 8 – 6

                             = 13 – 6

                             = 7 units

Question 14:

Using the method of integration find the area of the region bounded by lines:                           

   2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0.

Answer:

The given equations of lines are

2x + y = 4    …………1

3x – 2y = 6    ……….2

x – 3y + 5 = 0   …….3

The area of the region bounded by the lines is the area of ∆ABC.

Class_12_Application_Of_Integrals_Area_Under_Curve_15

AL and CM are the perpendiculars on x-axis.

So, Area (∆ABC) = Area (ALMCA) – Area (ALB) – Area (CMB)           

                             = ʃ1(x + 5)/3 dx - ʃ1(4 - 2x) dx - ʃ2(3x - 6)/2 dx

                             = (1/3)[x2/2 + 5x)]14 - [4x - x2]1- (1/2)[3x2/2 – 6x)]24

                             = (1/3)[8 +20 – 1/2 - 5] - [8 - 4 - 4 + 1]  - (1/2)[24 – 24 – 6 + 12]

                             = (1/3) * (45/2) - 1 – 6/2

                             = 15/2 – 3

                             = 7/2 units

Question 15:

Find the area of the region {(x, y): y2 ≤ 4x, 4x2 + 4y2 ≤ 9}.

Answer:

The area bounded by the curves {(x, y): y2 ≤ 4x, 4x2 + 4y2 ≤ 9} is represented as:

      Class_12_Application_Of_Integrals_Area_Under_Curve_16                                              

The points of intersection of both the curves are (1/2, √2) and (1/2, -√2)

The required area is given by OABCO.

It can be observed that area OABCO is symmetrical about x-axis.

So, Area OABCO = 2 * Area OBC

Now, Area OBCO = Area OMC + Area MBC

                                = ʃ01/2  2√x dx + (1/2)ʃ1/23/2  √(9 - 4x2) dx

                                = ʃ01/2  2√x dx + [9π/16 - √2/4 – (9/8) * sin-1 1/3]

                                = 2[x3/2/(3/2)]01/2  +  [9π/16 - √2/4 – (9/8) * sin-1 1/3]

                                = (4/3)* (1/2)3/2 + [9π/16 - √2/4 – (9/8) * sin-1 1/3]

                                = (4/3)* (1/2√2) + [9π/16 - √2/4 – (9/8) * sin-1 1/3]

                                = (4/3)* (1/2√2) * (√2/√2) + [9π/16 - √2/4 – (9/8) * sin-1 1/3]

                                 = (4√2)/(4 * 3) + [9π/16 - √2/4 – (9/8) * sin-1 1/3]

                                 = √2/3 + [9π/16 - √2/4 – (9/8) * sin-1 1/3] 

                                = 9π/16 - (9/8) * sin-1 1/3 + √2/12

So, Area OABCO = 2 * Area OBC

                              = 2 * [9π/16 - (9/8) * sin-1 1/3 + √2/12]

                              = 9π/8 - (9/4) * sin-1 1/3 + √2/6

                             = 9π/8 - (9/4) * sin-1 1/3 + (√2/6) * (√2/√2)

                             = 9π/8 - (9/4) * sin-1 1/3 + 2/6√2

                             = 9π/8 - (9/4) * sin-1 1/3 + 1/3√2 units

Question 16:

Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is                         

  A. – 9                                 B. -15/4                                    C. 15/4                               D.  17/4

Answer:

From the figure,

Class_12_Application_Of_Integrals_Area_Under_Curve_17

Required area = ʃ-2x3 dx

                          = [x4/4]-21                                              

                          = 1/4 – 16/4

                          = 1/4 - 4

                          = -15/4 units

Hence, the correct answer is option B.

 

Question 17:

The area bounded by the curve y = x|x|, x-axis and the ordinates x = –1 and x = 1 is given by

[Hint: y = x2 if x > 0 and y = –x2 if x < 0]                                                                                                       

A. 0                              B. 1/3                                C. 2/3                                D. 4/3

Answer:

From the figure,

Class_12_Application_Of_Integrals_Area_Under_Curve_18

Required area = ʃ-1x|x| dx

                           = ʃ-1x2 dx + ʃ01 x2 dx

                           = [x3/3]-10 + [x3/3]01                                             

                          = -(-1/3) + 1/3

                          = 1/3 + 1/3

                          = 2/3 units

Hence, the correct answer is option C.

Question 18:

The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is                                                

A. 4(4π - √3)/3            B. 4(4π + √3)/3               C. 4(8π - √3)/3               D. 4(8π + √3)/3

Answer:

Class_12_Application_Of_Integrals_Area_Under_Curve_19

The given equations are

x2 + y2 = 16    …………….1

y2 = 6x    ………………..2    

Area bounded by the circle and parabola

= 2[Area OADO + Area (ADBA)]

= 2[ʃ0√(6x) dx + ʃ2{√(16 – x2)} dx

= 2[√6{x3/2/(3/2)]02 + 2[x/2 * √(16 – x2) + (16/2) * sin-1 x/4]24               

= 2 * √6 * (2/3) * [x3/2]02 + 2[8 * π/2 - √(16 – 4) - 8 * sin-1 1/2]24

= 4√6 * (2√2) + 2[4π - √(12) - 8π/6]

= 16√3/3 + 8π - 4√3 - 8π/3

= (4/3)[4√3 + 6π - 3√3 - 2π]

= (4/3)[√3 + 4π]

= (4/3)[4π + √3]                   

Area of circle = π(r)2

                         = π(4)2

                         = 16π units

Now, required area = 16π - (4/3)[4π + √3]

                                    = (4/3)[4 * 3π - 4π - √3]

                                    = (4/3)[8π - √3] units      

Thus, the correct answer is C.

Question 19:

The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ x ≤ π/2                                       

  A. 2(√2 - 1)                        B. √2 – 1                   C. √2 + 1                             D. √2

Answer:

Class_12_Application_Of_Integrals_Area_Under_Curve_20

The given equations are y = cos x   ........1

and y = sin x    .........2

Required area = Area (ABLA) + area (OBLO)

                          = ʃ1/√2x dy + ʃ11/√2  x dy

                          = ʃ1/√2cos-1 y dy + ʃ11/√2  sin-1 y dy

Integrating by parts, we get

= [y * cos-1 y - √(1 – y2)]1/√21 + [y * sin-1 y - √(1 – y2)]11/√2 

= [cos-1 1 – (1/√2) * cos-1 (1/√2)  + √(1 – 1/2)] + [(1/√2) * sin-1 (1/√2)  + √(1 – 1/2) - 1]

= -π/4√2 + 1/√2 + π/4√2 + 1/√2 – 1

= 2/√2 – 1

= (√2 – 1) units

Thus, the correct answer is option B.

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