Application of Integrals - SelfStudys

206 CBSE Chapterwise-Topicwise Mathematics

AREA UNDER SIMPLE CURVES

8 Area of the region bounded by the curve y = f(x), x-axis and the lines x = a and x = b (b > a) is,

Area = y dx f x dxa

b

a

b

∫ ∫= ( )

8.1 Introduction

8.2 Area under Simple Curves

8Maximum weightage is of Area between Two curves.

8 Maximum LA type questions were asked from

Area between Two curves.

8 No VBQ, VSA and SA type questions were asked till now

8.3 Area between Two Curves

QUICK RECAP

8.1 8.2 8.3

60

70

Y

X

8 Area of the region bounded by the curve x = g(y), y-axis and the lines y = a and y = b (b > a) is,

Application of 08 Integrals

Topicwise Analysis of Last 10 Years’ CBSE Board Questions

207Application of Integrals

Area = x dy g y dya

b

a

b

∫ ∫= ( )

Y

X

8 Area of the region bounded by the curve y = f(x), some portion of which is above the x-axis and some below the x-axis is,

Area = f x dx f x dxa

c

c

b( ) ( )∫ ∫+

Y

X

AREA BETWEEN TWO CURVES

8 Area of the region between two curves y = f(x), y = g(x) and the lines x = a, x = b is,

Area ( ) ( ) , ( ) ( ) in ,= − ∫ f x g x dx f x g x a ba

b

Y

X

If f x g x( ) ( ) in [a, c] and f(x) g(x) in [c, b], where a < c < b , then

Area = − + −∫ ∫[ ( ) ( )] [ ( ) ( )]f x g x dx g x f x dxa

c

c

b

x = bx = cx = aO

y = g x( ) y = f x( )

y = f x( ) y = g x( )Y

X

8 Area of shaded portion as shown in �gure,

Area = +∫ ∫f x dx g x dxa

c

c

b( ) ( )

Y

X

8


PREVIOUS YEARS MCQSPrevious Years’ CBSE Board Questions

8.2 Area under Simple CurvesLA (6 marks)

1. Using integration, �nd the area of the region bounded by the line x – y + 2 = 0, the curve x y= and y – axis. (Foreign 2015)

2. Find the area of the region in the �rst quadrant enclosed by the y-axis, the line y = x and the circle x2 + y2 = 32, using integration.

(Delhi 2015C)3. Find the area of the region included between the

parabola 4y = 3x2 and the line 3x – 2y + 12 = 0.(AI 2015C, 2009)

4. Find the area of the region in the �rst quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32. (Delhi 2014)

5. Find the area of the smaller region bounded by

the ellipse x y2 2

9 41+ = and the line x y

3 21+ = .

(Foreign 2014)6. Using integration, �nd the area of the region

bounded by the curves :y = |x + 1| + 1, x = – 3, x = 3, y = 0

(Delhi 2014C)7. Using integration, �nd the area bounded by the

curve x2 = 4y and the line x = 4y – 2.(Delhi 2014C, 2013, 2013C, 2010)

8. Using intergration, �nd the area of the region in the �rst quadrant enclosed by the x – axis, the line y = x and the circle x2 + y2 = 18.

(AI 2014C)9. Find the area of the region bounded by the

parabola y = x2 and y = |x|. (AI 2013)10. Using integration, �nd the area of the region

bounded by the curves y = x2 and y = x.(Delhi 2013C)

11. Using integration, �nd the area of the region enclosed by the curves y2 = 4x and y = x.

(Delhi 2013C)12. Find the area of the region

{(x, y) : x2 + y2 , x + y } (AI 2012)

13. Draw the graph of y = |x + 1|and using integration, find the area below y = |x + 1|, above x - axis and between x = – 4 to x = 2. (Delhi 2012C)

14. Using integration, �nd the area of the region given by {(x, y) : x2 y |x|}

(AI 2012C, 2009C, Delhi 2011C)15. Sketch the graph of y = | x + 3 | and evaluate the

area under the curve y = | x + 3|,above x-axis and between x = – 6 to x = 0. (AI 2011)

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

(AI 2011C, Delhi 2010C)17. Using integration, �nd the area of the following

region : ( , ) :x y x y x− − 1 5 2

(Delhi 2010)18. Using integration, �nd the area of the following

region : ( , ) :x y x y x y2 2

9 41

3 2+ +

(Delhi 2010)

19. Find the area of the region included between the parabola y2 = x and the line x + y = 2.

(AI 2009)

8.3 Area between Two CurvesLA (6 marks)

20. Using integration �nd the area of the region {(x, y) : x2 + y2 2ax, y2 ax; x, y 0}

(Delhi 2016)21. Prove that the curves y2 = 4x and x2 = 4y divide

the area of the square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.

(AI 2016, 2015, Delhi 2009)22. Using integration, �nd the area of the region

bounded by the curves

y x x y x= − + − =4 4 02 2 2, and the x – axis.(Foreign 2016)

23. Using integration, �nd the area of the triangle formed by positive x-axis and tangent and normal to the circle x2 + y2 = 4 at ( , )1 3 .

(Delhi 2015)


24. Find the area of the region {(x, y) : y2 x, x2 + 4y2 }, using integration.

(AI 2015C, 2013, 2008C)

25. Using integration, �nd the area of the region bounded by the triangle whose vertices are (– 1, 2), (1, 5) and (3, 4). (AI 2014)

26. Using the method of integration, �nd the area of the region bounded by the lines 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0.

(AI 2014C, Delhi 2009)

27. Using integration, �nd the area of the triangle PQR, coordinates of whose vertices are P(2, 0), Q(4, 5) and R(6, 3). (AI 2014C)

28. Using integration, �nd the area of the region enclosed between the two circles x2 + y2 = 4 and (x – 2)2 + y2 = 4.

(Delhi 2013, 2008, AI 2013C, 2012C, 2010C)

29. Find the area of the region {(x, y) : y2 ax and x2 + y2 a2} using method of integration.

(AI 2013)

30. Find the area of the region enclosed between the two circles x2 + y2 = 9 and (x – 3)2 + y2 = 9.

(AI 2013C, Delhi 2009)

31. Using integration, �nd the area of the region enclosed between two circles x2 + y2 = 1 and (x – 1)2 + y2 = 1. (AI 2013C, Delhi 2007)

32. Using the method of integration, �nd the area of the region bounded by the lines : 3x – 2y + 1 = 0, 2x + 3y – 21 = 0 and x – 5y + 9 = 0.

(Delhi 2012)

33. Using the method of integration, �nd the area of the region bounded by the following lines :3x – y – 3 = 0, 2x + y – 12 = 0, x – 2y – 1 = 0

(Delhi 2012)

34. Using the method of integration, �nd the area of the region bounded by the following lines :

5x – 2y – 10 = 0, x + y – 9 = 0, 2x – 5y – 4 = 0(Delhi 2012)

35. Using integration, �nd the area of the triangle ABC where A is (2, 3), B is (4, 7) and C is (6, 2).

(Delhi 2012C)

36. Using integration, �nd the area of the region bounded by the two parabolas y2 = 4x and x2 = 4y. (Delhi 2012C)

37. Using integration, �nd the area of the circle x2 + y2 = 16, which is exterior to the parabola y2 = 6x. (AI 2012C, 2007)

38. Using integration, �nd the area of the triangular region whose sides have equations y = 2x + 1, y = 3x + 1 and x = 4. (Delhi 2011, AI 2011C)

39. Using integration, �nd the area of the region bounded by the triangle whose vertices are (2, 5), (4, 7) and (6, 2).

(Delhi 2011 C, AI 2010 C)

40. Using integration, �nd the area of the triangle ABC, coordinates of whose vertices are A(4, 1), B(6, 6) and C(8, 4). (AI 2010)

41. Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y. (AI 2010)

42. Using integration, �nd the area of the region {(x, y) : x2 + y2 x2 6y}. (Delhi 2010C)

43. Find the area of the region lying between the parabolas y2 = 4ax and x2 = 4ay, where a > 0.

(AI 2009)

44. Using integration, �nd the area of the region bounded by the triangle whose vertices are (1, 3), (2, 5) and (3, 4). (Delhi 2009 C)

45. Find the area of the region enclosed between the two curves (x – 6)2 + y2 = 36 and x2 + y2 = 36.

(Delhi 2009C)

46. Using integration, �nd the area of the region bounded by the parabola y2 = 4x and the circle 4x2 + 4y2 = 9. (Delhi 2008)

47. Using integration, �nd the area of the triangular region whose vertices are (1, 0), (2, 2) and (3, 1).

(AI 2008)

48. Using integration, �nd the area lying above x-axis and included between the circle x2 + y2 = 8x and the parabola y2 = 4x. (AI 2008)

49. Find the area of the region {(x, y) :y2 6x, x2 + y2 16} (Delhi 2008 C)


Detailed Solutions

1. We have curves x – y + 2 = 0 and x y= . x y= y = x2, which is a parabola with vertex at origin.From the given equations, we getx – x2 + 2 = 0 x – x + 1= x = 2 or x = –1x = 2 [Qx – 1, x is positive]When x = 2, y = 4So, the point of intersection is (2, 4)Required area

= + − ∫∫( )x dx x dx2 2

0

2

0

2

= + −∫( )x x dx2 2

0

2

= + −

x x x2 3

0

2

22

3

= + − =2 4 83

103

sq. units

2. �e given equation of the circle is x2 + y2 = 32 and the line is y = x�ese intersect at A(4, 4) in the �rst quadrant. �e required area is shown shaded in the �gure. Points

B(0, 4) and C 0 4 2,

Required area = Area BACB + Area OABO

= +∫ ∫x dy x dy14

4 2

20

4= − +∫ ∫32 2

4

4 2

0

4y dy y dy

= − +∫ ∫4 22 2

4

4 2

0

4y dy y dy

=−

+

+

−y y y y322

322 4 2 2

21

4

4 22

0

4

sin

=

+ −

+

+ −− −4 2 0

216 1

4 22

1612

12

4 01 1 2sin sin ( )

= − +

+ = −

=16

28 16

48 16

2 44

sq.units

3. �e given parabola is y x= 34

2 ...(i)

and the line is 3x – 2y + 12 = 0 ...(ii)Solving (i) and (ii), we get

3 2 34

12 02x x−

+ = 3 3

212 02x x− + =

6x – 3x2 + 24 = 0 x2 – 2x – 8 = 0 (x – 4) (x + 2) = 0 x = 4, – 2Putting values of x in (i), we get

y y= = − =34

4 12 34

2 32 2( ) ( )and =

Hence the line and parabola intersect at the points A (–2, 3) and B (4, 12).

Required area =+

−

−−∫∫

3 122

34

2

2

4

2

4 xdx x dx

= +

−

− −

12

32

1234 3

2

2

43

2

4xx x

= + − − − +

12

24 48 6 2434

643

83

[( ) ( )]

= + − = − =12

72 1834

24 45 18 27[ ] ( ) sq. units

4. We have curves,y = x ...(i) and x2 + y2 = 32 ...(ii)Curves (i) and (ii) intersect at (4, 4)�e region enclosed by y = x, x2 + y2 = 32 and x-axis in the �rst quadrant is shown below:


Required area = Area of region OBAO= Area of OBM + Area of region BMAB

= + = + −∫ ∫ ∫ ∫y dx y dx xdx x dx 0

4

4

32

0

42

4

3232

=

+ − +

−x x xx2

0

42 1

4

32

2 232 32

2 32sin

= 8 + [8 – (8 + 4)] = 4 sq. units.

5. We have x y2 2

9 41+ = ...(i) and x y

3 21+ = ...(ii)

Curve (i) is an ellipse of the form xa

yb

2

2

2

2 1+ = .

�at means its major axis is along x – axis. Also this ellipse is symmetrical about the x – axis.

Required Area = − − −∫ ∫23

3 23

32 2

0

3

0

3( ) ( )x dx x dx

= − +

− −

−

−23 2

992 3

23

32

2 1

0

3 2

0

3x

xx xsin

= +

− +

+ − − −2

30 9

21 0 9

20

13 0 91 1 2sin ( ) sin ( )

= −32

3 sq. units.

6. Here, y = |x + 1| +1

yx x

x x=

+ −− −

2 11

ifif

We now draw the lines : y = 0, x = 3, x = –3 andy = x + 2 if x −i y=−x if x < – iiLines (i) and (ii) intersect at (– 1, 1)

Required Area = − + +−

−

−∫ ∫( ) ( )x dx x dx3

1

1

32

= −

+ +

−

−

−

x x x2

3

1 2

1

3

2 22

= − − + − + +12

1 9 12

9 1 2 3 1( ) ( ) ( )

= 4 + 4 + 8 = 16 sq. units.7. �e given curve is x2 = 4y ...(i)�e given line is x = 4y – 2 ...(ii)

Putting 4y = (x + 2) from (ii) in (i), we get (x + 2) = x2

x2 – x – 2 = 0 (x – 2) (x + 1) = 0 x = 2, –1Putting x = 2 in (i), we get y = 1

Putting x = – 1 in (i), we get y = 14

�us the points of intersection of the given curve

and line are A −

1 14

, and B (2, 1)

Required Area

= +

− = + −

− − −∫ ∫ ∫

x dx x dx x x dx24 4 4

12 4

1

2 2

1

2 2

1

2

=

+ −

−

−−

14 2

12

14 3

2

1

2

12

3

1

2x x x[ ]

= − + + − +18

4 1 12

2 1 112

8 1[ ] [ ] [ ]

= + − = + −

38

32

34

32

14

1 12

=

=3

234

98

sq. units


8. Refer to answer 4.9. �e given curves are y = x2 ...(i)

y = x =−

xx

xx

,,if

if00

...(ii)

�eir points of intersection are A (1, 1), O(0, 0) and B (–1, 1).In view of symmetry, the required area

= − = −

∫2 2

2 32

0

1 2 3

0

1( )x x dx x x

= −

=2 1

213

13

sq.unit.

10. Refer to answer 9.11. �e given curves are y2 = 4x, y = x�ey intersect at O(0, 0) and A (4, 4).

(4, 4)

(4, 0)

Required area = − = −

∫ 4 2

3 2 20

4 3 2 2

0

4

x x dx x x/

/

= − = − =43

4 12

4 323

8 83

3 2 2/ sq.units.

12. �e given curves arex2 + y2 = 4 ...(i) and x + y = 2 ...(ii)

Reqd. area = − − − ∫ 4 22

0

2

x x dx( )

= − + − +

−x x x x x4

242 2

22

21

2

0

2

sin

= 0 + 2 sin–1 (1) – 4 + 2 – 0 = 2 − = −

2

2 2( ) sq.units.

13. Here y = |x + 1| =+ −

− − −

x xx x

1 11 1,,

ifif

�us we get two lines –x + y = 1 ...(i), x + y = –1...(ii)�eir graphs are as shown and the area to be calculated is shaded.

Hence, the required area

= + + − −− −

−

∫ ∫( ) ( )x dx x dx1 11

2

4

1

= +

− +

− −

−x x x x

2

1

2 2

4

1

2 2

= + − −

− −

− −

( ) ( )2 2 12

1 12

1 8 4

= + =92

92

9 sq.units.

14. Refer to answer 9.



17. We have curves, y = | x – 1 | and y x= −5 2

y xx xx x

= − =−

− −

| |( )

11 11 1

and y2 = 5 – x2 x2 + y2 = 5


�e rough sketch of the circle and the line is

Required area

= − − − − − −− −∫ ∫ ∫5 1 12

1

2

1

1

1

2x dx x dx x dx( ) ( )

= − +

+ −

− −

−

−

−

x x x

x x x x

25 5

2 5

2 2

2 1

1

2

2

1

1 2

1

sin

22

= +

− − + −

− −1 52

25

1 52

15

1 1sin sin

+ −

− +

− − − −

12

1 12

1 2 2 12

1( )

= − + +

− −12

52

25

15

1 1sin sin

= − + +

− −12

52

15

15

1 1cos sin

= − + =−

12

52 2

5 24

sq. units

18. Refer to answer 5.19. We have, y2 = x and x + y = 2 Solving these two equations, we get y2 + y – 2 = 0 (y + 2)(y – 1) = 0 y = – 2, 1 When y = – 2, x = 4 and when y = 1, x = 1

Points of intersection are A (1, 1) and B (4, –2). Required area

= − − = − −− − −∫ ∫ ∫( ) ( )2 22

12

2

12

2

1y dy y dy y y dy

= − −

−

22 3

2 3

2

1

y y y = − −

− − − +

2 1

213

4 42

83

= − =8216

92

sq. units.

20. Let R = {(x, y) : x2 + y2 2ax, y2 ax ; x, y 0} R = R1 R2 R3where R1 = {(x, y) : x2 + y2 2ax},R2 = {(x, y) : y2 ax} and R3 = {(x, y) : x 0, y 0}Region R1 : (x – a)2 + y2 = a2 represents a circle with centre at (a, 0 ) and radius a.Region R2 : y2 = ax represents a parabola with vertex at (0, 0) and its axis along x-axis.Region R3 : x 0, y 0 represents the �rst quadrant. R = R1 R2 R3 is the shaded portion in the �gure.Since, given curves are x2 + y2 = 2ax and y2 = ax So, point of intersection of the curves are (0, 0) and (a, a).

Required area = − − − ∫ a x a ax dxa

2 2

0( )

= − − − +

−

−

−

12

223

2 2

21 3 2

0

( ) ( )

sin /

x a a x a

a x aa

axa

= −

− −

−23

12

13 2 2 1a a a/ sin

= − − −

= − +

23 2 2

23 4

2 2 2 2a a a a

= −

4

23

2a sq. units

21. We have y2 = 4x =x y2

4

Also, x y y y22 2

44

4=

=


y4 = 64y y = 0 or y = 4When y = 0, x = 0 and when y = 4, x = 4So, the points of intersection are O(0, 0) and P(4, 4)Let A1, A2, A3 be the area denoted in the �gure.We need to prove A1 = A2 = A3.

Ax

dxx

1

2 3

0

4

0

4

414 3

63

= =

=

∫ sq. units

A x x dx2

2

0

44

4= −

∫

= −

=4

3 12163

3 23

0

4

x x/ sq. units

Ay

dyy

3

2

0

4 3

0

4

414 3

163

= =

=∫ sq. units

�erefore, A A A1 2 3163

= = = sq. units

�us y2 = 4x and x2 = 4y divide the area of square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.

22. �e given curves are y x= −4 2 andx y x2 2 4 0+ − =

y x x y= − + =4 42 2 2 ...(i)�is represents a circle with centre O(0, 0) and radius = 2 units. x2 + y2 – 4x = 0 (x – 2)2 + y2 = 4 ...(ii)�is represents a circle with centre B(2, 0) and radius = 2 units.Solving (i) and (ii), we get (x – 2)2 = x2 x = 1y2 = 3 y = ± 3

�us, the given curves intersect at A 1 3, Required area = Area of the region OABO

= − − + −∫ ∫4 2 420

1 21

2x dx x dx

= − − − + −

−12 2 4 2

42

22

2 1

0

1

x xxsin

+ − +

−12

4 42 2

2 1

1

2x x xsin

= − + −

− + −− −3

22 1

20 2 11 1sin [ sin ( )]

+ + − +

− −0 2 1

32

2 12

1 1sin ( ) sin

= − − + + − − 32

26

22

22

32

26

= −

43

3 sq. units

23. Given equation of circle is x2 + y2 = 4Di�erentiate w.r.t. ‘x’ on both sides, we get

2 2 0x y dydx

+ =

= −dydx

xy

= −dydx ( , )1 3

13

Equation of tangent at ( , )1 3 is

y x− = − −3 13

1( ) + =x y3 4

Equation of normal at ( , )1 3 is

y x− = −3 3 1( ) − =3 0x y

So, point of intersection of x y+ =3 4

and 3 0 1 3x y A− = is ( , )

Required area = area of OAB

= 3 430

1

1

4x dx x dx∫ ∫+

−


=

+ −

= + − −

−

32

13

42

32

143

4 11

3 24 1

2

0

1 2

1

4

2 2

xx

x

( ) ( )

= + − 32

123

12 3

15 = =6 3

32 3 sq. units

24. Let R = {(x, y) : y2 ≤ 4x, 4x2+ 4y2 ≤ 9}= {(x, y): y2 ≤4x }(x, y) : 4x2 + 4y2 ≤ 9 } = R1 R2.

Let 4x2+ 4y2 = 9 ...(i)

x2+ y2 = 32

2

and y2 = 4x ...(ii)R1 is the region lying inside y2 = 4x

R2 is the region lying inside x2 + y2 = 94

Curves (i) and (ii) intersect at A 12

2,

and

B 12

2,−

As both the curves are symmteric. about x – axis. Reqd. area = 2 (Area of the shaded region above x – axis) = 2 (Area OADO + ADCA)

= +

−

=

∫ ∫2 2 32 2 2

3 20

12 2

2

12

3 23 2xdx x dx x

//

/

0

12

+

−

+

−

x xx

32

23 2

2 3 2

22

21

12

3 2

( / ) sin/

/

=

+ − −

− −2 4

312

98

1 14

2 98

13

3 21 1

/sin ( ) sin

= − + −

−2 4

31

2 22

4916

98

13

1 sin

= + −

−2

698

94

13

1 sin sq.units.

25. Let A (–1, 2), B (1, 5) and C (3, 4)

Eq. of AB is y x− = −+

−5 5 21 1

1( ) y x= +32

72

Eq. of BC is y x− = −−

−4 4 53 1

3( ) y x= − +12

112

Eq. of AC is y – 2 = 4 23 1

1−+

+( )x y x= +12

52

Area of the reqd. triangular region, ABC= Area of trap. ALMB + Area of trap. BMNC – Area of trap. ALNC

= +

+ −

+

− +

− −∫ ∫∫3

272 2

112 2

521

1

1

3

1

3x dx x dx x dx

= +

+ −

+

− +

− −

34

72 4

112 4

52

2

1

1 2

1

3 2

1

3x x x x x x

= +

− −

+

− +

−

− +

34

72

34

72

94

332

14

112

− +

+ −

94

152

14

52

= 7 – 2 + 11 – 2 – 10 = 4 sq. units.

26. �e given lines are 2x + y = 4 ...(i) 3x – 2y = 6 ...(ii) and x – 3y + 5 = 0 ...(iii)Solving (i) and (ii), we get x = 2, y = 0Solving (ii) and (iii), we get x = 4, y = 3Solving (i) and (iii), we get x = 1, y = 2

Y

XO

32

=6

x–

y

B (4, 3)A(1, 2)

C(2, 0) QP

2 = 4x + y

x y +–3 5 = 0

Required area = area (ABC) = area (ABQP) – area (APC) – area (BCQ)


= + − − − −∫ ∫∫

x dx x dx x dx53

4 2 3 62

1

4

2

4

1

2( )

= +

− − − −

13 2

5 4 12

32

62

1

42

12

2

2

4x x x x x x[ ]

= + − +

− − − −1

38 20 1

25 8 4 4 1( ) [( ) ( )]

− − − −12

24 24 6 12[( ) ( )]

= − −

− − − + =1

328 1

25 4 3 1

20 6 7

2[ ] [ ] sq. units


28. �e given circles areC1 : x2 + y2 = 4 ...(i) and C2: (x – 2)2 + y2 = 4 ...(ii)Eliminating y from (i) and (ii), we get4 – x 2 = 4 – (x – 2)2 4x = 4 x = 1 Putting x = 1 in (i), we get y2 = 3 y = ± 3 Points of intersection of the two circles areA( , )1 3 and B( , )1 3−

Y

X

Y

X

Required area = 2(area AOCA) = 2(area AODA+ area ADCA)

= − − + −∫ ∫2 4 2 2 42

0

12

1

2( )x dx x dx

= − − − + −

−2 22

4 2 42

22

2 1

0

1x x x( ) sin

+−

+

−242

42 2

21

1

2x x xsin

= − + −

− −

+ − −

2 3

22

62

22

23

22

6

= − + −

= − +

2 3 2 2

32 3 3 4

3 sq. units

29. Let R = { (x, y) : y2 ≤ 6 ax and x2 + y2 ≤ 16 a2}Let us draw the curvesy2 = 6ax ...(i) and x2+ y2 = 16a2 ...(ii)

From (i) and (ii),x2 + 6ax = 16 a2 (x + 8a) (x – 2a) = 0 x = 2a(Note : x ≠ – 8a as curve (i) lies in 1st and 4th quadrants only).Curves intersect at pointB (2a, 2 3a ) and C (2a, – 2 3a )Now in view of symmetry, the required area

= + −

∫ ∫2 6 16

0

22 2

2

4axdx a x dx

a

a

a

=

+ −+

−

2 63 2

162

162 4

3 2

0

2

2 2 21

a x

x a x a xa

a/

/

sin

2

4

a

a

= −

2 23

6 2 2 03 2a a( )/

+ − −

− −8 1 2 12

28 1

22 1 2 1a a a asin ( ) sin

= + − − 16 33

162

4 3 166

2 2 2 2a a a a

= +

4 33

163

2 a sq.units.



32. �e given lines are 3x – 2y + 1 = 0 ...(i), 2x + 3y – 21 = 0 ...(ii)and x – 5y + 9 = 0 ...(iii)


Solving (i) and (ii), we get x =3, y = 5Solving (ii) and (iii), we get x =6, y = 3Solving (i) and (iii), we get x =1, y = 2

Area of ∆ABC= area PQCA + area QRBC – area PRBA

= + + − − +∫ ∫ ∫

3 12

21 23

95

1

3

3

6

1

6x dx x dx x dx

= +

+ −

− +

34 2 7

3 1095

2

1

3 2

3

6 2

1

6

x xx x x x

= + − +

+ − − −27

432

34

12

42 12 21 3{ ( )}

− + − +

185

545

110

95

= − + − − 334

54

12 725

1910

= + − =7 12 252

132

sq.units.

33. Refer to answer 32.34. Refer to answer 32.

35. Here A (2, 3), B (4, 7) and C (6, 2)

Eq. of AB is y x− = −−

−3 7 34 2

2( ) y =x–

Eq. of BC is y x− = −−

−7 7 24 6

4( ) y x= − +52

17

Eq. of AC is y x− = −−

−3 2 36 2

2( ) y x= − +4

72

�e required area of ABC

= − + − +

− − +

∫∫ ∫( )2 1 5

217

4724

6

2

4

2

6x dx x dx x dx

= − + − +

− − +

x x x x x x2

24 2

4

6 2

2

654

178

72

= [16 – 4 – (4 – 2)] +[ – 45 + 102 –(– 20 + 68)]

− − + − − +

92

21 12

7

= 10 + 9 – 10 = 9 sq. units.


37. We have y2 = 6x which is a parabola and x2 + y2 = 16 which is a circle with centre at (0, 0) and radius 4.Solving both, we get x2 + 6x – 16 = 0 (x + 8)(x – 2) = 0 x = 2 (Q x = – 8 is not possible)

2,2 3A

2 2C

�e required area

= − −

−

∫ ∫2 16 62

4

2

0

2x dx x dx

= − +

−

−

−2

216

162 4

623

2 1

4

23 2

0

2xx

x xsin ( ) /

= +

− −

−

−2 2 3 86

8 1 2 23

6 2 21 sin ( )

= + + −4 3 83

8 163

3 = −323

4 33

= −43

8 3( ) sq. units

38. �e given lines are y = 2x + 1 ...(i)y = 3x + 1 ...(ii) and x = 4 ...(iii)Solving (i) and (ii), we get x = 0, y = 1Solving (ii) and (iii), we get x = 4, y = 13Solving (iii) and (i), we get x = 4, y = 9


Required area = area (ABC) = area (OABD) – area (OACD)

= + − +∫∫( ) ( )3 1 2 10

4

0

4x dx x dx

= +

− + 3

2

2

0

42

04x x x x

= +

− +

− + − +3

216 4 0 0 16 4 0 0( ) ( ) [( ) ( )]

= 28 – 20 = 8 sq. units


40. Here A(4, 1), B(6, 6) and C(8, 4).

Equation of AB is y – 1 = 52

(x – 4) y = 52x – 9

Equation of BC is y – 6 = – 1(x – 6) y = 12 – x

Equation of AC is y – 1 = 34

(x – 4) y = 34x – 2

Required area

= −

+ − − −

∫ ∫ ∫

52

9 1234

24

6

6

8

4

8xdx x dx

xdx( )

=

−

+ −

−

−

52 2

9 122

34 2

22

4

6 2

6

8 2x x x x x x

4

8

= [(45 – 54) – (20 – 36)] + [(96 – 32) – (72 – 18)] – [(24 – 16) – (6 – 8)]= – 9 + 16 + 64 – 54 – 8 – 2 = 7 sq. units.

41. �e given circle is 4x2 + 4y2 = 9 i.e.

x y2 2 94

+ = which has centre (0, 0) and radius 32

.

�e given parabola is x2 = 4y which is symmetrical about positive y-axis with vertex (0, 0).Solving both the equations, we get

4 94

4 16 9 02 2y y y y+ = + − =

+ − = = −( )( ) ,2 9 2 1 0 92

12

y y y

�us, y = 12

Q y −

92

Required area = area (OABC)= 2 [area (ADOA) + area (DABD)]

= + −

∫ ∫2 2 9

40

1 22

1 2

3 2y dy y dy

/

/

/

=

4

3 2

3 2

0

1 2y / /

/+ − +

−2

294

98 3 2

2 1

1 2

3 2y y ysin

/ /

/

= + +

− +

− −83

12 2

2 0 98

1 14

2 98

13

1 1sin ( ) sin

= − + − −2 23

22

94 2

94

13

1sin

= + −

−26

98

94

13

1sin


43. �e equations of the curves are y2 = 4ax ...(i) and x2 = 4ay ...(ii)

Putting y xa

=2

4 from (ii) in (i), we have

xa

ax2 2

44

= x4 = 64a3x

x(x3 – 64a3) = 0 x = 0 or x = 4ax = 0 y = 0 and x = 4a y = 4a


So, the given curves intersect at (0, 0) and (4a, 4a).

So, required area

= −

∫ 2

4

2

0

4ax

xa

dxa

= −

43 12

32

3

0

4a

xx

a

a

= −4

34

412

32

3aa

aa

( )( )

= 163

2a sq. units

44. Let the points be A(1, 3), B(2, 5) and C(3, 4)Equation of AB is y – 3 = 2 (x – 1) y = 2x + 1Equation of BC is y – 5 = – 1(x – 2) y = 7 – x

Equation of AC is y – 3 = 12

(x – 1) y = 12

(x + 5)

Required area

= + + − −+

∫ ∫ ∫( ) ( )2 1 75

21

2

2

3

1

3x dx x dx

xdx

= + + −

− +

x x x x x x2

12 2

2

3 2

1

3

72

12 2

5

= − + −

− − − +

− +

( )6 2 21 92

14 2 12

92

15 12

5

= + − − =4 332

12 12

14 32

[ ] sq. units

45. Refer to answer 28.46. Refer to answer 24.47. Refer to answer 44.48. We have x2 + y2 = 8x and y2 = 4xNow, x2 + y2 = 8x (x – 4)2 + y2 = (4)2

�e centre of circle is (4, 0) and radius is 4.�e point of intersection of circle and parabola are 0(0, 0) and A(4, 4) above x-axis.

�e required area = + − −∫ ∫2 16 40

42

4

8xdx x dx( )

=

+ −− −

2 2

34

216 4

32

0

4

2x x x( ) ( )

+ −

−8 44

1

4

8sin x

= + = + = +43

8 82

323

4 43

8 3 ( ) sq. units


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Application of Integrals - SelfStudys