NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

NCERT
SOLUTIONS
CLASS - 12th

aglase .co

Class : 12th
Subject : Maths
Chapter : 8
Chapter Name : Application of integrals

Q1 Find the area of the region bounded by the curve y 2
= x and the lines x = 1, x = 4 and the x-axis in the rst quadrant.

The area of the region bounded by the curve, y = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.
2

4

Area of ABCD = ∫ ydx

= ∫ √xdx

3

2
= [ ]
3

2 1

2 3 3

= [(4) 2 − (1) 2 ]
3

2
= [8 − 1]
3

14
= units
3

Page : 365 , Block Name : Exercise 8.1

Q2 Find the area of the region bounded by y 2
= 9x , x = 2, x = 4 and the x-axis in the rst quadrant.

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

4

Area of ABCD = ∫ ydx
2

4

= ∫ 3√xdx
2

4
3
= 3[ ]
2
2

3
4

= 2[x 2 ]
2

3 3

= 2 [(4) 2 − (2) 2 ]

= 2[8 − 2√2]

= (16 − 4√2) units

Page : 365 , Block Name : Exercise 8.1

Q3 Find the area of the region bounded by x 2
= 4y , y = 2, y = 4 and the y-axis in the
rst quadrant.

the area of the region bounded by x = 4y, y = 2, y = 4 and the y-axis is the area ABCD.
2

4

Area of ABCD = ∫ xdy
2

4

= ∫ 2√ydy
2

4

= 2∫ √ydy
2

3

2
= 2[ ]
2
2

4 3 3

= [(4) 2 − (2) 2 ]
3

4
= [8 − 2√2]
3

32 − 8√2
= ( ) units
3

Page : 366 , Block Name : Exercise 8.1

Q4 Find the area of the region bounded by the ellipse
2 2
x y
+ = 1
16 9

2 2

The given equation of the ellipse, , can be represented as
x y
+ = 1
16 9

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

4

Area of OAB = ∫ ydx
0

4 2
x
= ∫ √1 − dx
16
0

4
3
= ∫ √16 − x2 dx
4
0
4
3 x 16 x
= [ √16 − x2 + sin
−1
]
4 2 2 4
0

3
−1 −1
= [2√16 − 16 + 8 sin (1) − 0 − 8 sin (0)]
4

3 8π
= [ ]
4 2

3
= [4π]
4

= 3π

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

Page : 366 , Block Name : Exercise 8.1

Q5 Find the area of the region bounded by the ellipse
2 2
x y
+ = 1
4 9

The given equation of the ellipse can be represented as

2 2
x y
+ = 1
4 9

2
x
⇒ y = 3√1 −
4

It can be observed that the ellipse is symmetrical about x-axis and y-axis.
∴ Area bounded by ellipse = 4 × Area OAB
2

∴ Area of OAB = ∫ ydx
0

2 2
3
x
= ∫ √1 − dx [ Using (1)]
4
0

2
3
= ∫ √4 − x2 dx
2
0
2
3 x 4 x
= [ √4 − x2 + sin
−
]
2 2 2 2
0

3 2π
= [ ]
2 2

3π
=
2
3π
Therefore, area bounded by the ellipse =4 × = 6π units
2

Page : 366 , Block Name : Exercise 8.1

Q6 Find the area of the region in the rst quadrant enclosed by x-axis, line x = √3y and the circle x 2
+ y
2
= 4

The area of the region bounded by the circle,x 2
+ y
2
= 4, x = √3y , and the x-axis is the area OAB.

The point of intersection of the line and the circle in the rst quadrant is (√3, 1) .
Area OAB = Area ΔOCA + Area ACB
1 1 √3
Area of OAC = × OC × AC = × √3 × 1 = … (i)
2 2 2
2

Area of ABC = ∫ ydx
√3

2

= ∫ √4 − x2 dx

√3

2
x 4 x
= [ √4 − x2 + sin
−1
]
2 2 2
√5

π √3 √3
−1
= [2 × − √4 − 3 − 2 sin ( )]
2 2 2

√3 1
= [π − − 2( )]
2 3

√3 2π
= [π − − ]
2 3

π √3
= [ − ] . . . (ii)
3 2

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

√3π 3√2 π
+ = units
2 2 3

Page : 366 , Block Name : Exercise 8.1

Q7 Find the area of the smaller part of the circle x 2
+ y
2
= a
2
cut off by the line
a
x =
√2

2 2 2
The area of the smaller part of the circle, x + y = a , cut off by the line,

area ABCDA.

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

π

Area of ABC = ∫ ydx
a
π

= ∫ √a2 − x2 dx

√2

a
2
x a x
= [ √a2 − x2 + sin
−1
]
2 2 a a

√2

2 2 2
a π a a a 1
−1
= [ ( ) − √a2 − − sin ( )]
2 2 2 2
2 √2 √2

2 2
a π a a a π
= − ⋅ − ( )
4 2 4
2 √2 √2

2 2 2
a π a a π
= − −
4 4 8
2
a π
= [π − 1 − ]
4 2
2
a π
= [ − 1]
4 2
2 2
a π a π
⇒ Area ABCD = 2 [ ( − 1)] = ( − 1)
2 2 2 2
2 2 2 a
Therefore, the area of smaller part of the circle, x + y = a , cut off by the line, x =
√2

2
a π
( − 1) units.
2 2

Page : 366 , Block Name : Exercise 8.1

Q8 The area between x = y and x = 4 is divided into two equal parts by the line x = a, nd the value of a.
2

{ The line, } x=a \text { , divides the area bounded by the parabola and } x=4 \text { into two equal }} \\ {\text { parts. }} \\ {\therefore \text { Area OAD

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

Area OED = ∫ ydx
0
π

= ∫ √xdx
0
a
3
= [ ]
2
0

2 3

= (a) 2 . . . (i)
3
4

Area of EF CD = ∫ √xdx
0

3

2
= [ ]
3

2 0

2 3

= [8 − a 2 ] … (ii)
3

From (1) and (2), we obtain

3 3
2 2
(a) 2 = [8 − (a) 2 ]
3 3

3

⇒ 2 ⋅ (a) 2 = 8

3

⇒ (a) 2 = 4

2

⇒ a = (4) 3

2

Therefore, the value of a is (4) 3

Page : 366 , Block Name : Exercise 8.1

Q9 Find the area of the region bounded by the parabola y = x 2
and y = |x|

2
Find the area of the region bounded by the parabola y = x and y = |x|

Answer

2
The area bounded by the parabola, x = y, and the line, y = |x|, can be represented as

The given area is symmetrical about y -axis.

∴ Area OACO = Area ODBO

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

Area of OACO = Area ΔOAB − Area OBACO
1 1 1
∴ Area of ΔOAB = × OB × AB = × 1 × 1 =
2 2 2
1
3
2 x 1
Area of OBACO = ∫ ydx = ∫ x dx = [ ] =
0 3 3
0

Therefore,required area
⇒ Area of OACO = Area of ΔOAB − Area of OBACO

1 1
= −
2 3

1
=
6

1 1
= 2[ ] =
6 3

Page : 366 , Block Name : Exercise 8.1

Q10 Find the area bounded by the curve x = 4y and the line x = 4y – 2
2

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

Answer

2
The area bounded by the curve, x = 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.

1
A are (−1, )
4

Coordinates of point

Coordinates of point B are (2, 1)

We draw AL and BM perpendicular to x -axis.

It can be observed that,

Area OBAO = Area OBCO + Area OACO … (i)

Then, Area OBCO = Area OMBC - Area OMBO
2 2 2
x + 2 x
= ∫ dx − ∫ dx
4 4
0 0

2 2
2 3
1 x 1 x
= [ + 2x] − [ ]
4 2 4 3
0 0

1 1 8
= [2 + 4] − [ ]
4 4 3

3 2
= −
2 3

5
=
6

Similarly, Area OACO = Area OLAC - Area OLAO
∞ 2
x+2 x
= ∫ dx − ∫ dx
−1 4 −1 4

0 0
2 3
1 x 1 x
= [ + 2x] − [ ]
4 2 4 3
−1

2 3
1 (−1) 1 (−1)
= − [ + 2(−1)] − [− ( )]
4 2 4 3

1 1 1
= − [ − 2] −
4 2 12

1 1 1
= − −
2 8 12

7
=
24

Therefore, required area = ( 5

6
+
7
) =
9

8
24

Page : 366 , Block Name : Exercise 8.1

Q11 Find the area of the region bounded by the curve y = 4x and the line x = 3 2

The region bounded by the parabola, y 2
= 4x, and the line, x = 3, is the area OACO.

The area OACO is symmetrical about x -axis.

∴ Area of OACO = 2 (Area of OAB)

3

Area OACO = 2 [∫ ydx]
0

= 2∫ 2√xdx

3
3
x
= 4[ ]
2
0

8 3

= [(3) 2 ]
3

= 8 √3

Therefore,the reqiured area is 8√3

Page : 366 , Block Name : Exercise 8.1

Q12 Area lying in the rst quadrant and bounded by the circle x 2
+ y
2
= 4 and the lines x = 0 and x = 2 is
A. π
π
B.
2

π
C.
3

π
D.
4

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

represented as

2

∴ Area OAB = ∫ ydx
0

2

= ∫ √4 − x2 dx

0

2
x 4 x
= [ √4 − x2 + sin
−1
]
2 2 2
0

π
= 2( )
2

= π units

Thus,the correct answer is A.

Page : 366 , Block Name : Exercise 8.1

Q13 Area of the region bounded by the curve y 2
= 4x, y-axis and the line y = 3 is
A. 2

9
B.
4

9
C.
3

9
D.
2

The bounded by the curve, y 2
= 4x, y -axis, and y = 3 is represented as

3

∴ Area OAB = ∫ xdy
0

3 2
y
= ∫ dy
4
0

3
3
1 y
= [ ]
4 3
0

1
= (27)
12
9
= units
4

Thus the correct answer is B.

Page : 366 , Block Name : Exercise 8.1

Q1 Find the area of the circle 4x 2
+ 4y
2
= 9 which is interior of the parabola x
2
= 4y

Answer. The required area is represented by the shaded area OBCDO.

2 2 2
Solving the given equation of circle, 4x + 4y = 9, and parabola, x = 4y, we obtain the

1 1
point of intersection as (√2, ) and D (−√2, )
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 2 2
(9 − 4x ) x
= ∫ √ dx − ∫ √ dx
0
4 4

√2 √2
1 1
= ∫ √9 − 4x2 dx − ∫
2
x dx
2 0
4 0

√2 √2
3
1 9 2x 1 x
2 −1
= [x√9 − 4x + sin ] − [ ]
4 2 3 4 3
0 0

1 9 2 √2 1
−1 3
= [√2√9 − 8 + sin ] − (√2)
4 2 3 12

√2 9 2 √2 √2
−1
= + sin −
4 8 3 6

√2 9 2 √2
−1
= + sin
12 8 3

1 √2 9 2 √2
−1
= ( + sin )
2 6 4 3

Therefore, the required area OBCDO is

1 √2 9 −1 2√2 √2 9 −1 2√2
(2 × [ + sin ]) = [ + sin ] units
2 6 4 3 6 4 3

Page : 371 , Block Name : Exercise 8.2

Q2 Find the area bounded by curves (x − 1) 2
+ y
2
= 1 and x
2
+ y
2
= 1

2 2 2 2
The area bounded by the curves, (x − 1) + y = 1 and x + y = 1, is represented by

the shaded area as

2 2 2 2
On solving the equations, (x − 1) + y = 1 and x + y = 1, we obtain the point of

1 √3
intersection as A ( , )
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.

1
The coordinates of M are ( , 0)
2

⇒ Area OCAO = Area OMAO+Area MCAM
1
1
2
2
= [∫ √1 − (x − 1) dx + ∫ √1 − x2 dx]
1
0
2

1
1
x − 1 1 2
x 1
2 −1 −1
= [ √1 − (x − 1) + sin (x − 1)] + [ √1 − x2 + sin x]
2 2 2 2 1
0
2

2
⎡ 1 1 1 1 1 ⎤
−1 −1
= − √1 − (− ) + sin ( − 1) − sin (−1) +
⎣ 4 2 2 2 2 ⎦

2
1 −1 1 1 1 −1 1
[ sin (1) − √1 − ( ) − sin ( )]
2 4 2 2 2

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

√3 π π π π 1 π √3 1 π
= [− − + + − ] + [ ( ) − − ( )]
4 12 4 4 12 2 2 8 2 6

√3 π π
= [− − + ]
4 6 2

2π √3
= [ − ]
6 4

√3 √3
Therefore ,required area OBCAO = 2 × (
2π 2π
− ) = ( − ) units
6 4 3 2

Page : 371 , Block Name : Exercise 8.2

Q3 Find the area of the region bounded by the curves y = x 2
+ 2, y = x, x = 0 and x = 3

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

the shaded area OCBAO as

Then, Area OCBAO = Area ODBAO - Area ODCO

3 3
2
= ∫ (x + 2) dx − ∫ xdx
0 0

3 3
3 2
x x
= [ + 2x] − [ ]
3 2
0 0

9
= [9 + 6] − [ ]
2

9
= 15 −
2
21
= units
2

Page : 371 , Block Name : Exercise 8.2

Q4
Using integration finds the area of the region bounded by the triangle whose vertices are

(−1, 0), (1, 3) and (3, 2)

BL and CM are drawn perpendicular to x -axis.

It can be observed in the following figure that,

Area (ΔACB) = Area (ALBA) + Area (BLM CB) − Area (AM CA) … (1)

Equation of line segment AB is

3−0
y − 0 = (x + 1)
1+1

3
y = (x + 1)
2

1
+ 2
3 3 x 3 1 1
∴ Area (ALBA) = ∫ (x + 1)dx = [ + x] = [ + 1 − + 1] = 3 units
−1 2 2 2 2 2 2
−1

Equation of line segment BC is

2−3
y − 3 = (x − 1)
3−1

1
y = (−x + 7)
2

3
3 2
1 1 x 1 9 1
∴ Area (BLMCB) = ∫ (−x + 7)dx = [− + 7x] = [− + 21 + − 7] = 5 units
1 2 2 2 2 2 2
1

Equation of line segment AC is

2−0
y − 0 = (x + 1)
3+1

1
y = (x + 1)
2

3
3 2
1 1 x 1 9 1
∴ Area (AMCA) = ∫ (x + 1)dx = [ + x] = [ + 3 − + 1] = 4 units
2 1 2 2 2 2 2
−1

Therefore, from equation (1), we obtain

Area (ΔABC) = (3 + 5 − 4) = 4 units

Page : 371 , Block Name : Exercise 8.2

Q5
Using integration find the area of the triangular region whose sides have the equations y

= 2x + 1, y = 3x + 1 and x = 4 .

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 (ΔACB) = Area (OLBAO) -Area (OLCAO)

4 4
= ∫ (3x + 1)dx − ∫ (2x + 1)dx
0 0

4 4
2 2
3x 2x
= [ + x] − ( + x]
2 2
0 0

= 28 − 20

= 8 units

Page : 371 , Block Name : Exercise 8.2

Q6
2 2
Smaller area enclosed by the circle x + y = 4 and the line x + y = 2 is

A. 2(n − 2)

B. π − 2

C. 2π − 1

D. 2(n + 2)

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

represented by the shaded area ACBA as

It can be observed that,

Area ACBA = Area OACBO - Area (\DeltaOAB)

2 2
= ∫ √4 − x2 dx − ∫ (2 − x)dx
0 0

2 2
2
x 4 x x
= [ √4 − x2 + sin
−1
] − [2x − ]
2 2 2 2
0 0

π
= [2 ⋅ ] − [4 − 2]
2

= (π − 2) − [4 − 2]

= (π − 2) units

Thus, the correct answer is B.

Page : 372 , Block Name : Exercise 8.2

Q7 Area lying between the curve y 2
= 4x and y = 2x is
2
A.
3

1
B.
3
1
C.
4

3
D.
4

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

area OBAO as

The points of intersection of these curves are 0(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 (ΔOCA)− Area(OCABO)
4 2
= ∫ 2xdx − ∫ 2√xdx
0 0

3 1
1
2
x x2
= 2[ ] − 2[ ]
2 3
0 2 0

4
= ∣
∣1 −
∣
∣
3

1
= ∣
∣−
∣
∣
3

1
= units
3

Thus, the correct answer is B.

Page : 372 , Block Name : Exercise 8.2

Q1
Find the area under the given curves and given lines:

2
(i) y = x , x = 1, x = 2 and x -axis

4
(ii) y = x , x = 1, x = 5 and x -axis

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

2

Area ADCBA = ∫ ydx

2
= ∫ x dx

2
3
x
= [ ]
3
1

8 1
= −
3 3

7
= units
3

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

5

4
Area ADCBA = ∫ x dx

5
5
x
= [ ]
5
1

5
(5) 1
= −
5 5

1
4
= (5) −
5

1
= 625 −
5

= 624.8 units

Page : 375 , Block Name : Miscellaneous Exercise

Q2 Find the area between the curve y = x and y = x 2

Answer. The required area is represented by the shaded area OBAO as

2
The points of intersection of the curves, y = x and y = x , is A(1, 1) .

We draw AC perpendicular to x -axis.

∴ Area (OBAO) = Area (ΔOCA) − Area (OCABO) … (i)
t
2
= ∫ xdx − ∫ x dx
0

1 1
2 3
x x
= [ ] − [ ]
2 3
0 0

1 1
= −
2 3

1
= units
6

Page : 375 , Block Name : Miscellaneous Exercise

Q3
2
Find the area of the region lying in the first quadrant and bounded by y = 4x , x = 0, y

= 1 and y = 4

2
The area in the first quadrant bounded by y = 4x , x = 0, y = 1, and y = 4 is
represented by the shaded area ABCDA as

1

∴ Area ABCD = ∫ xdx
x

√y
= ∫ dx
2
3 4

1 y2
= [ ]
2 3

2

1 3

= [(4) 2 − 1]
3

1
= [8 − 1]
3

7
= units
3

Page : 375 , Block Name : Miscellaneous Exercise

0
Q4 Sketch the graph of y = |x + 3| and evaluate ∫ −6
|x + 3|dx

The given equation is y = |x + 3|

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

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

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

0 −3 0

∴ ∫ (x + 3)dx = − ∫ (x + 3)dx + ∫ (x + 3)dx
−6 −6 −3

−3 0
2 2
x x
= −[ + 3x] + [ + 3x]
2 2
−6 −3

2 2 2
(−3) (−6) (−3)
= − [( + 3(−3)) − ( + 3(−6))] + [0 − ( + 3(−3))]
2 2 2

9 9
= − [− ] − [− ]
2 2

= 9

Page : 375 , Block Name : Miscellaneous Exercise

Q5 Find the area bounded by curve y = sin x between x = 0 and x = 2π

Answer. The graph of y = sin x can be drawn as

∴ Required area = Area OABO + Area BCDB
π 2π
∣ ∣
= ∫ sin xdx + ∫ sin xdx
0 ∣ π ∣

x 2π
= [− cos x] + | − cos x]π |
0

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

= 1 + 1 + |(−1 − 1)|

= 2 + | − 2|

= 2 + 2 = 4 units

Page : 375 , Block Name : Miscellaneous Exercise

Q6 Find the area enclosed between the parabola y = 4ax and the line y = mx
2

Answer. The area enclosed between the parabola, \(y^{2} = 4ax, and the line, y = mx, is represented by the shaded area OABO as

4d 4a
The points of intersection of both the curves are (0, 0) and ( , )
m
2 m

we draw AC perpendicular to x -axis.

∴ Area OABO = Area OCABO – Area (ΔOCA)

3
2
4 4a 2
m 4a
= √a( ) − [( ) ]
3 2 2
m 2 m

2 2
32a m 16a
= − ( )
3 4
3m 2 m

2 2
32a 8a
= −
3 3
3m m
2
8a
= units
3
3m

Page : 375 , Block Name : Miscellaneous Exercise

Q7 Find the area enclosed by the parabola 4y = 3x 2
and the line 2y = 3x + 12

Answer. The area enclosed between the parabola 4y = 3x 2
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)
4 1 2
1 3x
= ∫ (3x + 12)dx − ∫ dx
2 4
−2 −2

4 4
2 3
1 3x 3 x
= [ + 12x] − [ ]
2 2 4 3
−2 −2

1 1
= [24 + 48 − 6 + 24] − [64 + 8]
2 4

1 1
= [90] − [72]
2 4

= 45 − 18

= 27 units

Page : 375 , Block Name : Miscellaneous Exercise

Q8
2 2
x y
Find the area of the smaller region bounded by the ellipse + = 1 and the line
9 4

x y
+ = 1
3 2

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

is represented by the shaded region BCAB as
2 y y
x x
+ = 1 and the line, + = 1
9 4 3 2

∴ Area BCAB = Area (OBCAO) – Area (OBAO)
3 3
2
x x
= ∫ 2√1 − dx − ∫ 2 (1 − ) dx
9 3
0 0

3 3
2 2
= [∫ √9 − x2 dx] − ∫ (3 − x)dx
3 3
0 0

3 3
2
2 x 9 x 2 x
= [ √9 − x2 + sin
−1
] − [3x − ]
3 2 2 3 3 2
0 0

2 9 π 2 9
= [ ( )] − [9 − ]
3 2 2 3 2

2 9π 9
= [ − ]
3 4 2

2 9
= × (π − 2)
3 4
3
= (π − 2) units
2

Page : 375 , Block Name : Miscellaneous Exercise

Q9 Find the area of the smaller region bounded by the Find the area of the smaller region bounded by the ellipse
2 2
x y x y
+ = 1 and the line + = 1
2 2 a
a b b

Answer. The area of the smaller region bounded by the ellipse,
2 2
y y
x
2
+ 2
= 1 and the line
x

a
+
b
= 1 is represented by the shaded region BCAB as
a b

∴ Area BCAB = Area (OBCAO) – Area (OBAO)
a a
2
x x
= ∫ b√1 − dx − ∫ b (1 − ) dx
2
0 a b
a

0 a
b b
= ∫ √a2 − x2 dx − ∫ (a − x)dx
a a
0 0
a a
2 2
b x a x x
= [{ √a2 − x2 + sin
−1
} − {ax − } ]
a 2 2 a 2
0 0

2 2
b a π a
2
= [{ ( )} − {a − }]
a 2 2 2

2 2
b a π a
= [ − ]
a 4 2

Page : 375 , Block Name : Miscellaneous Exercise

Q10 Find the area of the region enclosed by the parabola x = y, the line y = x + 2 and x-axis 2

Answer. The area of the region enclosed by the parabola, x = y, the line, y = x + 2, and x-axis is represented by the shaded region OACO as
2

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

∴ Area OABCO = Area (BCA) + Area COAC
1 0
2
= ∫ (x + 2)dx + ∫ x dx
2 −1

−1 0
2 3
x x
= [ + 2x] + [ ]
2 3
−2 −1

2 2 3
(−1) (−2) (−1)
= [ + 2(−1) − − 2(−2)] + [− ]
2 2 3

1 1
= [ − 2 − 2 + 4 + ]
2 3

5
= units
6

Page : 375 , Block Name : Miscellaneous Exercise

Q11
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 = 11]

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.
∴ Area ADCB = 4 × Area OBAO
d

= 4∫ (1 − x)dx
0

1
2
x
= 4(x − )
2

1
= 4[1 − ]
2
0

1
= 4( )
2

= 2 units

Page : 375 , Block Name : Miscellaneous Exercise

Q12 Find the area bounded by curves
2
{(x, y) : y ≥ x and y = |x|}

Answer. The area bounded by the curves,
{(x, y) : y ≥ x and y = |x|}, is represented by the shaded region as
2

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

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

2
= 2 [∫ xdx − ∫ x dx]

1 1
2 3
x x
= 2 [[ ] − [ ] ]
2 3
0 0

1 1
= 2[ − ]
2 3

1 1
= 2[ ] = units
6 3

Page : 376 , Block Name : Miscellaneous Exercise

Q13 Using the method of integration nd 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 ΔABC are A (2, 0), B (4, 5), and C (6, 3).

Equation of line segment AB is
Equation of line segment AB is

5−0
y − 0 = (x − 2)
4−2

2y = 5x − 10

5
y = (x − 2) . . . (i)
2

Equation of line segment BC is
3−5
y − 5 = (x − 4)
6−4

2y − 10 = −2x + 8

2y = −2x + 18

y = −x + 9 . . . (ii)

Equation of line segment ca is

0−3
y − 3 = (x − 6)
2−6

−4y + 12 = −3x + 18

3
y = (x − 2) . . . (iii)
4

Area (ΔABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)
4 6 6
5 3
= ∫ (x − 2)dx + ∫ (−x + 9)dx − ∫ (x − 2)dx
2 4
2 4 2

4 6 6
2 2 2
5 x −x 3 x
= [ − 2x] + [ + 9x] − [ − 2x]
2 2 2 4 2
2 4 2

5 3
= [8 − 8 − 2 + 4] + [−18 + 54 + 8 − 36] − [18 − 12 − 2 + 4]
2 4
3
= 5 + 8 − (8)
4

= 13 − 6

= 7 units

Page : 376 , Block Name : Miscellaneous Exercise

Q14 Using the method of integration nd 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 … (i)
3x – 2y = 6 … (ii)

And, x – 3y + 5 = 0 … (iii)

The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on x-axis.
Area (ΔABC) = Area (ALMCA) – Area (ALB) – Area (CMB)
4 2 4
x + 5 3x − 6
= ∫ ( ) dx − ∫ (4 − 2x)dx − ∫ ( ) dx
3 2
2

4 4
2 2
1 x 2 1 3x
2
= [ + 5x] − [4x − x ] − [ − 6x]
1
3 2 2 2
1 2

1 1 1
= [8 + 20 − − 5] − [8 − 4 − 4 + 1] − [24 − 24 − 6 + 12]
3 2 2

1 45 1
= ( × ) − (1) − (6)
3 2 2

15
= − 1 − 3
2
15 15 − 8 7
= − 4 = = units
2 2 2

Page : 376 , Block Name : Miscellaneous Exercise

Q15 Find the area of the region {(x, y) : y 2
≤ 4x, 4x
2
+ 4y
2
≤ 9}

Answer. The area bounded by the curves, {(x, y) : y 2
≤ 4x, 4x
2
+ 4y
2
≤ 9} is represented as

The points of intersection of both the curves are
1 1
( , √2) and ( , −√2)
2 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
3
1
2 1
= ∫ 2√xdx + ∫ √9 − 4x2 dx
1 2
0
2

3
1
2 1
2 2
= ∫ 2√xdx + ∫ √(3) − (2x) dx
1 2
0
2

Page : 376 , Block Name : Miscellaneous Exercise

Q16 Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is

A. − 9
15
−
B. 4

15

B. 4

17
D.
4

1

Required area = ∫ ydx
2

1

3
= ∫ x dx
2

1
4
x
= [ ]
4
−2

4
1 (−2)
= [ − ]
4 4

1 15
= ( − 4) = − units
4 4

1 15
= ( − 4) = − units
4 4

Thus, the correct answer is B.

Page : 376 , Block Name : Miscellaneous Exercise

Q17
The area bounded by the curve y = x|x|, x -axis and the ordinates x = −1 and x = 1 is

given by

2 2
[Hint: y = x if x > 0 and y = −x if x < 0]

A. 0

1
B.
3

2
C.
3

4
D.
3

1

Required area = ∫ ydx
−1
∞

= ∫ x|x|dx
−1

0 1

2 2
= ∫ x dx + ∫ x dx
−1 0

0 1
3 3
x x
= [ ] + [ ]
3 3
−1 0

2
= − units
3

2
= units
3

Thus, the correct answer is C.

Page : 376 , Block Name : Miscellaneous Exercise

Q18 The area of the circle x 2
+ y
2
= 16 exterior to the parabola y
2
= 6x
4 4 4 4
(A) (4π − √3) (B) (4π + √3) (C) (8π − √3) (D) (8π + √3)
3 3 3 3

Answer. The given equations are
2 2
x + y = 16 … (i)

2
y = 6x … (ii)

Area bounded by the circle and parabola
= 2[Arca(OADO) + Area(ADBA)]

2 +

= 2 [∫ √16xdx + ∫ √16 − x2 dx]

0 2

3 2
4
⎡ x2 ⎤ x 16 x
= 2 √6{ } + 2[ √16 − x2 + sin
−1
]
⎣ 2 ⎦ 2 2 4
2
0

2
2 3 π 1
−1
= 2 √6 × [x 2 ] + 2 [8 ⋅ − √16 − 4 − 8 sin ( )]
3 0 2 2

4√6 π
= (2√2) + 2 [4π − √12 − 8 ]
3 6

16√3 8
= + 8π − 4√3 − π
3 3

4
= [4√3 + 6π − 3√3 − 2π]
3

4
= [√3 + 4π]
3

2
Area of circle = n(r)

= 16n units

4
∴ Required area = 16π − [4π + √3]
3

4
= [4 × 3π − 4π − √3]
3

4
= (8π − √3) units
3

Thus, the correct answer is C

Page : 376 , Block Name : Miscellaneous Exercise

Q19 The area bounded by the y-axis, y = cos x and y = sin x
When

π
0 ≤ x ≤ .
2

2(√2−1)
A.

B. √2 − 1

C. √2 + 1

D. √2

Answer. The given equations are
y = cos x … (i)
And, y = sin x … (ii)

Required area = Area (ABLA) + area (OBLO)
4 1

= ∫ xdy + ∫ xdy
1 0
1
d
2
−1 −1
= ∫ cos ydy + ∫ sin xdy
1 0

Integrating by parts, we obtain
1
1
−1 √2
−1 2 2
= [y cos y − √1 − y ] + [x sin x + √1 − x ]
1
0
√2

−1 1 −1 1 1 1 −1 1 1
= [cos (1) − cos ( ) + √1 − ] + [ sin ( ) + √1 − − 1]
√2 √2 2 √2 √2 2

−π 1 π 1
= + + + − 1
4√2 √2 4√2 √2

2
= − 1
√2

= √2 − 1 units

Thus, the correct answer is B.
dt
Put 2x = t ⇒ dx =
2

3 1
When x = , t = 3 and when x = ,t = 1
2 2

1

1 3
√(3)2 − (t)2 dt
2
= ∫ 2√xdx + ∫
0 4 1
1
3
2 3
x2 1 t 9 t
= 2[ ] + [ √9 − t2 + sin
−1
( )]
2 4 2 2 3
1

3

2 1 2
1 3 9 3 1 9 1
2 −1 2 −1
= 2[ ( ) ] + [{ √9 − (3) + sin ( )} − { √9 − (1) + sin ( )}]
3 2 4 2 2 3 2 2 3

2 1 9 1 9 1
−1 −1
= + [{0 + sin (1)} − { √8 + sin ( )}]
4 2 2 2 3
3 √2

√2 1 9π 9 1
−1
= + [ − √2 − sin ( )]
3 4 4 2 3

√2 9π √2 9 1
−1
= + − − sin ( )
3 16 4 8 3

9π 9 1 √2
−1
= − sin ( ) +
16 8 3 12

√2
Therefore , the required area is [2 × ( 9π

16
−
9

8
sin
−1
(
1

3
) +
12
)] =
9π

8
−
9

4
sin
−1
(
1

3
) +
1
units
3√2

Page : 376 , Block Name : Miscellaneous Exercise