2 solution 1 · class xii chapter 8 ± application of integrals maths _____ printed from...

Class XII Chapter 8 – Application of Integrals Maths

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Book Name: NCERT Solutions

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.

Solution 1:

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.

Area of ABCD = 1

1ydx

1

1xdx

43

2

1

3

2

x

3 3

2 22

4 13

28 1

3

14 units

3

Question 2:

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


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Solution 2:

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

ABCD.

Area of ABCD = 1

2ydx

1

2

43

2

2

3

332

xdx

x

43

2

2

2 x

3 3

2 22 4 2

28 2 2

3

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.

Solution 3:


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

ABCD.

Area of ABCD = 1

2xdx

1

2

1

2

2

2

ydx

ydx

43

2

2

322

y

3 3

2 24

4 23

48 2 2

3

32 8 2 units

3

Question 4:

Find the area of the region bounded by the ellipse 2 2x

116 9

y

Solution 4:

The given equation of the ellipse, 2 2x

116 9

y , can be represented as


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It can be observed that the ellipse is symmetrical about x-axis and y-axis. ∴ Area bounded by ellipse = 4 × Area of OAB

Area of ABCD = 4

0ydx

24

03 1

16

xdx

42

0

316

4x dx

4

2 1

0

3 1616 sin

4 2 2 4

x xx

1 132 16 16 8sin 1 0 8sin 0

4

3 8

4 2

34

4

3

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

Question5:

Find the area of the region bounded by the ellipse 2 2

14 9

x y

Solution 5:

The given equation of the ellipse can be represented as


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2 2

14 9

x y

2

3 1 ... 14

xy

It can be observed that the ellipse is symmetrical about x-axis and y-axis. ∴ Area bounded by ellipse = 4 × Area OAB

2

0

22

0

22

0

Area of OAB= ydx

3 1 Using (1)4

34

2

x

x

2

2 1

0

3 44 sin

2 2 2 2

3 2

2 2

3

2

x xx

Therefore, area bounded by the ellipse = 3

4 6 units2

Question 6:

Find the area of the region in the first quadrant enclosed by x-axis, line 3x y and the circle

2 2 4x y


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Solution 6:

The area of the region bounded by the circle, 2 2 4x y , 3x y , and the x-axis is the area

OAB.

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

Area OAB = Area ∆OCA + Area ACB

Area of OAC = 1 1 3

3 12 2 2

OC AC …

Area of ABC = 2

3ydx

22

3

2

2 1

3

1

4

44 sin

2 2 2

3 32 4 3 2sin

2 2 2

x dx

x xx

32

2 3

3 2

2 3

3

3 2

…

Therefore, area enclosed by x-axis, the line 3x y , and the circle 2 2 4x y in the first

quadrant = 3 3

2 3 2 3

units


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Question 7:

Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line, 2

ax .

Solution 7:

The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, 2

ax , is the area

ABCDA.

It can be observed that the area ABCD is symmetrical about x-axis. ∴ Area ABCD = 2 × Area ABC

Area of ABC = 2

a

a ydx

2 2

2

22 2 1

2

sin2 2

a

a

a

a

a x dx

x a xa x

a

2 2 22 1 1

sin2 2 2 22 2 2

a a a aa

2 2

.4 2 42 2 2

a a a a

2 2 2

2

2

4 4 8

14 2

14 2

a a a

a

a


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2 2

2 1 14 2 2 2

a aArea ABCD

Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off all the line, 2

ax , is

2

12 2

a

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.

Solution 8: The line, x = a, divides the area bounded by the parabola and x = 4 into two equal parts. ∴ Area OAD = Area ABCD

It can be observed that the given area is symmetrical about x-axis. ⇒ Area OED = Area EFCD

Area OED = 0

a

ydx

0

a

xdx

3

2

0

3

2

a

x

3

22

3a …(1)


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Area of EFCD = 4

0xdx

43

2

0

3

2

x

3

22

8 ... 23

a

From (1) and (2), we obtain

3 3

2 2

3

2

3

2

3

2

2 28

3 3

2. 8

4

4

a a

a

a

a

Therefore, the value of a is 2

34 .

Question 9:

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

Solution 9:

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. ∴ Area OACO = Area ODBO


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

Area of OACO = Area ∆OAB – Area OBACO

13

1 12

0 00

1 1 1Area of OAB= 1 1

2 2 2

1Area of OBACO = ydx x dx

3 3

OB AB

x

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

= 1 1

2 3

= 1

6

Therefore, required area = 1 1

26 3

units.

Question 10:

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

Solution 10:

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

OBAO.

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

Coordinates of point A are 1

1,4

.

Coordinates of point B are (2, 1).

We draw AL and BM perpendicular to x-axis.

It can be observed that,


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Area OBAO = Area OBCO + Area OACO … (1) Then, Area OBCO = Area OMBC – Area OMBO

22 2

0 0

2 22 3

0 0

2

4 4

1 12

4 2 4 3

1 1 82 4

4 4 3

x xdx dx

x xx

3 2

2 3

5

6

Similarly, Area OACO = Area OLAC – Area OLAO 2

0 0

1 1

0 02 3

1 1

2

4 4

1 12

4 2 4 3

x xdx dx

x xx

31 11 1

2 14 2 4 3

1 1 12

4 2 2

1 1 1

2 8 12

7

24

Therefore, required area = 5 7 9

6 24 8

units

Question 11:

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

Solution 11:

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


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The area OACO is symmetrical about x-axis. ∴ Area of OACO = 2 (Area of OAB)

Area OACO = 3

02 ydx

3

0

33

2

0

2 2

342

xdx

x

3

28

33

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. 2

C. 3

D. 4


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Solution 12:

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

as

2

0

22

0

2

2 1

0

Area OAB=

4

44 sin

2 2 2

22

units

ydx

x dx

x xx

Thus, the correct answer is A.

Question 13:

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

A. 2

B. 9

4

C. 9

3

D. 9

2

Solution 13:

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


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3

0

23

0

33

0

Area OAB =

4

1

4 4

127

12

9 units

4

xdy

ydy

y

Thus, the correct answer is B.


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EXERCISE- 8.2

Question 1:

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

Solution 1:

The required area is represented by the shaded area OBCDO.

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

intersection as B 1

2,2

and D 1

2,2

.

It can be observed that the required area is symmetrical about y-axis. ∴ Area OBCDO = 2 × Area OBCO

We draw BM perpendicular to OA.

Therefore, the coordinates of M are 2,0 .

Therefore, Area OBCO = Area OMBCO – Area OMBO

2 22 2

0 0

2 22 2

0 0

9 4

4 4

1 19 4

2 4

x xdx

x dx x dx

22 32 1

0 0

31

1 9 2 19 4 sin

4 2 3 4 3

1 9 2 2 12 9 8 sin 2

4 2 3 12

x xx x


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1

1

1

2 9 2 2 2sin

4 8 3 6

2 9 2 2sin

12 8 3

1 2 9 2 2sin

2 6 4 3

Therefore, the required area OBCDO is

1 11 2 9 2 2 2 9 2 22 sin sin units

2 6 4 3 6 4 3

Question 2:

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

Solution 2:

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

area as

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

as A 1 3

,2 2

and B 1 3

,2 2

It can be observed that the required area is symmetrical about x-axis. ∴ Area OBCAO = 2 × Area OCAO

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

The coordinates of M are 1

,02

Area OCAO Area OMAO Area MCAM


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1

12 221

02

1 1 1x dx x dx

1

122 1 2 1

10

2

1 1 11 1 sin 1 1 sin

2 2 2 2

x xx x x x

2 2

1 1 1 11 1 1 1 1 1 1 1 1 11 sin 1 sin 1 sin 1 1 sin

4 2 2 2 2 2 4 2 2 2

3 1 1 1 3 1

8 2 6 2 2 2 2 8 2 6

3

4 12 4 4 12

3

4 6 2

2 3

6 4

Therefore, required area OBCAO = 2 3 2 3

2 units6 4 3 2

Question 3:

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

Solution 3:

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

area OCBAO as


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Then, Area OCBAO = Area ODBAO – Area ODCO

3 32

0 0

3 33 2

0 0

2

23 2

x dx xdx

x xx

99 6

2

915

2

21units

2

Question 4:

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

(1, 3) and (3, 2).

Solution 4:

BL and CM are drawn perpendicular to x-axis.

It can be observed in the following figure that,

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

Equation of line segment AB is

3 00 1

1 1

31

2

y x

y x

1

21

11

3 3 3 1 1Area ALBA 1 1 1 3units

2 2 2 2 2 2

xx dx x


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Equation of line segment BC is

2 33 1

3 1

17

2

y x

y x

3

23

11

1 1 1 9 1Area BLMCB 7 7 21 7 5units

2 2 2 2 2 2

xx dx x

Equation of line segment AC is

2 00 1

3 1

11

2

y x

y x

3

23

11

1 1 1 9 1Area AMCA 1 3 1 4units

2 2 2 2 2 2

xx dx x

Therefore, from equation (1), we obtain

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.

Solution 5: 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).


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


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4 4

0 03 1 2 1x dx x dx

4 42 2

0 0

3 2

2 2

x xx x

24 4 16 4

28 20

8units

Question6:

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) Solution 6: The smaller area enclosed by the circle, x2 + y2 = 4, and the line, x + y = 2, is represented by the

shaded area ACBA as


Area ACBA = Area OACBO – Area (∆OAB)

2 2

2

0 04 2x dx x dx

2 2

2 1

0 0

44 sin 2

2 2 2 2

x x xx x

2. 4 22

2 units



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Question 7: Area lying between the curve y2 = 4x and y = 2x is

A.2

3 C.

1

4

B. 1

3 D.

3

4

Solution 7: The area lying between the curve, y2 = 4x and y = 2x, is represented by the shaded area OBAO

as

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) 1 1

0 02 2xdx xdx

13

12 2

0

0

2 232

2

x x

41

3

1

3

1 units

3



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Miscellaneous EXERCISE

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

Solution 1: i. The required area is represented by the shaded area ADCBA as

Area ADCBA = 2

1ydx

22

1x dx

23

13

x

8 1

3 3

7 units

3

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

Area of ADCBA = 5

4

1x dx

55

15

x


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55 1

5 5

4 15

5

1625

5

624.8 units

Question 2: Find the area between the curves y = x and y = x2

Solution 2: The required area is represented by the shaded area OBAO as

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

We draw AC perpendicular to x-axis. ∴ Area (OBAO) = Area (∆OCA) – Area (OCABO) … (1) 1 1

2

0 0xdx x dx

1 12 3

0 02 3

x x

1 1

2 3

1 units

6


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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

Solution 3:

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

4

1

4

1

Area ABCD=

4

2

xdx

dx

43

2

1

1

32

2

y

3

21

4 13

18 1

3

7 units

3

Question 4:

Sketch the graph of 3y x and evaluate 0

63x dx

Solution 4:

The given equation is 3y x


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

On plotting these points, we obtain the graph 3y x of as follows.

It is known that, 3 0 for -6 x -3 and 3 0 for -3 x 0x x

0 3 0

6 6 33 3 3x dx x dx x dx

3 02 2

6 3

3 32 2

x xx x

2 2 2

3 6 33 3 3 6 0 3 3

2 2 2

9 9

2 2

9

Question 5:

Find the area bounded by the curve y = sin x between x = 0 and x = 2π Solution 5:

The graph of y = sin x can be drawn as


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∴ Required area = Area OABO + Area BCDB

2

0sin sinxdx xdx

2

0cos cosx x

cos cos 0 cos 2 cos

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

Solution 6:

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

shaded area OABO as


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The points of intersection of both the curves are (0, 0) and 2

4 4,

a a

m m

.

We draw AC perpendicular to x-axis. ∴ Area OABO = Area OCABO – Area (∆OCA) 2 2

4 4

0 02

a a

m maxdx mxdx

2

2

4

3 422

0

0

32

2

2 2

23 2

2

4 4 4

3 2

a

m a

mx xa m

a m aa

m m

2 2

3 2

2 2

3 3

32 16

3 2

32 8

3

a m a

m m

a a

m m

2

3

8 units

3

a

m

Question 7:

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

Solution 7:

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

the shaded area OBAO as

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

We draw AC and BD perpendicular to x-axis. ∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO)

2

1 4

2 2

1 33 12

2 4

xx dx dx

4 42 3

2 2

1 3 312

2 2 4 3

x xx

1 124 48 6 24 64 8

2 4


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1 190 72

2 4

45 18

27 units

Question 8:

Find the area of the smaller region bounded by the ellipse 2 2

19 4

x y and the line 1

3 2

x y

Solution 8:

The area of the smaller region bounded by the ellipse, 2 2

19 4

x y , and the line, 1

3 2

x y , is

represented by the shaded region BCAB as ∴ Area BCAB = Area (OBCAO) – Area (OBAO)

23 3

0 0

3 32

0 0

2 1 2 19 3

2 29 3 1

3 3

x xdx dx

x dx dx


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33 2

2 1

0 0

2 9 29 sin 3

3 2 2 3 3 2

x x xx x

2 9 2 99

3 2 2 3 2

2 9 9

3 4 2

2 92

3 4

32 units

2

Question 9:

Find the area of the smaller region bounded by the ellipse 2 2

2 21

x y

a b and the line 1

x y

a b

Solution 9:

The area of the smaller region bounded by the ellipse, 2 2

2 21

x y

a b , and the line, 1

x y

a b , is

represented by the shaded region BCAB as ∴ Area BCAB = Area (OBCAO) – Area (OBAO)

2

20 01 1

a ax xb dx b dx

a a


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2 2

0 0

a ab ba x dx a x dx

a a

2 22 2 1

0 0

sin2 2 2

a a

b x a x xa x ax

a a

2 22

2 2 2

b a aa

a

2 2

2

4 2

12 2

b a a

a

ba

a

12 2

24

ab

ab

Question 10:

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

Solution 10:

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


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

1 0

2

2 12x dx x dx

1 02 3

2 1

22 3

x xx

2 2 31 2 1

2 1 2 22 2 3

1 12 2 4

2 3

5 units

6

Question 11:

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

[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x – y = 11]

Solution 11:

The area bounded by the curve, 1x y , is represented by the shaded region ADCB as


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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. ∴ Area ADCB = 4 × Area OBAO

1

04 1 x dx

12

0

42

xx

14 1

2

14

2

2 units

Question 12:

Find the area bounded by curves 2, : and y= xx y y x

Solution 12:

The area bounded by the curves, 2, : and y= xx y y x , is represented by the shaded region

as


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

Required area = 2 [ Area (OCAO) – Area (OCADO)]

1 12

0 02 xdx x dx

1 12 3

0 0

22 3

x x

1 12

2 3

1 12 units

6 3

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)

Solution 13:

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


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Equation of line segment AB is

5 00 2

4 2

2 5 10

52 ... 1

2

y x

y x

y x

Equation of line segment BC is

3 55 4

6 4y x

2 10 2 8

2 2 18

9 ... 2

y x

y x

y x

Equation of line segment CA is

0 33 6

2 6y x

4 12 3 18

4 3 6

y x

y x

32 ... 3

4y x

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

4 6 6

2 4 2

5 32 9 2

2 4x dx x dx x dx


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4 6 62 2 2

2 4 2

5 32 9 2

2 2 2 4 2

x x xx x x

5 38 8 2 4 18 54 8 36 18 12 2 4

2 4

35 8 8

4

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

Solution 14:

The given equations of lines are

2x + y = 4 … (1) 3x – 2y = 6 … (2) And, x – 3y + 5 = 0 … (3)

The area of the region bounded by the lines is the area of ∆ABC. AL and CM are the perpendiculars on x-axis.

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

4 2 4

1 1 2

5 3 64 2

3 2

x xdx x dx dx


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4 42 2

22

11 2

1 1 35 4 6

3 2 2 2

x xx x x x

1 1 18 20 5 8 4 4 1 24 24 6 12

3 2 2

1 45 11 6

3 2 2

151 3

2

15 15 8 74 units

2 2 2

Question 15:

Find the area of the region 2 2 2, : 4 .4 4 9x y y x x y

Solution 15:

The area bounded by the curves, 2 2 2, : 4 .4 4 9x y y x x y , is represented as

The points of intersection of both the curves are 1 1

, 2 and , 22 2

.

The required area is given by OABCO.

It can be observed that area OABCO is symmetrical about x-axis. ∴ Area OABCO = 2 × Area OBC

Area OBCO = Area OMC + Area MBC


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1 3

22 21

02

1 32 2

2 21

02

12 9 4

2

12 3 2

2

xdx x dx

xdx x dx

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

Solution 16:

Required area = 1

2ydx

13

2

14

24

x dx

x


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41 ( 2)

4 4

1 154

4 4units


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

Solution 17:

Required area = 1

1ydx

1

1

0 12 2

1 0

x x dx

x dx x dx

0 13 3

1 03 3

x x


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1 1

3 3

2 units

3

Thus, the correct answer is C.

Question 18:

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

A. 44 3

3

B. 44 3

3

C. 48 3

3

D. 48 3

3

Solution 18:

The given equations are

x2 + y2 = 16 … (1) y2 = 6x … (2)

Area bounded by the circle and parabola

2 4

2

0 2

2 Area OADO Area ADBA

2 16 16xdx x dx


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23

42

2 1

2

0

162 6 2 16 sin

3 2 2 4

2

x x xx

23

12

0

2 12 6 2 8. 16 4 8sin

3 2 2x

4 62 2 2 4 12 8

3 6

16 3 88 4 3

3 3

44 3 6 3 3 2

3

43 4

3

44 3 units

3

Area of circle = π (r)2

= π (4)2

= 16π units

4Required area = 16 - 4 3

3

44 3 4 3

3

48 3 units

3

Thus, the correct answer is C.

Question 19:

The area bounded by the y-axis, y = cos x and y = sin x when 02

x

A. 2 2 1

B. 2 1

C. 2 1

D. 2


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Solution 19:

The given equations are

y = cos x … (1) And, y = sin x … (2)

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

11

21

02

1 11 1

1 1

2 2

cos sin

xdy xdy

ydy xdy

Integrating by parts, we obtain 1

11 2 1 2 2

10

2

cos 1 sin 1y y y x x x

1 1 11 1 1 1 1 1cos 1 cos 1 sin 1 1

2 22 2 2 2

1 11

4 2 2 4 2 2

21

2

2 1units


Put 2x = t dx = 2

dt

When x = 3

2, t = 3 and when x =

1

2, t = 1

1

3 2 22

0 1

12 3

4xdx t dt


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1

233

22 1

1

0

1 92 9 sin

3 4 2 2 3

2

x t tt

3

2 2 21 12 1 1 3 9 3 1 9 12 9 3 sin 9 1 sin

3 2 4 2 2 3 2 2 3

1 12 1 9 1 9 10 sin 1 8 sin

4 2 2 2 33 2

1

1

2 1 9 9 12 sin

3 4 4 2 3

2 9 2 9 1sin

3 16 4 8 3

19 9 1 2sin

16 8 3 12

Therefore, the required area is1 19 9 1 2 9 9 1 1

2 sin sin units16 8 3 12 8 4 3 3 2

2 solution 1 · class xii chapter 8 ± application of integrals maths _____ printed from...

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