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Class XII Chapter 8 – Application of Integrals Maths

______________________________________________________________________________

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Top tutor.

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:

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.

Solution 2:

Study Materials NCERT Solutions for Class 6 to 12 (Math & Science)

Revision Notes for Class 6 to 12 (Math & Science)

RD Sharma Solutions for Class 6 to 12 Mathematics

RS Aggarwal Solutions for Class 6, 7 & 10 MathematicsImportant Questions for Class 6 to 12 (Math & Science)

CBSE Sample Papers for Class 9, 10 & 12 (Math & Science)

Important Formula for Class 6 to 12 Math

CBSE Syllabus for Class 6 to 12

Lakhmir Singh Solutions for Class 9 & 10Previous Year Question Paper

CBSE Class 12 Previous Year Question Paper

CBSE Class 10 Previous Year Question Paper

JEE Main & Advanced Question Paper

NEET Previous Year Question Paper

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______________________________________________________________________________


Top tutor.

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:


______________________________________________________________________________


Top tutor.

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:


______________________________________________________________________________


Top tutor.

Area of ellipse = 4 × Area of OAB

Area of OAB = 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

=

3

44

=

3=

area of ellipse = 4 × 3π = 12π units

Question5:

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

14 9

x y+ =

Solution 5:


______________________________________________________________________________


Top tutor.

2 2

14 9

x y+ =

2

3 14

xy = −

Area of 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

− = − +

=

=

area of 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+ =

Solution 6:


______________________________________________________________________________


Top tutor.

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

= −

area enclosed by x-axis 3x y= , and 2 2 4x y+ = in first quadrant =

3 3

2 3 2 3

+ − = units

Question 7:

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

ax = .

Solution 7:


______________________________________________________________________________


Top tutor.

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

= − −

= − −

= −

2 2

Area 2 1 14 2 2 2

a aABCD

= − = −

area of smaller part of x2 + y2 = a2 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.


______________________________________________________________________________


Top tutor.

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

∴ Area OAD = Area ABCD

Area OED = Area EFCD

Area OED = 0

a

ydx

0

a

xdx=

3

2

0

3

2

a

x

=

( )3

22

3a=

Area of EFCD = 4

0xdx

43

2

0

3

2

x

=

3

22

83

a

= −


______________________________________________________________________________


Top tutor.

( ) ( )

( )

( )

( )

3 3

2 2

3

2

3

2

3

2

2 28

3 3

2. 8

4

4

a a

a

a

a

= −

=

=

=

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:

Area OACO = Area ODBO

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

Area of OACO = Area ∆OAM – Area OCAM

13

1 12

0 00

1 1 1Area of OAM= 1 1

2 2 2

1Area of OACM = ydx x dx

3 3

OM AM

x

= =

= = =

⇒ Area of OACO = 1 1

2 3− =

1

6

required area = 1 1

26 3

=

.


______________________________________________________________________________


Top tutor.

Question 10:

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

Solution 10:

Coordinates of point A 1

1,4

−

.

Coordinates of point B (2, 1).

draw AL and BM perpendicular to x-axis.

Area OBAO = Area OBCO + Area OACO

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

= −

=

Area OACO = Area OLAC – Area OLAO


______________________________________________________________________________


Top tutor.

20 0

1 1

0 02 3

1 1

2

4 4

1 12

4 2 4 3

x xdx dx

x xx

− −

−

+= −

= + −

( )( )

( )3

1 11 12 1

4 2 4 3

1 1 12

4 2 2

1 1 1

2 8 12

− − = − + − − −

= − − −

= − −

7

24=

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:

OACO is symmetrical about x-axis.

Area of OACO = 2 (Area of OAB)

Area OACO = 3

02 ydx


______________________________________________________________________________


Top tutor.

3

0

33

2

0

2 2

342

xdx

x

=

=

( )3

28

33

8 3

=

=

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

Solution 12:


______________________________________________________________________________


Top tutor.

2

0

22

0

2

2 1

0

Area OAB =

4

44 sin

2 2 2

22

units

ydx

x dx

x xx

−

= −

= − +

=

=

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:


______________________________________________________________________________


Top tutor.

( )

3

0

23

0

33

0

Area OAB =

4

1

4 4

127

12

9 units

4

xdy

ydy

y

=

=

=

=

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

Solution 1:


______________________________________________________________________________


Top tutor.

Solving 4x2 + 4y2 = 9 and x2 = 4y, point of intersection B 1

2,2

and D 1

2,2

−

.

required area is symmetrical about y-axis.

Area OBCDO = 2 × Area OBCO

draw BM perpendicular to OA.

coordinates of M are ( )2,0 .

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 −

−

= − + −

= − + −

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

−

−

−

= + −

= +

= +

required area OBCDO is


______________________________________________________________________________


Top tutor.

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:

solving (x – 1)2 + y2 = 1 and x2 + y2 = 1, point of intersection A 1 3

,2 2

and B 1 3

,2 2

−

required area is symmetrical about x-axis.

Area OBCAO = 2 × Area OCAO

join AB, intersects OC at M

AM is perpendicular to OC.

coordinates of M 1

,02

Area OCAO Area OMAO Area MCAM = +

( )1

12 221

02

1 1 1x dx x dx

= − − + −

( ) ( )

11

22 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

= − + − − − + − −


______________________________________________________________________________


Top tutor.

3

4 12 4 4 12

3

4 6 2

= − − + + −

= − − +

2 3

6 4

= −

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:

Area OCBAO = Area ODBAO – Area ODCO

( )3 3

2

0 0

3 33 2

0 0

2

23 2

x dx xdx

x xx

= + −

= + −

9

9 62

915

2

21units

2

= + −

= −

=


______________________________________________________________________________


Top tutor.

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 perpendicular to x-axis.

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

Equation of 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

−−

= + = + = + − + =

Equation of 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 AC is

( )

( )

2 00 1

3 1

11

2

y x

y x

−− = +

+

= +


______________________________________________________________________________


Top tutor.

( ) ( )3

23

11

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

2 2 2 2 2 2

xx dx x

−−

= + = + = + − + =

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: vertices of triangle A(0, 1), B(4, 13), and C (4, 9).

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

( ) ( )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:


______________________________________________________________________________


Top tutor.

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

= − −

= −


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:

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

draw AC perpendicular to x-axis


______________________________________________________________________________


Top tutor.

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

= −

= −

=


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.

Area ADCBA = 2

1ydx

22

1x dx=


______________________________________________________________________________


Top tutor.

23

13

x =

8 1

3 3

7 units

3

= −

=

ii.

Area of ADCBA = 5

4

1x dx

55

15

x =

( )5

5 1

5 5= −

( )4 1

55

1625

5

624.8 units

= −

= −

=

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

Solution 2:


______________________________________________________________________________


Top tutor.

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

draw AC perpendicular to x-axis.

∴ Area (OBAO) = Area (∆OCA) – Area (OCABO) 1 1

2

0 0xdx x dx= −

1 12 3

0 02 3

x x = −

1 1

2 3

1 units

6

= −

=

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:


______________________________________________________________________________


Top tutor.

4

1

4

1

Area ABCD=

4

2

xdx

dx

=

43

2

1

1

32

2

y

=

( )3

21

4 13

= −

1

8 13

7 units

3

= −

=

Question 4:

Sketch the graph of 3y x= + and evaluate 0

63x dx

−+

Solution 4:

.

( ) ( )3 0 for -6 x -3 and 3 0 for -3 x 0x x+ +


______________________________________________________________________________


Top tutor.

( ) ( ) ( )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 23 6 3

3 3 3 6 0 3 32 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:

∴ 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:


______________________________________________________________________________


Top tutor.

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

Solution 6:

points of intersection of curves are (0, 0) and 2

4 4,

a a

m m

.

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:


______________________________________________________________________________


Top tutor.

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

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 1

24 48 6 24 64 82 4

= + − + − +

1 1

90 722 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:


______________________________________________________________________________


Top tutor.

∴ 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

= − − −

= − − −

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:

∴ Area BCAB = Area (OBCAO) – Area (OBAO)


______________________________________________________________________________


Top tutor.

2

20 01 1

a ax xb dx b dx

a a

= − − −

( )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:


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Top tutor.

point of intersection of x2 = y and 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 3

1 2 12 1 2 2

2 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:


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Top tutor.

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

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:


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Top tutor.

required area is symmetrical about y-axis.

Required area = 2 [ Area (OCAO) – Area (OCADO)] 1 1

2

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:


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Top tutor.

Equation of AB is

( )

( )

5 00 2

4 2

2 5 10

52

2

y x

y x

y x

−− = −

−

= −

= −

Equation of BC is

( )3 5

5 46 4

y x−

− = −−

2 10 2 8

2 2 18

9

y x

y x

y x

− = − +

= − +

= − +

Equation of CA is

( )0 3

3 62 6

y x−

− = −−

4 12 3 18

4 3 6

y x

y x

− + = − +

= −

( )3

24

y 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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Top tutor.

4 6 62 2 2

2 4 2

5 32 9 2

2 2 2 4 2

x x xx x x

−= − + + − −

5 3

8 8 2 4 18 54 8 36 18 12 2 42 4

= − − + + − + + − − − − +

( )3

5 8 84

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:

AL and CM are 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

+ − = − − −

4 42 2

22

11 2

1 1 35 4 6

3 2 2 2

x xx x x x

= + − − − −

1 1 1

8 20 5 8 4 4 1 24 24 6 123 2 2

= + − − − − − + − − − +


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Top tutor.

( ) ( )1 45 1

1 63 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:

points of intersection of curves are 1 1

, 2 and , 22 2

−

.

required area is OABCO.

area OABCO is symmetrical about x-axis.

∴ Area OABCO = 2 × Area OBC

Area OBCO = Area OMC + Area MBC

( ) ( )

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

= + −

= + −


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Top tutor.

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

−

−

=

=

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


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Top tutor.

[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

−

= +

1 1

3 3

2 units

3

= − − +

=

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 −


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Top tutor.

B. ( )44 3

3 +

C. ( )48 3

3 −

D. ( )48 3

3 +

Solution 18:

Area

( ) ( )2 4

2

0 2

2 Area OADO Area ADBA

2 16 16xdx x dx

= +

= + −

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

= + − −

= + − −


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Top tutor.

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

= − −

= −

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

Solution 19:

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


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Top tutor.

11

21

02

1 11 1

1 1

2 2

cos sin

xdy xdy

ydy xdy− −

= +

= +

11

1 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

= −

= −


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= + −

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

−

−

= + − −

= + − −


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Top tutor.

19 9 1 2sin

16 8 3 12

− = − +

required area is1 19 9 1 2 9 9 1 1

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

− − − + = − +

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Starting at just 11 - AWS

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