Class 12 Chapter 8 - Applications of Integrals

EXERCISE 8.1
1. Find the area of the region
bounded by the curve
y2 = x and the lines x = 1, x = 4
and the x-axis.
Sol. Equation of the curve
(rightward parabola) is y2 = x
∴ y = x ...(i)
(For branch of the parabola above x-axis)
∴ Required area (as shown shaded in the figure)
4 4
= ∫1 y dx = ∫1 x dx (... From (i) y = x)
4
 3
 2
x  3 3
4  1 2 2
= ∫1 x1/ 2 dx =
3
=
3
(4 − 12 )
2
2 3 2+1 2 1 1 1
(4 4 − 1 1) [... 2 + 1+ 1
=
3 x = x 2 = x2 2 = x 2 = x . x 2 = x x ]
2 2 2 14
= (4(2) − 1(1)) = (8 – 1) = × 7 = sq. units.
3 3 3 3
MathonGo 1
 Class 12 Chapter 8 - Applications of Integrals

3
Note. x 2 = x x .
Remark. Equation of the curve is y2 = x.
∴ y = – x for branch of the
parabola below the x-axis.
The reader is within his or her
rights to find the required area as
shown shaded in the figure in the
x=4
remark as ∫ x =1 y dx taking

y = – x.
2. Find the area of the region bounded by y2 = 9x, x = 2, x = 4
and the x-axis in the first quadrant.
Sol. Equation of (rightward
parabola) curve is y2 = 9x
∴ y = 9x = 3 x ...(i)
for branch of curve in first
quadrant.
∴ Required (shaded) area
bounded by curve y 2 = 9x,
(vertical lines x = 2, x = 4),
and x-axis in first quadrant
4 4
= ∫2 y dx = ∫2 3 x dx
(By (i))
4
 3
 2
1 x  2 3
4  2 3
= 3 ∫2 x 2 dx = 3
3
= 3   [4 2 − 2 2 ]
3
2
3
= 2(4 4 – 2 2 ) [... x 2 = x x ]

= 2(8 – 2 2 ) = (16 – 4 2 ) sq. units.

3. Find the area of the region bounded by x2 = 4y, y = 2, y = 4
and the y-axis in the first quadrant.
Sol. Equation of (upward parabola)
curve is x2 = 4y
∴ x = 4y = 2 y ...(i)
for branch of curve in first
quadrant.
∴ Required (shaded) area
bounded by curve x 2 = 4y,
(Horizontal lines y = 2, y = 4)
and y-axis in first quadrant

MathonGo 2
 Class 12 Chapter 8 - Applications of Integrals

4 4
= ∫2 x dy = ∫2 2 y dy (By (i))
4
 3
 2
1 y 
4  2
= 2∫ y 2 dy = 2
3
2
2
4 3 3 4 3
= (4 2 − 22 ) = (4 4 – 2 2 ) (... x2 = x x )
3 3
4  32 − 8 2 
= (4(2) – 2 2 ) =  3
 sq. units.
3  
4. Find the area of the region bounded by the ellipse
x2 y2
+ = 1.
16 9
Sol. Equation of ellipse is
x2 y2
+ = 1 ...(i)
16 9
Here a2(= 16) > b2(= 9)
y2 x2
From (i), = 1 –
9 16
16 − x 2
=
16
9
⇒ y2 =
(16 – x2)
16
3 2
⇒
4 16 − x
y = ...(ii)
for arc of ellipse in first quadrant.
Ellipse (i) is symmetrical about x-axis.
(... On changing y → – y in (i), it remains unchanged).
Ellipse (i) is symmetrical about y-axis.
(... On changing x → – x in (i), it remains unchanged)
Intersections of ellipse (i) with x-axis ( y = 0)
x2
Putting y = 0 in (i), = 1 ⇒ x2 = 16 ⇒ x = ± 4
16
∴ Intersections of ellipse (i) with x-axis are (4, 0) and (– 4, 0).
Intersections of ellipse (i) with y-axis (x = 0)
y2
Putting x = 0 in (i), = 1 ⇒ y2 = 9 ⇒ y = ± 3.
9
∴ Intersections of ellipse (i) with y-axis are (0, 3) and (0, – 3).

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 Class 12 Chapter 8 - Applications of Integrals

∴ Area of region bounded by ellipse (i)

= Total shaded area
= 4 × Area OAB of ellipse in first quadrant
4
= 4 ∫0 y dx (... At end B of arc AB of ellipse;
x = 0 and at end A of arc AB; x = 4)
4 3
= 4 ∫0 4
16 − x2 dx [By (ii)]
4
4 x 2 2 42 −1 x 
∫0 = 3  2 4 − x + 2 sin 4 
2 2
= 3 4 − x dx
  0

 x a2 x
∵

∫ a2 – x 2 dx =
2
a2 – x 2 +
2
sin –1 
a 

4   8π 
= 3  16 − 16 + 8 sin −1 1 − (0 + 8 sin −1 0)  = 3 0 +
2   2 
 π −1 π −1 
∵ sin 2 = 1 ⇒ sin 1 = 2 and sin 0 = 0 ⇒ sin 0 = 0 
 
= 3(4π) = 12π sq. units.
Remark. We can also find area of this ellipse as
3
4 ∫0 x dy

5. Find the area of the region bounded by the ellipse

x2 y2
+ = 1.
4 9
Sol. Equation of the ellipse is

x2 y2
+ = 1 ...(i)
4 9
Here a2(= 4) < b2(= 9)
y2 x2 4 − x2
From (i), = 1 – =
9 4 4
9 3 2
⇒ y2 =
(4 – x2) ⇒
2 4−x
y= ...(ii)
4
for arc of ellipse in first quadrant. Clearly ellipse (i) is
symmetrical about x-axis and y-axis both.
[ . . . On changing y to – y in (i) or x to – x in (i) keep it
unchanged]
Intersections of ellipse (i) with x-axis ( y = 0)
x2
Putting y = 0 in (i), = 1 ⇒ x2 = 4 ⇒ x = ± 2
4

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 Class 12 Chapter 8 - Applications of Integrals

∴ Intersections of ellipse (i) with x-axis are (2, 0) and (– 2, 0)

Intersections of ellipse (i) with y-axis (x = 0)
y2
Putting x = 0 in (i), = 1 ⇒ y2 = 9 ⇒ y = ± 3
9
∴ Intersections of ellipse (i) with y-axis are (0, 3) and (0, – 3).
∴ Area of region bounded by ellipse (i)
= Total shaded area
= 4 × area OAB of ellipse in first quadrant
2
= 4 ∫0 y dx (... At end B of arc AB of ellipse x = 0
and at end A of arc AB, x = 2)
2 3
= 4 ∫0 2
4 − x2 dx (By (ii))
2
3 2 x 22 x
22
− x 2
+ sin − 1 
= 4 .
2 ∫0 22 − x 2 dx 
= 6 2
 2 2  0
 x a2 x
∵

∫ a2 – x 2 dx =
2
a2 – x 2 +
2
sin –1 
a 
2 
= 6  4 − 4 + 2 sin −1 1 − 0 − 2 sin −1 0 
2 
 π 
= 6 0 + 2 . − 0  = 6π sq. units.
 2 
6. Find the area of the region in the first quadrant enclosed
by x-axis, line x = 3 y and the circle x2 + y2 = 4.
Sol. Step I. To draw the graphs
and shade the region whose
area we are to find.
Equation of the circle is
x2 + y2 = 4 = 22 ...(i)
We know that eqn. (i)
represents a circle whose centre
is (0, 0) and radius is 2.
Equation of the given line is
x = 3y
1
⇒ y = x ...(ii)
3
We know that equation (ii) being of the form y = mx where m =
1
= tan 30° = tan θ ⇒ θ = 30° represents a straight line
3
passing through the origin and making angle of 30° with x-axis.
We are to find area of shaded region OAB in first quadrant
(only).

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Class 12 Chapter 8 - Applications of Integrals

Step II. Let us solve (i) and (ii) for x and y to find their
points of intersection.
x x2
Putting y = from (ii) in (i), x2 + = 4
3 3
⇒ 3x2 + x2 = 12 ⇒ 4x2 = 12 ⇒ x2 = 3
⇒ x = ± 3
1
For x = 3 , from (ii), y = 3 3 =1
1
For x = – 3 , from (ii), y = 3
(– 3) = – 1
∴ The two points of intersections of circle (i) and line (ii) are
A( 3 , 1) and D(– 3 , – 1).
Step III. Now shaded area OAM between segment OA of line (ii)
and x-axis
3
= ∫0 y dx (... At O, x = 0 and at A, x = 3)
3 1
= ∫0 3
x dx [By (ii)]

3
1  x2  1 3  3 3
=   =  2 − 0 = = sq. units ...(iii)
3 2
 0 3   2 3 2
Step IV. Now shaded area AMB between arc AB of circle and
x-axis
2
= ∫ 3
y dx (... at A, x = 3 and at B, x = 2)
2
= ∫ 3
22 − x 2 dx (From (ii), y2 = 22 – x2 ⇒ y = 22 − x 2 )
2
x 22 x
=  22 − x2 + sin −1 
2 2 2 3

 x a2 x
∵

∫ a2 – x 2 dx =
2
a2 – x 2 +
2
sin –1 
a 
2  3 3 
=  4 − 4 + 2 sin −1 1 −  4 − 3 + 2 sin −1 
 2  2 2  

π 3 π  π 3 π 3
= 0 + 2 . – – 2 . ∵ sin = ⇒ = sin −1 
2 2 3  3 2 3 2 
3 2π 2π 3 3π − 2π 3
= π – – = π – – = –
2 3 3 2 3 2
π 3
=  3 − 2  sq. units. ...(iv)
 
MathonGo 6
 Class 12 Chapter 8 - Applications of Integrals

Step V. Required shaded area OAB

= Area OAM + Area AMB

3 π 3 π
+  3 − 2  =
= sq. units. [By (iii) and (iv)]
2   3
7. Find the area of the smaller part of the circle x2 + y2 = a2 cut
a
off by the line x = .
2
Sol. Given: Equation of the circle is
x2 + y2 = a2 ...(i)
∴ y2 = a2 – x2
∴ y= a2 − x 2 ...(ii)
for arc BM of circle in Ist
quadrant.
We know that equation (i)
represents a circle whose centre
is origin (0, 0) and radius a.
Clearly, circle (i) is symmetrical both about x-axis and y-axis.
a
We also know that graph of (vertical) line x = is parallel
2
a
to y-axis at a distance (< a) to the right of origin.
2
∴ Area of smaller part of the circle x2 + y2 = a2 cut off by the
a
line x = = Area ABMC = 2 × Area ABM
2
a
= 2 ∫ a y dx
2
a
[... At point B (point of vertical line BC) x =
2
and at point M, x = radius a)
a
= 2 ∫ a a2 − x2 dx (By (ii))
2
a
x 2 a2 2 x
= 2  a –x + sin –1 
 2 2 a  a
2

  a a 
a a 2  a 2
a 2 
= 2  a2 − a2 + sin −1 1 −  2 a2 − + sin −1 2  
 2 2  2 2 2 a  
 a2 π a a2 a2 1 
= 2 0 + − − sin−1 
 2 2 2 2 2 2 2 

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 Class 12 Chapter 8 - Applications of Integrals

 πa2 a a a2 π   π 1 1 π
= 2  4 − −  ∵ sin = ⇒ sin −1 = 
 2 2 2 2 4   4 2 2 4
 πa2 πa2 a2 
= 2  4 − 8 − 4  [... 2
2 2 = ( 2 ) = 2]
 
π π 1  2π − π − 2 
= 2a2  − −  = 2a2  
 4 8 4   8 
2 2
a a π 
= (π – 2) =  − 1  sq. units.
4 2 2 
Note. It may be clearly noted that in this question No. 7 we were
not to find only area AMB or only area AMC because x-axis is
not given to be a boundary of the region in question whose area
is required.
We have drawn x-axis here only as a line of reference because
without drawing x-axis and y-axis as lines of reference, we can’t
draw any graph.
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.
Sol. Equation of the curve (rightward
parabola) is
x = y2 i.e., y2 = x ...(i)
From (i), y = x ...(ii)
for arc OAC of parabola in first
quadrant.
We know that equation (i) represents a right-ward parabola with
symmetry about x-axis.
(... Changing y to – y in (i) keeps it unchanged)
Given: Area bounded by parabola (i) and vertical line x = 4 is
divided into two equal parts by the vertical line x = a.
⇒ Area OAMB = Area AMBDNC.
a 4
⇒ 2 ∫0 y dx = 2 ∫a y dx

(For multipliction by 2 on each side,

see Note above after solution of Q. No. 7)
1
Dividing by 2 and putting y = x = x 2 from (ii),
1 1
a 4
∫0 x 2 dx = ∫a x 2 dx

( )
3 a
x2 0 (x )
3 4
2
a 2 3 2 3 3
⇒ 3 = 3 ⇒ [ a 2 − 0] = [4 2 − a2 ]
3 3
2 2

MathonGo 8
 Class 12 Chapter 8 - Applications of Integrals

2 3 3
Dividing both sides by , = 4 4 – a2
3 a2
3 3
2
Transposing, 2 a2 = 8 ⇒ a2 = 4 ⇒ a =.
43
9. Find the area of the region bounded by the parabola y = x2
and y = | x | .
Sol. The required area is the area included between the parabola
y = x2 and the modulus function
 x, if x ≥ 0
y = | x | = 
− x, if x ≤ 0
We know that, the graph of the modulus function consists of two
rays (i.e., half lines y = x for x ≥ 0 and y = – x for x ≤ 0) passing
through the origin and at right angles to each other. The half line
y = x if x ≥ 0 has slope 1 and hence makes an angle of 45° with
positive x-axis.
y = x2 represents an upward parabola with vertex at origin.
The graphs of the two functions y = x 2 and y = | x | are
symmetrical about the y-axis.
[... Both equations remain unchanged on
changing x to – x as | – x | = | x | ]
Let us first find the area between the parabola
y = x2 ...(i)
and the ray y = x for x ≥ 0 ...(ii)
To find limits of integration, let us solve (i) and (ii) for x.
Putting y = x2 from (i) in (ii), we have x2 = x
2
or x – x = 0 or x(x – 1) = 0∴ x = 0 or x = 1
For y = | x |
Table of values
y = x if x ≥ 0 y = – x if x ≤ 0
x 0 1 2 x 0 – 1 – 2
y 0 1 2 y 0 1 2
Y

B B

X′

Y′
MathonGo 9
 Class 12 Chapter 8 - Applications of Integrals

Area between parabola (i) and x-axis between limits

x = 0 and x = 1
1
1 1  x3  1
= ∫ 0 y dx = ∫ 0 x dx =  3  =
2
...(iii)
 0 3
Area between ray (ii) and x-axis,
1
1 1  x2  1
= ∫ 0 y dx = ∫ 0 x dx =  2  = ...(iv)
 0 2
∴ Required shaded area in first quadrant
= Area between ray y = x for x ≥ 0 and x-axis
– Area between parabola (i) and x-axis in first quadrant
= Area given by (iv) – Area given by (iii)
1 1 1
= – = sq. units
2 3 6
1
Similarly, shaded area in second quadrant = sq. units.
6
∴ Total area of shaded region in the above figure
1 1 1 1
= + = 2 × = sq. units.
6 6 6 3
10. Find the area bounded by the curve x 2 = 4y and the
line x = 4y – 2.
Sol. Step I. Graphs and region of Integration.
Equation of the given curve is x2 = 4y ...(i)
We know that eqn. (i) reprsents an upward parabola symmetrical
about y-axis
[... on changing x to – x in (i), eqn. (i) remains unchanged]

Y
x = 4y – 2
2
x = 4y D(2, 1)

1
– 1,
4 B 1
0,
C 2
A E
X′ X
(– 2, 0) M O N

Y′

Equation of the given line is

x = 4y – 2 ...(ii)
x+2
⇒ x + 2 = 4y ⇒ y =
4

MathonGo 10
Class 12 Chapter 8 - Applications of Integrals

Table of values for x = 4y – 2

x 0 – 2
1
y 0
2
We are to find the area of the shaded region shown in the
adjoining figure.
Step II. To find points of intersections of curve (i) and line
(ii), let us solve (i) and (ii) for x and y.
x2
Putting y = from (i) in (ii),
4
x2
x = 4 . – 2 ⇒ x = x2 – 2 ⇒ – x2 + x + 2 = 0
4
or x2 – x – 2 = 0
⇒ 2
x – 2x + x – 2 = 0 or x(x – 2) + (x – 2) = 0
or (x – 2)(x + 1) = 0
∴ Either x – 2 = 0 or x + 1 = 0
i.e., x = 2 or x = – 1
x2 4
For x = 2, from (i), y= = = 1 ∴ (2, 1)
4 4
x2 1  1
For x = – 1, from (i), y = = ∴  − 1,  .
4 4  4
∴ The two points of intersection of parabola (i) and line (ii) are
 1
C  − 1,  and D(2, 1).
 4
Step III. Area CMOEDN between parabola (i) and x-axis
2 2 x2  x2 
= ∫ −1 y dx = ∫ −1 4
dx ∵ From (i) y =


4 

(x ) 3 2
−1 1 3 1
= = (2 − (− 1)3 ) = (8 – (– 1))
12 12 12
1 9 3
= (8 + 1) = = sq. units ...(iii)
12 12 4
Step IV. Area of trapezium CMND between line (ii) and x-axis
2 2 x+2 1 2
= ∫ −1 y dx = ∫ −1 4
dx =
4 ∫ − 1 ( x + 2) dx
2
1  x2  1 4  1 
=  + 2 x  =  2 + 4 −  2 − 2
4  2 − 1 4    

1 1 1 1
= 2+4− +2 = 8−
4 2 4 2

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 Class 12 Chapter 8 - Applications of Integrals

1 16 − 1 1  15  15
= =   = sq. units. ...(iv)
4 2 4  2  8
∴ Required shaded area
= Area given by (iv) – Area given by (iii)
= Area of trapezium CMND – Area (CMOEDN)
15 3 15 − 6 9
= – = = sq. units.
8 4 8 8
11. Find the area of the region bounded by the curve y2 = 4x
and the line x = 3. Y
Sol. Equation of the (parabola) curve is
2
y = 4x ...(i)
1
A

∴ y = 4x = 2 x2 ...(ii) M
X′ X
for arc OA of parabola in first O
quadrant. (x = 0)
B 2
We know that equation (i) y =4x
represents a rightward parabola
with symmetry about x-axis. Y′ x=3
(... Changing y to – y in (i), keeps it unchanged)
∴ Required shaded area OAMB.
(See Note after solution of example 7)
= 2(Area OAM)
1
3 3
= 2 ∫0 y dx = 2 ∫0 2x 2 dx (By (ii))

= 4
(x )
3 3
2
0
= 4 .
2 3
[ 3 2 – 0] =
8
3 3 = 8 3 sq. units.
3 3 3
2
12. Choose the correct answer:
Area lying in the first quadrant and bounded by the
circle x2 + y2 = 4 and the lines x = 0 and x = 2 is
π π π
(A) π (B) (C) (D) .
2 3 4
Sol. Equation of the circle is
x2 + y2 = 4 = 22 ...(i)
Y
We know that equation (i)
represents a circle whose centre B x=0
is origin and radius is 2. x=2
∴ y2 = 22 – x2
O A
∴ 22 − x 2
y = ...(ii) X′
2
X
(2, 0)
for arc AB of the circle in first
quadrant.
∴ Required area lying in the
first quadrant bounded by the
Y′

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 Class 12 Chapter 8 - Applications of Integrals

circle x 2 + y 2 = 4 and the

(vertical) lines x = 0 and
(tangent line) x = 2.
2 2
= ∫0 y dx = ∫0 22 − x 2 dx By (ii))
2
x 22 x
= 
2
2 −x + 2
sin −1 
2 2 2 0

 x a2 x
∵

∫ a2 – x 2 dx =
2
a2 – x 2 +
2
sin –1 
a 
2
= 4 − 4 + 2 sin–1 1 – (0 + 2 sin–1 0)
2
π
– 0 – 0 = π sq. units.
= 0 + 2 .
2
[... sin 0 = 0 ⇒ sin–1 0 = 0]
∴ Option (A) is the correct answer.
13. Choose the correct answer:
Area of the region bounded by the curve y2 = 4x, y-axis and
the line y = 3 is Y
9
(A) 2 (B)
4 y=3
M A
9 9
(C) (D) .
3 2 X′ X
O
Sol. Equation of the curve (rightward
parabola) is
y2 = 4x ...(i)
∴ Required area of the region
bounded by parabola (i), y-axis and Y′
the (horizontal) line y = 3
= Area OAM
3
= ∫0 x dy ...(ii)
. .
[ . For arc OA of the parabola (i), at point O,
y = 0 and at point A, y = 3]
y2
Putting x = from (i) in (ii), required area
4
3 y2
= ∫0 4
dy

3
1  y3  1 27 9
=   = −0 = sq. units
4  3 0 4 3 4
∴ Option (B) is correct answer.

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 Class 12 Chapter 8 - Applications of Integrals

Exercise 8.2
1. Find the area of the circle 4x2 + 4y2 = 9 which is interior to
the parabola x2 = 4y.
Sol. Step I. Let us draw graphs and shade the region of
integration.
Given: Equation of the circle is 4x2 + 4y2 = 9
2
9 3
Dividing by 4, x2 + y2 =
=   ...(i)
4 2
We know that this equation (i) represents a circle whose centre is
3
(0, 0) and radius (x2 + y2 = r2)
2
Equation of parabola is x2 = 4y ...(ii)
Y 2
x = 4y
3
0,
2
D
– 2, 1
2
M B
A 1
2,
2
X′ X
N O H

2
2 2 3
x +y =
2

Y′

(eqn. (ii) represents an upward parabola symmetrical about

y-axis)
Step II. Let us solve eqns. of circle (i) and parabola (ii) for
x and y to find their points of intersection.
9
Putting x2 = 4y from (ii) in (i), we have 4y + y2 =
4
Multiplying by L.C.M. (=4),
16y + 4y2
=9 or 4y2 + 16y – 9 = 0
⇒ 2
4y + 18y – 2y – 9 = 0 ⇒ 2y(2y + 9) – 1(2y + 9) = 0
⇒ (2y + 9)(2y – 1)
=0
∴ Either 2y + 9
=0 or 2y – 1 = 0
⇒ 2y
=– 9 or 2y = 1
9 1
⇒ y =– or y =
2 2
9  9 
For y = – , from (i) x2 = 4y = 4  −  = – 18
2  2

MathonGo 14
Class 12 Chapter 8 - Applications of Integrals

which is impossible because square of a real number can never be

negative.
1 1
For y = , from (i), x2 = 4y = 4 × = 2
2 2
∴ x = ± 2
∴ Points of intersections of circle (i) and parabola (ii) are
 1  1
A  − 2,  and B  2,  .
 2   2
Step III. Area OBM = Area between parabola (ii) and y-axis
1
= ∫ 0
2 x dy

1
(... at O, y = 0 and at B, y = )
2
1
From (ii), putting x = 4 y = 2 y = 2 y2 ,
1

1 1
( )3
y2
2

Area OBM = ∫ 02 2 y 2 dy = 2 .
3
0

2
 3  3
2  1  2 − 0  4 1 1
= 2 .     = [... x 2 = x x ]
3  2   3 2 2

2 1 2 x
= = ...(iii) ∵ = x
3 2 3 x
Step IV. Now area BDM = Area between circle (i) and y-axis
3
1 3
= ∫ 12 x dy [... At point B, y = and at point D, y = ]
2 2
2

2 2
3 3 2
From (i), putting x2 =   – y2 i.e., x = 2 − y ,
2  
3
 3
2  2
3 2  2   
3 y 3  2  sin −1 y 
= 
2
∫
2 − y +
 2  − y dy
2
2  2 
= 1
  2  3 
2   2 
    1
2

 y a2 y
∵ ∫ a2 − y2 dy =
2
a 2 − y2 +
2
sin −1 
a 

3   1 
2 2
3 3 3 9 2 1 9 1 9  
−1 2
= 2 − 2 + sin   – 
–1 − + sin   
4     8  3  4 4 4 8  3  
2   2  
MathonGo 15
 Class 12 Chapter 8 - Applications of Integrals

3  9 1 8 9 1
=  × 0 + sin–1 1 –  4 + sin −1 
 4  8  4 8 3 

9 π 1 9 –1
1
=
8
×
2
–
4 2 – 8 sin 3

9π 2 9 1
= – – sin–1 ...(iv)
16 4 8 3
Step V. ∴ Required shaded area (of circle (i) which is interior to
parabola (ii)) = Area AOBDA
= 2(Area OBD) = 2[Area OBM + Area MBD]
 2  9π 2 9 −1 1 

= 2  3 +  16 − 4 − 8 sin 3  
   
(By (iii)) (By (iv))
  1 1  9π 9 1
= 2 2 − + − sin −1 
  3 4  16 8 3
4 −3 9π 9 1
= 2 2   + – sin–1
 12  8 4 3
 2 9π 9 −1 1  2 9 π −1 1 
=  6 + 8 − 4 sin 3  = +  − sin 3 
  6 4 2 

2 9 1
= + cos–1 sq. units.
6 4 3

 −1 −1 π
Ans ∵ sin x + cos x = 
 2 
2 9 1
Remark: = + sin–1 1− (... cos–1 x = sin–1 1 − x2 )
6 4 9

2 9 8  2 9 2 2
= + sin–1 =  6 + sin–1 3  sq. units.
6 4 9  4 
Note: The equation (x–α)2 + (y–β)2 = r2 represents a circle whose
centre is (α, β) and radius is r.
2. Find the area bounded by the curves (x – 1)2 + y2 = 1 and
x2 + y2 = 1.
Sol. The equations of the two circles are
x2 + y2 = 1 ...(i)
and (x – 1)2 + y2 = 1 ...(ii)
The first circle has centre at the origin and radius 1. The second
circle has centre at (1, 0) and radius 1. Both are symmetrical
about the x-axis. Circle (i) is symmetrical about y-axis also.
For points of intersections of circles (i) and (ii), let us solve
equations (i) and (ii) for x and y.

MathonGo 16
 Class 12 Chapter 8 - Applications of Integrals

Y
1, 3
2 2 2 2
(x – 1) + y = 1
2 2
x +y =1 A

X′ X
O D , 0 C (1, 0)
(0, 0)

B
1, 3
–
2 2
Y′
From (i), y2 = 1 – x2
Putting y = 1 – x in eqn. (ii), (x – 1)2 + 1 – x2 = 1
2 2

or x2 + 1 – 2x + 1 – x2 = 1
1
or – 2x + 1 = 0 ∴ x =
2
1 1 3
Putting x = , y2 = 1 – x2 = 1 – =
2 4 4
3 3
∴ y = ±
= ±
4 2
∴ The two points of intersections of circles (i) and (ii) are
1 3  1 3
 ,  and  , − 
2 2  2 2 
From (i), y2 = 1 – x2;
∴ 1 − x 2 in first quadrant.
y =
From (ii), y2 = 1 – (x – 1)2 and
∴ 1 − ( x − 1) 2 in first quadrant.
y =
Required area OACBO (area enclosed between the two circles)
(shown shaded)
= 2 × Area OAC
= 2 [Area OAD + Area DAC]
 1/ 2 1 
= 2  ∫ 0 y of circle (ii) dx + ∫ 1/ 2 y of circle (i) dx 
 
 1/ 2 1
1 − x 2 dx 
= 2 ∫ 0 1 − ( x − 1) 2 dx + ∫ 1/ 2 

 
1/ 2

1 
 
2 2
  ( x − 1) 1 − ( x − 1) + 1 sin −1 ( x − 1)  +  x 1 − x + 1 sin −1 x 
= 2   
2 2 2 2
  0  1/ 2 

 x a2 x
∵ ∫ a2 – x 2 dx =
2
a2 – x 2 +
2
sin –1 
a 

MathonGo 17
 Class 12 Chapter 8 - Applications of Integrals

 1 3 −1  1   −1 −1  1 3 −1 1 

= − 2 4 + sin  − 2   − {sin (− 1)} + sin 1 −  2 4 + sin 2 
     
3 π π π  2π
3 3
π
= – – + + – =  3 − 2  sq. units.
–
4 6 2 2  4 6 
3. Find the area of the region bounded by the curves
y = x2 + 2, y = x, x = 0 and x = 3.
Sol. Equation of the given curve is y = x2 + 2 ...(i)
2
or x = y – 2
It is an upward parabola ( . . . An equation of the form x2 = ky,
k > 0 represents an upward parabola).
Eqn. (i) contains only even powers of x and hence remains
unchanged on changing x to – x in (i).
∴ The parabola (i) is symmetrical about y-axis.
Parabola (i) meets y-axis (its line of symmetry) i.e. x = 0 in (0, 2)
[put x = 0 in (i) to get y = 2]
∴ Vertex of the parabola is (0, 2).
Equation of the given line is y = x ...(ii)
We know that it is a straight line passing through the origin and
having slope 1 i.e., making an angle of 45° with x-axis.
Table of values for the line y = x
x 0 1 2
y 0 1 2

Also the required area is given to be bounded by the vertical

lines x = 0 to x = 3.
∴ Limits of integration
are given to be x = 0 to
x = 3.
Area bounded by parabola
(i) namely y = x2 + 2, the x- B
axis and the ordinates x = A
0 to x = 3 is the area
OACD and
3 3
= ∫0 y dx = ∫0 ( x2 + 2) dx

3
 x3 
=  3 + 2 x  = (9 + 6) – 0 = 15 ...(iii)
 0
Area bounded by line (ii) namely y = x, the x-axis and the
ordinates x = 0, x = 3 is

MathonGo 18
 Class 12 Chapter 8 - Applications of Integrals

3
3 3  x2 
area OAB and = ∫0 y dx = ∫0 x dx =  
 2 0
9 9
=
– 0 = ...(iv)
2 2
∴ Required area (shown shaded) i.e., area OBCD
= area OACD – area OAB
= Area given by (iii) – Area given by (iv)
9 21
= 15 – = sq. units.
2 2
Remark: On solving Eqns (i) and (ii) for x we get imaginary
values of x and hence curves (i) and (ii) don’t intersect.
4. Using integration, find the area of the region bounded by
the triangle whose vertices are (– 1, 0), (1, 3) and (3, 2).
Sol. Given: Vertices of triangle Y
are A(– 1, 0), B(1, 3) and
C(3, 2). B(1, 3)
∴ Equation of line AB is
C(3, 2)
3−0
y – 0 = (x – (– 1))
1 − (− 1)
X′ X
 y2 – y1  A O L M
 y – y1 = ( x – x1 )  (–1, 0)
 x 2 – x1 
3
or y =
(x + 1)
2 Y′
∴ Area of ∆ABL = Area bounded by this line AB and x-axis
1
= ∫ −1 y dx

(... At point A, x = – 1 and at point B, x = 1)

1 3 3 1
= ∫ −1 2
( x + 1) dx =
2 ∫ − 1 ( x + 1) dx
1
3  x2  3  1  1 
=  + x  =  2 + 1  −  2 − 1  
2  2 − 1 2    

3  3  1  3 3 + 1 3 4
− −   =
2  2  2  
= = . = 3 ...(i)
2 2 2 2 2
Again equation of line BC is
2−3
y – 3 = (x – 1)
3−1
1  x −1 6− x +1
⇒ y–3=– (x – 1) ⇒ y = 3 –   =
2  2  2
7−x 1
⇒ y = = (7 – x)
2 2
∴ Area of trapezium BLMC = Area bounded by line BC and x-axis

MathonGo 19
 Class 12 Chapter 8 - Applications of Integrals

3 3 1
= ∫1 y dx = ∫1 2
(7 − x) dx

3
1  x2  1  9  1 
=  7 x −  = 21 − 2 −  7 − 2  
2  2 1 2   
1  9 1 1  42 − 9 − 14 + 1  1
= 21 − 2 − 7 + 2  =  2  = (20)
2   2   4
= 5 ...(ii)
Again equation of line AC is
2−0 2
y – 0 = (x – (– 1)) ⇒ y = (x + 1)
3 − (− 1) 4
1
⇒ y = (x + 1)
2
∴ Area of ∆ACM = Area bounded by line AC and x-axis
3
3 3 1 1  x2 
+ x 
= ∫−1 y dx = ∫ −1 2
( x + 1) dx =
2

 2 − 1

1 9 1  1 9 1 
 + 3 −  2 − 1 = + 3 − + 1
2  2
=
2 2   2 
1 9 + 6 − 1 + 2 16
2   = 4 = 4
= ...(iii)
2 
We can observe from the figure that required area of ∆ABC
= Area of ∆ABL + Area of Trapezium BLMC – Area of ∆ACM
= 3 + 5 – 4 = 4 sq. units.
By (i) By (ii) By (iii)
5. Using integration, find the area of the triangular region
whose sides have the equations y = 2x + 1, y = 3x + 1 and
x = 4.
Sol. Equation of one side of triangle is y = 2x + 1 ...(i)
Equation of second side of triangle is y = 3x + 1 ...(ii)
Third side of triangle is x = 4. ...(iii)
It is a line parallel to y-axis at a distance 4 to right of y-axis.
Let us solve (i) and (ii) for x and y.
Eqn. (ii) – eqn. (i)
gives x = 0.
Put x = 0 in (i), y = 1.
∴ Point of intersection of lines (i) and (ii) is A(0, 1)
Putting x = 4 from (iii) in (i), y = 8 + 1 = 9
∴ Point of intersection of lines (i) and (iii) is B(4, 9).
Putting x = 4 from (iii) in (ii), y = 12 + 1 = 13.

MathonGo 20
 Class 12 Chapter 8 - Applications of Integrals

∴ Point of intersection of lines Y

(ii) and (iii) is C(4, 13).
Area between line (ii) i.e., line
AC and x-axis C(4, 13)
4 4
= ∫0 y dx = ∫0 (3 x + 1) dx
[By (ii)]
4
 3 x2 
=  2 + x 
 0

3
2x +
= 24 + 4 = 28 sq. units ...(iv) B(4, 9)
Area between line (i) i.e., line

y=
AB and x-axis

1+
2x
4 4
= ∫0 y dx = ∫0 (2 x + 1) dx

y=
(x )
4
2
= +x [By (i)]
0
= 16 + 4 = 20 sq. units ...(v)
A(0, 1)
∴ Area of triangle ABC =
Area given by (iv) X′ X
O
– Area given by (v)
= 28 – 20 = 8 sq. units.
6. Choose the correct Y′
answer:
Smaller area enclosed by the circle x 2 + y 2 = 4 and the
line x + y = 2 is
π – 2)
(A) 2(π (B) π – 2 (C) 2π π – 1 π + 2).
(D) 2(π
Sol. Step I. Equation of circle is x2 + y2
= 4 = 22 ...(i)
Y
∴ y2 = 22 – x2
B(0, 2)
∴ y= 22 − x 2 ...(ii) C
2 A(2, 0)
for arc AB of the circle in first
quadrant. X′ X
O 2
We know that eqn. (i) represents a
x+y=2
circle whose centre is origin and
radius is 2. 2
x +y =4
2

Equation of the line is x + y = 2 ...(iii) Y′

Table of values
x 0 2
y 2 0
∴ Graph of equation (iii) is the straight line joining the points
(0, 2) and (2, 0).
The region for required area is shown as shaded in the figure.

MathonGo 21
 Class 12 Chapter 8 - Applications of Integrals

Step II. From the graphs of circle (i) and straight line (iii), it is
clear that points of intersections of circle (i) and straight line
(iii) are A(2, 0) and B(0, 2).
Step III. Area OACB, bounded by circle (i) and coordinate axes in
first quadrant
2 2
= ∫ y dx = ∫ 22 − x2 dx (... From (ii), y = 22 − x 2 )
0 0
2
x 22 x
=  2 22 − x 2 + sin −1 
 2 2 0
 x a2 x
∵

∫ a2 − x 2 dx =
2
a2 − x 2 +
2
sin −1 
a 
2 
=  4 − 4 + 2 sin −1 1  – (0 + 2 sin–1 0)
2 
 π
= 0 + 2   – 2(0) = π ...(iv)
2
Step IV. Area of triangle OAB, bounded by straight line (iii) and
co-ordinate axes
2 2
= ∫0 y dx = ∫0 (2 − x) dx (... From (iii), y = 2 – x)
2
 x2 
=  2 x − 2  = (4 – 2) – (0 – 0) = 2 ...(v)
 0
Step V. ∴ Required shaded area
= Area OACB given by (iv) – Area of triangle OAB by (v)
= (π – 2) sq. units.
∴ Option (B) is the correct answer.
7. Choose the correct answer:
Area lying between the curves y2 = 4x and y = 2x is
2 1 1 3
(A) (B) (C) (D) .
3 3 4 4
Sol. Step I. Equation of one curve
(parabola) is
y2 = 4x ...(i)
1

∴ y = 4x = 2 x = 2 x 2 ...(ii)
for arc of the parabola in first
quadrant.
We know that eqn. (i) represents a
rightward parabola symmetrical
about x-axis.
Equation of second curve (line) is y = 2x ...(iii)
We know that y = 2x represents a straight line passing through
the origin.
We are required to find the area of the shaded region.

MathonGo 22
Class 12 Chapter 8 - Applications of Integrals

II. Let us solve (i) and (iii) for x and y.

Putting y = 2x from (iii) in (i), we have
4x2 = 4x ⇒ 4x2 – 4x = 0 ⇒ 4x(x – 1) = 0
∴ Either 4x = 0 or x – 1 = 0
0
i.e., x = = 0 or x = 1
4
When x = 0, from (ii), y = 0 ∴ point is O(0, 0)
When x = 1, from (ii), y = 2x = 2 ∴ point is A(1, 2)
∴ Points of intersections of circle (i) and line (ii) are O(0, 0) and
A(1, 2).
III. Area OBAM = Area bounded by parabola (i) and x-axis
1 1
1 1
= ∫0 y dx = ∫0 2x 2 dx [... From (ii) y = 2 x 2 ]

( ) 4
3 1
x2 0 4
= 2 = (1 – 0) = ...(iv)
3 3 3
2
IV. Area of ∆OAM = Area of bounded by line (iii) and x-axis
1 1
= ∫0 y dx = ∫0 2x dx (... From (iii) y = 2x)
1
 x2 
2 1
( )
= 2  2  = x 0 = 1 – 0 = 1
 0
...(v)

V. ∴ Required shaded area OBA

= Area OBAM – Area of ∆ OAM
4 4−3 1
= – 1 = = sq. units.
3 3 3
(By (iv)) (By (v))
∴ Option (B) is the correct answer.

MathonGo 23
 Class 12 Chapter 8 - Applications of Integrals

MISCELLANEOUS EXERCISE
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.
Sol. (i) Equation of the curve
(parabola) is
y = x2 i.e., x2 = y ...(i)
It is an upward parabola
symmetrical about y-axis.
[... Changing x to – x in (i)
keeps it unchanged]
Required area bounded by
curve (i) y = x2, vertical lines

MathonGo 24
 Class 12 Chapter 8 - Applications of Integrals

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

x 3 2
8 1 7
=   = − = sq. units.
 3 1 3 3 3
(ii) Equation of the curve is y = x4 ...(i)
⇒ y ≥ 0 for all real x (... Power of x is Even (4))
Curve (i) is symmetrical about y-axis.
[... On changing x to – x in (i), eqn. (i) remains unchanged]
Clearly, curve (i) passes through the origin because for
x = 0, from (i) y = 0.
Table of values for curve y = x4 for x = 1 to x = 5 (given)
x 1 2 3 4 5
y 1 24 = 16 34 = 81 44 = 256 54 = 625
Y

(5, 625)

(1, 1)
X′ X
O
x=1 x=5

Y′
Required shaded area between the curve y = x4, vertical
lines x = 1, x = 5 and x-axis
5 5
= ∫1 y dx = ∫1 x 4 dx (By (i))

5
 x5  55 15 3125 − 1 3124
=   = – = =
 5 1 5 5 5 5

3124 × 2
= = 624.8 sq. units.
10
2. Find the area between the curves y = x and y = x2.
Sol. Step I. To draw the graphs and region of integration.
Equation of one curve (straight
Y
line) is
y = x ...(i)
We know that graph of eqn. (i) A(1, 1)
is a straight line passing
through origin. B
Equation of second curve X′ O M
X
(parabola) is
y = x2 or x2 = y ...(ii)
We know that equation (ii) represents an Y′

MathonGo 25
 Class 12 Chapter 8 - Applications of Integrals

upward parabola with symmetry about y-axis.

Step II. Let us find points of intersections of curves (i) and (ii) by
solving them for x and y.
Putting y = x from (i) in (ii), we have
x = x2 or x – x2 = 0 or x(1 – x) = 0
∴ Either x = 0 or 1 – x = 0 i.e., x = 1.
When x = 0, from (i) y = 0 ∴ Point is O(0, 0)
When x = 1, from (i), y = 1 ∴ Point is A(1, 1)
∴ Points of intersections of line (i) and parabola (ii) are O(0, 0)
and A(1, 1).
Step III. Area of triangle OAM
= Area bounded by line (i) and x-axis
1 1
= ∫0 y dx = ∫0 x dx (... From (i) y = x)
1
 x2  1 1
=  2  = – 0 = ...(iii)
 0 2 2
Step IV. Area OBAM = Area bounded by parabola (ii) and x-axis
1 1
= ∫0 y dx = ∫0 x 2 dx (... From (ii) y = x2)
1
 x3  1 1
=  3  = – 0 = ...(iv)
 0 3 3
∴ Required shaded area OBA between line (i) and parabola (ii)
= Area of ∆ OAM – Area OBAM
1 1 3−2 1
= – = = sq. units.
2 3 6 6
(By (iii)) (By (iv))
3. Find the area of the region lying in first quadrant and
bounded by y = 4x 2, x = 0, y = 1 Y
2
and y = 4. y = 4x
x=0
Sol. Equation of the (parabola) curve is y=4
y = 4x2
y
⇒ x2 = ...(i) y=1
4 X′ X
O (0, 0)
y y
∴ x = = ...(ii)
4 2
For branch of parabaola in first quadrant. Y′
We know that this equation represents an upward parabola with
symmetry about y-axis.
∴ Required shaded area of the region lying in first quadrant
bounded by parabola (i), x = 0 (⇒ y-axis) and the horizontal
lines y = 1 and y = 4 is

MathonGo 26
 Class 12 Chapter 8 - Applications of Integrals

4 4 y
∫1 x dy = ∫1 2
dy (By (ii))

1 4
1
1 ( )
3 4
y2 1 1 2 3 3
=
2 ∫1 y 2 dy =
2 3
=
2
.
3 [4 2 − 12 ]
2
1 1 7
(4 4 – 1) =
= (8 – 1) = sq. units.
3 3 3
4. Sketch the graph of y = | x + 3 | and evaluate
0
∫ – 6 | x + 3|dx .
Sol. Equation of the given curve is
y = | x + 3 | ...(i)
We know that from (i),
y = | x + 3 | ≥ 0 for all real x.
∴ Graph of curve is only above the x-axis i.e., in first and
second quadrants only.
From (i), y = | x + 3 | = x + 3 if x + 3 ≥ 0 i.e., if x ≥ – 3 ...(ii)
and y = | x + 3 | = – (x + 3) if x + 3 ≤ 0 i.e., if x ≤ – 3 ...(iii)
Table of values for y = x + 3 for x ≥ – 3
x – 3 – 2 – 1 0
y 0 1 2 3
Table of values for y = – (x + 3) for x ≤ – 3
x – 3 – 4 – 5 – 6
y 0 1 2 3
∴ Graph of curve (i) is as shown in the following figure.

∴ Graph of y = | x + 3 | is L-shaped consisting of two rays

above the x-axis at right angles to each other.
0 −3 0
Now, ∫ – 6 | x + 3| dx = ∫ − 6 | x + 3| dx + ∫ − 3 | x + 3| dx
 ..
 . On putting expression x + 3 within modulus equal to zero,

b c b 
we get x = – 3 and ∫ a f ( x) dx = ∫ a f ( x) dx + ∫ c f ( x) dx 

MathonGo 27
 Class 12 Chapter 8 - Applications of Integrals

−3 0
= ∫−6 − ( x + 3) dx + ∫ − 3 ( x + 3) dx
(By (iii) because on (By (ii) because on
(– 6, – 3), x < – 3 ⇒ x + 3 < 0) (– 3, 0), x > – 3 ⇒ x + 3 > 0)
−3 0
 x2   x2 
= –  2 + 3 x  +  2 + 3 x 
 − 6  − 3

9   9 
= –  − 9 − (18 − 18)  + 0 −  − 9  
2   2 
9 9 18
= – + 9 + 0 + 0 – + 9 = 18 – = 18 – 9 = 9 sq. units.
2 2 2
5. Find the area bounded by the curve y = sin x between x = 0
and x = 2ππ.
Sol. Equation of the curve is y = sin x ...(i)
Let us draw the graph of y = sin x from x = 0 to x = 2π
Now we know that y = sin x ≥ 0 for 0 ≤ x ≤ π i.e., in first
and second quadrants
and y = sin x ≤ 0 for π ≤ x ≤ 2π i.e., in third and fourth
quadrants.
Y
15
x= π
A 2 5

B (x = π) D (x = 2π)
X′ X
O x= 0

C x = 3π
2
Y′
dy
To find points where tangent is parallel to x-axis, put = 0.
dx
π 3π
⇒ cos x = 0 ⇒ x = , x =
2 2
Table of values for curve y = sin x
between x = 0 and x = 2π π
π 3π
x 0 π 2π
2 2
y 0 1 0 – 1 0
[... sin nπ = 0 for every integer n
3π
and sin = sin 270° = sin (180° + 90°) = – sin 90° = –1]
2

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 Class 12 Chapter 8 - Applications of Integrals

Required shaded area = Area OAB + Area BCD

π 2π
= ∫0 y dx + ∫π y dx

[Here we will have to find area OAB and Area BCD separately
because y = sin x ≥ 0 for 0 ≤ x ≤ π
and y = sin x ≤ 0 for π ≤ x ≤ 2π]
Putting y = sin x from (i),
π 2π
= ∫0 sin x dx + ∫π sin x dx

π 2π
= − (cos x )0 + − (cos x )π
= | – (cos π – cos 0) | + | – (cos 2π – cos π) |
= | – (– 1 – 1) | + | – (1 + 1) |
[... cos nπ = (– 1)n for every integer n
putting n = 1, 2; cos π = – 1, cos 2π = 1]
= 2 + 2 = 4 sq. units.
6. Find the area enclosed by the parabola y2 = 4ax and the
line y = mx.
Sol. Step I. To draw the graphs and shade the region of
integration.
Equation of the parabola is
y2 = 4ax ...(i)
(It is rightward parabola with
symmetry about x-axis)
From (i),
1
y = 4ax = 2 a x 2 ...(ii)
for arc ODA of parabola in first
quadrant.
Equation of the line is y = mx ...(iii)
We know that eqn. (iii) represents a straight line passing through
the origin.
Step II. To find points of intersections of curves (i) and (iii), let
us solve (i) and (iii) for x and y
Putting y = mx from (iii) in (i),
m2x2 = 4ax ⇒ m2x2 – 4ax = 0
⇒ 2
x(m x – 4a) = 0
⇒ Either x = 0 or m2x – 4a = 0 i.e., m2x = 4a
4a
⇒ x = 0 or x =
m2
When x = 0, from (iii) y = 0 ∴ Point is O(0, 0)
4a 4a 4a
When x = , y = m . =
m2 m2 m
∴ Second point of intersection of parabola (i) and line (iii) is

MathonGo 29
 Class 12 Chapter 8 - Applications of Integrals

 4a 4a 
A 2 , .
m m 
Step III. Area ODAM = Area parabola (i) and x-axis
4a
y dx ∵ At O, x = 0 and at A, x =
4a 
= ∫ 0
m2


m2 
1

Putting y = 2 a x 2 from (ii),

4a

4a 1 ( )
3
x2
m2
0
= ∫ 0m2 2 a x2 dx = 2 a 3
2
3
4 a  4a  2 a 4a 4a . . 3
=  2 = 4 3 [ . x2 = x x ]
3 m  m2 m2
4 a 4a a 32 a2
= 2 2 = ...(iv)
3 m m 3 m3
Step IV. Area of ∆OAM = Area between line (iii) and x-axis
4a
= ∫ 0
m2 y dx

4a
Putting y = mx from (iii), ∫ 0m2 mx dx
4a
 x 2  m2  2 
m  4 a  − 0 

= m    =  2 
 2 0 2  m  

m 16a2 8a2
.= 4 = ...(v)
2 m m3
Step V. Required shaded area
= Area ODAM given by (iv)
– Area of ∆OAM given by (v)
32 a2 8a2 a2  32 
=  − 8
3 – 3 =
3m m m3  3 
a2  32 − 24  a2 8 8 a2
= 3  3  = 3 . = .
m   m 3 3 m3
2
7. Find the area enclosed by the parabola 4y = 3x and the line
2y = 3x + 12.
Sol. Equation of the parabola is 4y = 3x2 ...(i)
4
or x2 = y
3
It is an upward parabola with vertex at the origin and is
symmetrical about y-axis.
Equation of the line is

MathonGo 30
 Class 12 Chapter 8 - Applications of Integrals

2y = 3x + 12 ...(ii)
Putting y = 0 in (ii), x = – 4
∴ (– 4, 0) is a point on
line (ii)
Putting x = 0 in (ii), y = 6
∴ (0, 6) is also a point on
line (ii).
Joining the points (– 4, 0)
and (0, 6) we get the graph
of line (ii).
To find the points of
intersections, let us solve
eqns. (i) and (ii), for x
and y.
3 x + 12
Putting y = from (ii) in (i), we have
2
2(3x + 12) = 3x2 or 3x2 – 6x – 24 = 0 or x2 – 2x – 8 = 0
or (x – 4) (x + 2) = 0 ∴ x = 4, – 2.
3 x + 12 3 x + 12
When x = 4, y = = 12 ; When x = – 2, y = = 3.
2 2
∴ The points of intersection are B(4, 12) and C(– 2, 3).
3
Area bounded by the line (ii) namely 2y = 3x + 12 or y = x + 6,
2
the x-axis and the ordinates x = – 2, x = 4 is ABCD
4
4 43  3 2 
= ∫−2 ∫ − 2  2 x + 6  dx =  4 x + 6 x  − 2
y dx =
= (12 + 24) – (3 – 12) = 45 ...(iii)
3 2
Area bounded by the curve (i) namely y = x , the x-axis and the
4
ordinates x = – 2, x = 4 is (area CDO + area OAB)
4
4 4 3 2  3 x3 
= ∫−2 y dx = ∫−2 x dx =  . 
4  4 3  − 2
1
= [64 – (– 8)] = 18. ...(iv)
4
∴ Required area (shown shaded)
= (Area under the line – Area under the curve) between the
lines x = – 2 and x = 4
= Area given by (iii) – Area given by (iv)
= 45 – 18 = 27 sq. units.
8. Find the area of the smaller region bounded by the ellipse
x2 y2 x y
+ = 1 and the line + = 1.
9 4 3 2

MathonGo 31
 Class 12 Chapter 8 - Applications of Integrals

Sol. Step I. Equation of the Y

ellipse is
B (0, 2)
x2 y2
+ = 1 ..(i) D
9 4 2
A′ O A(3, 0)
Clearly, ellipse (i) is X′ X
(– 3, 0) 3 3
symmetrical about both 2
axes. B′(0, – 2)
Intersections of ellipse
(i) with x-axis.
Put y = 0 in (i), Y′
2
x
= 1 ⇒ x2 = 9 ∴ x = ± 3
9
∴ Intersections of ellipse (i) with x-axis are A(3, 0) and
A′(– 3, 0).
Similarly, intersections of ellipse (i) with y-axis (putting x = 0 in
(i)) are B(0, 2) and B′(0, – 2).
x y
Equation of the line is + = 1
3 2
y x 3 − x
⇒ = 1 – ⇒ y = 2  ...(ii)
2 3  3 
Table of values
x 0 3
y 2 0
∴ Graph of line (ii) is the line joining the points (0, 2) and (3, 0).
We have shaded the smaller region whose area is required.
Step II. From the graphs, it is clear that points of intersections
of ellipse (i) and straight line (ii) are A(3, 0) and B(0, 2).
Step III. Area OADB = Area between ellipse (i) (arc AB of it)
and x-axis
3  y2 x2 9 − x2
= ∫0 y dx
 From (i), 4 = 1 – 9 = 9

4 2
⇒ y2 = (9 – x2) ⇒ y = 9 − x2 
9 3 
(At point B, x = 0 and at point A, x = 3)
3 2 2 3
= ∫0 3
9 − x 2 dx =
3 ∫0 32 − x 2 dx

3
2 x 32 − x 2 +
32 x
sin −1 
= 
3  2 2 3  0

 x a2 x
∵ ∫ a2 − x 2 dx =
2
a2 − x 2 +
2
sin −1 
a 

2 3 9  9 
=  9 − 9 + sin −1 1 −  0 + sin −1 0  
3 2 2  2 
MathonGo 32
 Class 12 Chapter 8 - Applications of Integrals

2  9 π  2 9π 3π
0 + . − 0 =
3 
= . = ...(iii)
2 2  3 4 2
Step IV. Area of triangle OAB = Area bounded by line AB and
x-axis
3 3 2
= ∫0 y dx = ∫0 3
(3 − x) dx [From (ii)]
3
2  x2  2  9  2 9
=  3 x −  = 9 −  − 0 = . =3 ...(iv)
3  2 0 3  2  3 2
Step V. ∴ Required shaded area
= Area OADB – Area OAB
3π
= – 3
2
(By (iii) (By (iv)
π  3
= 3  − 1 = (π – 2) sq. units.
2  2
9. Find the area of the smaller region bounded by the ellipse
x 2 y2 x y
2
+ 2 = 1 and the line + = 1.
a b a b
x 2 y2
Sol. Step I. Equation of the ellipse is + = 1 ...(i)
a2 b2
Ellipse (i) is symmetrical
about both the axes.
Intersections of ellipse
( i) w i t h x - a x i s ( y = 0 )
are A(a, 0) and A′(– a, 0). A'
Intersections of ellipse (i)
with y-axis (x = 0) are
B(0, b) and B′(0, – b)
Again equation of chord
AB is
x y
+ = 1 ...(ii)
a b
Table of Values
x 0 a
y b 0
y2 x2 a2 − x 2
Step II. From equation (i), 2 = 1 – 2 =
b a a2
b2 ( a2 − x 2 ) b
∴ y2 = 2 ∴ y =
a a2 − x 2 (in first quadrant)
a
Area between arc AB of the ellipse and x-axis (in first quadrant)

MathonGo 33
 Class 12 Chapter 8 - Applications of Integrals

a a b b a
= ∫0 y dx = ∫0 a
a2 − x 2 dx =
a ∫0 a2 − x2 dx
a
b x a2 x  a2 
=  a2 − x2 + sin −1  = b 0 + sin −1 1 − (0 + 0) 
a  2 2 a  0 a  2 
b a2 π πab
= . . = ...(iii)
a 2 2 4
y x a−x
Step III. From equation (ii), = 1 – =
b a a
b
∴ y = (a – x)
a
∴ Area between chord AB and x-axis
a a b b a
= ∫ 0 y dx = ∫ 0 ( a − x) dx = ∫0 (a − x) dx
a a
a
b  ax − x 
2
b  2 a 
2
  =  a − 
=
a  2  0 a  2 

b a2 1
= . = ab ...(iv)
a 2 2
Step IV. ∴ Area of smaller region bounded by ellipse (i) and
straight line (ii)
= Area between arc AB and chord AB
= Area given by (iii) – Area given by (iv)
πab ab ab
= – = (π – 2).
4 2 4
10. Find the area of the region enclosed by the parabola x2 = y,
the line y = x + 2 and x-axis.
Sol. Step I. Equation of the parabola is x2 = y ...(i)
We know that equation
(i) represents an upward
parabola with symmetry
about y-axis.
(... Changing x to – x in
(i), keeps it unchanged)
Equation of the line is
y = x + 2 ...(ii)
Table of values
x 0 – 2
y 2 0
∴ Graph of line (ii) is the
line joining the points (0, 2) and (– 2, 0).
Step II. Let us solve (i) and (ii) for x and y
Putting y = x + 2 from (ii) in (i),
MathonGo 34
 Class 12 Chapter 8 - Applications of Integrals

x2 = x + 2
2
or x – x – 2 = 0 or x2 – 2x + x – 2 = 0
or x(x – 2) + 1(x – 2) = 0
⇒ (x – 2)(x + 1) = 0
∴ Either x – 2= 0 or x + 1 = 0
i.e., x = 2 or x = – 1.
When x = 2, from (i), y = x2 = 22 = 4 ∴ Point is (2, 4)
When x = – 1, from (i), y = (– 1)2 = 1 ∴ Point is (– 1, 1).
∴ The two points of intersections of parabola (i) and line (ii)
are A(– 1, 1) and B(2, 4).
Step III. Area ALODBM = Area bounded by parabola (i) and x-
axis
2 2
= ∫ −1 y dx = ∫ −1 x 2 dx [... From (i) y = x2]
2
 x3  8  −1 8 1 9
=  3  = –   = + = = 3 ...(iii)
 − 1 3  3  3 3 3
Step IV. Area of trapezium ALMB = Area bounded by line (ii)
and x-axis
2
= ∫ − 1 ( x + 2) dx [... From (ii) y = x + 2]
2
 x2  1  1
=  2 + 2 x  = 2 + 4 –  − 2 = 6 – + 2
 − 1  2  2
1 15
= 8 – = ...(iv)
2 2
Step V. ∴ Required shaded area = Area of trapezium ALMB
– Area ALODBM
15 9
= – 3 = sq. units.
2 2
11. Using the method of integration, find the area bounded by
the curve | x | + | y | = 1.
Sol. Given: Equation of the curve Y
(graph) is
| x | + | y | = 1 ...(i) B(0, 1)
1 x
Curve ( i ) i s s y m m e t r i c a l = +
y y
about x-axis. + =
. . x 1
[ . On changing y to – y in – A(1, 0)
O
eqn. (i), it remains unchanged X′ X
C(– 1, 0)
as we know that | – y | = | y |] x
+ =
1
Similarly, curve (i) is y y
= –
– x
symmetrical about y-axis. 1
D(0, – 1)
We know that, for first
quadrant; x ≥ 0 and y ≥ 0 Y′
⇒ | x | = x and | y | = y
∴ Eqn. (i) becomes x + y = 1 ...(ii)
which is the equation of a straight line.
MathonGo 35
 Class 12 Chapter 8 - Applications of Integrals

Table of values
x 0 1
y 1 0
∴ Graph of x + y = 1 is the straight line joining the points (0, 1)
and (1, 0).
We know that for second quadrant, x ≤ 0 and y ≥ 0
⇒ | x | = – x and | y | = y
∴ Equation (i) becomes – x + y = 1
which represents a straight line.
Table of values
x 0 – 1
y 1 0
∴ Graph of – x + y = 1 is the straight line joining the points
(0, 1) and (– 1, 0).
We know that for third quadrant, x ≤ 0 and y ≤ 0.
⇒ | x | = – x and | y | = – y
∴ Eqn. (i) becomes – x – y = 1 or x + y = – 1
which represents a straight line.
Table of values
x 0 – 1
y – 1 0
∴ Graph of x + y = – 1 is the straight line joining the points
(0, – 1) and (– 1, 0).
We know that for fourth quadrant x ≥ 0 and y ≤ 0.
⇒ | x | = x and | y | = – y
⇒ Equation (i) becomes x – y = 1 which again represents a
straight line.
Table of values
x 0 1
y – 1 0
∴ Graph of x – y = 1 is the straight line joining the points (0, – 1)
and (1, 0).
∴ Graph of Eqn. (i) is the square ABCD.
∴ Area bounded by curve (i)
= Area of square ABCD
= 4 × ∆OAB
= 4 × Area bounded by line (ii) namely x + y = 1
and the coordinate axes
1 1
= 4 ∫0 y dx = 4 ∫0 (1 − x) dx
[... x + y = 1 ⇒ y = 1 – x]
1
 x2   1  1
= 4  x − 2  = 4  1 −  − 0  = 4 × = 2 sq. units.
 0  2  2
12. Find the area bounded by the curves
{(x, y) : y ≥ x2 and y = | x | }
Sol. It is same as Q. No. 9, Exercise 8.1, page 558.

MathonGo 36
 Class 12 Chapter 8 - Applications of Integrals

13. Using the method of integration find the area of the triangle
whose vertices are A(2, 0), B(4, 5) and C(6, 3).
Sol. Vertices of the given triangle are Y
A(2, 0), B(4, 5) and C(6, 3).
Now, equation of side AB is B(4, 5)
5−0
y – 0 = (x – 2)
4−2
C(6, 3)
( y – y1 )
y – y1 = 2 ( x – x1 )
x2 – x1
5 X′ X
⇒ y= (x – 2) O A(2, 0) L M
2
∴ Area of ∆ALB bounded by
line AB and x-axis Y′
4
4 4 5 5  x2 
= ∫2 y dx = ∫2 2
( x − 2) dx =
2

 2
− 2 x 
2
5 5
= [(8 – 8) – (2 – 4)] = (0 + 2)
2 2
5
= × 2 = 5 sq. units. ...(i)
2
Again equation of side BC is
3−5 ( y – y1 )
y – 5 = y – y1 = 2 ( x – x1 )
6 − 4 (x – 4) x2 – x1
⇒ y – 5 = – (x – 4)
⇒ y = 5 – x + 4 = 9 – x
∴ Area of trapezium BLMC bounded by line BC and x-axis
6
6 6  x2 
= ∫4 = ∫4
y dx =  9 x − 2 
(9 − x) dx
 4
= | 54 – 18 – (36 – 8) | = | 36 – 36 + 8 |
= 8 ...(ii)
Again equation of line AC is
3−0 3
y – 0 = (x – 2) ⇒ y = (x – 2)
6−2 4
∴ Area of ∆AMC bounded by line AC and x-axis
6 6 3
= ∫2 y dx = ∫2 4
( x − 2) dx

6
 x2
3  3

= − 2 x  = [18 – 12 – (2 – 4)]
42 2 4
3 3
= (6 + 2) = × 8 = 6 sq. units. ...(iii)
4 4
We observe from the above figure that

MathonGo 37
 Class 12 Chapter 8 - Applications of Integrals

Area of ∆ABC = Area of ABL + Area of trapezium BLMC

– Area of AMC
= 5 + 8 – 6
(by (i)) (by (ii)) (by (iii))
= 7 sq. units.
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
Sol. Equation of one line is 2x + y = 4 ...(i)
Equation of second line is 3x – 2y = 6 ...(ii)
Equation of third line is x – 3y + 5 = 0 ...(iii)
Let ABC be triangle (region) bounded by the given lines (i), (ii),
(iii). Let us find point of intersection A of lines (i) and (ii) i.e.
solve (i) and (ii) for x and y. Eqn (i) × 2 + Eqn (ii) gives 4x + 2y
+ 3x – 2y = 8 + 6
or 7x = 14 or x = 2
Putting x = 2 in (i) 4 + y = 4 ∴ y = 0
∴ point A is (2,0)
Let us find point of intersection B of lines (ii) and (iii) i.e., solve
(ii) and (iii) for x and y.
Eqn. (ii) – 3 × eqn. (iii) gives
3x – 2y – 6 – 3(x – 3y + 5) = 0
i.e., 3x – 2y – 6 – 3x + 9y – 15 = 0
or 7y – 21 = 0 ⇒ 7y = 21
⇒ y = 3
Putting y = 3 in (ii), 3x – 6 = 6 ⇒ 3x = 12 ⇒ x = 4
∴ Point B is (4, 3).
Let us find point of intersection C of lines (i) and (iii) i.e., solve
(i) and (iii) for x and y.
Eqn. (i) – 2 × eqn. (iii) gives
2x + y – 4 – 2 (x – 3y + 5) = 0
⇒ 2x + y – 4 – 2x + 6y – 10 = 0
⇒ 7y – 14 = 0 ⇒ 7y = 14 ⇒ y = 2
Putting y = 2 in (i), 2x + 2 = 4 or 2x = 2 or x = 1.
∴ Point C is (1, 2)
∴ Vertices A, B, C of
triangle (region) ABC are A(2,
0), B(4, 3) and C(1, 2).
Join of A and C is the graph
of line (i) 2x + y = 4.
( ∵ (i) intersects (ii) at A and
(iii) at C)
Similarly A B and BC.
Now area of ∆ACM bounded
by line (i) i.e., AC and x-axis.
2
= ∫1 y dx

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 Class 12 Chapter 8 - Applications of Integrals

(At point C, x = 1 and at point A, x = 2)

Putting y = 4 – 2x from (i),
2
2  2 x2 
∫1 (4 − 2 x) dx  4 x − 
= =
 2 1
= (8 – 4) – (4 – 1) = 4 – 3 = 1 ...(iv)
Now area of ∆ABL, bounded by line (ii) i.e., AB and x-axis
4 4 3
= ∫2 y dx = ∫2 2
( x − 2) dx
[... From (ii), – 2y = – 3x + 6
−1 3
⇒ y = (– 3x + 6) = (x – 2)]
2 2
4
3  x2  3
=  − 2 x  = | (8 – 8) – (2 – 4) |
2  2 2 2
3
= (2) = 3 ...(v)
2
Now area of trapezium CMLB bounded by line (iii) i.e., BC and
x-axis
4 4 1
= ∫1 y dx = ∫1 3
( x + 5) dx

1
[... From (iii), x + 5 = 3y ⇒ y = (x + 5)]
3
4
1  x2  1  1 
=  + 5 x  = 8 + 20 −  2 + 5  
3  2 1 3   
1  11  1  56 − 11  1  45 
=  28 − 2  =  2  =  
3   3   3  2 
15
= ...(vi)
2
∴ Required area of region (triangle) bounded by the three given
lines
= Area of trapezium CLMB – Area of ∆ACM
– Area of ∆ABL
15
= – 1 – 3
2
(by (vi)) (by (iv)) (by (v))
15 7
= – 4 = sq. units.
2 2
15. Find the area of the region
{(x, y) : y2 ≤ 4x and 4x2 + 4y2 ≤ 9}
Sol. The required area is the area common to the interiors of the

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Class 12 Chapter 8 - Applications of Integrals

parabola y2 = 4x ...(i)
[Parabola (i) is a rightward parabola with vertex at origin and is
symmetrical about x-axis.]
and the circle 4x2 + 4y2 = 9 ...(ii)
Dividing every term of eqn. (ii) by 4,
2
9 3
x2 + y2 = =  
4 2
3
which is a circle whose centre is origin and radius is .
2
This circle is symmetrical about both the axes.
To find the points of intersection, let us solve (i) and
(ii) for x and y. 1 
A  , 2
Putting y 2 = 4x from (i) in 2 
(ii), we have
4x2 + 16x – 9 = 0
− 16 ± 256 + 144
∴ x =
8
 − b ± b2 − 4 ac 
∵ x = 
 2a 
 
− 16 ± 20 1 9
= = , –
8 2 2
9
When x = – , from (i),
2
y2 = – 18 is
negative and hence y is imaginary and hence impossible.
9
Therefore x = – is rejected.
2
1 1
When x = , from (i), y2 = 4x = 4 × = 2
2 2
∴ y = ± 2
∴ The two points of intersection of parabola (i) and circle (ii) are
1  1 
A  , 2  and B =  , − 2  .
 2   2 

For the parabola (i), y = 2 x in the first quadrant.

9
For the circle (ii), 4y2 = 9 – 4x2 or y2 = – x2
4

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 Class 12 Chapter 8 - Applications of Integrals

9
or y = − x 2 in first quadrant.
4
Required area OADBO (Area of the circle which is interior to the
parabola) (shaded)
= 2 × Area OADO = 2 [Area OAC + Area CAD]
= 2 [Area between parabola (i) and x-axis in first quadrant
+ Area between circle (ii) and x-axis in first quadrant]
 1/ 2 3/2 9 
= 2 * ∫ 0 2 x dx + ∫ 1/ 2 4 − x dx 
2
[Area = ∫ y dx]
 
 1
3
 3 2  9 2 9 2 
 x −x
 x 2  x  
= 2  2 .  + 4 + 4 sin −1  
 3  2 2 3 

 2 0  2  1 
 2
 x a2 x
∵ ∫ a – x dx =
2 2
a2 – x 2 + sin –1 
 2 2 a 
 1 
4 2
1 9 −1 2 9 −1 1    1 3 / 2 1 1 1 
× + sin 1 − − sin
= 2  3 2 2 8 2 8 3
 ∵   = = 
  2  2 2 2 2 
 
 2 9 π 2 9 −1 1  9π 9 1 2
= 2  3 + 8 . 2 − 4 − 8 sin 3  = – sin–1 + .
  8 4 3 6

2 2 1
∵ − =
3 4 6
16. Choose the correct answer:
Area bounded by the curve y = x 3 , the x-axis and the
ordinates x = – 2 and x = 1 is
– 15 15 17
(A) – 9 (B) (C) (D) .
4 4 4
Sol. Equation of the curve is y = x3 ...(i)
Let us draw the graph of curve (i) for values of x from x = – 2
to x = 1.
Table of Values for y = x3
x – 2 – 1 0 1
y – 8 – 1 0 1
We are to find the area of the total shaded region.
We will have to find the two shaded areas OBN and OAM
separately because from the table,

*Limits of integration for parabola are x = 0 to x of point of

intersection and for circle are x of point of intersection to x = radius of
circle.
MathonGo 41
 Class 12 Chapter 8 - Applications of Integrals

y = x3 ≤ 0 for – 2 ≤ x ≤ 0 for the region OBN

and y = x3 ≥ 0 for 0 ≤ x ≤ 1 for the region OAM
0 Y
Now area of region OBN = ∫−2 y dx
A (1, 1)
0
= ∫−2 x 3 dx (By (i))
N
0 X′ X
 x4  O M
=  
 4 − 2 (–1, –1)

16 B
= 0− = | – 4 | = 4 ...(ii)
4 (– 2, – 8) Y′
1
Again area of region OAM = ∫0 y dx

1
= ∫0 x 3 dx (By (i))
1
 x4  1 1
=  4  = −0 = ...(iii)
 0 4 4
Adding areas (ii) and (iii), the total required shaded area
1 16 + 1 17
= 4 + = = sq. units
4 4 4
∴ Option (D) is the correct answer.
17. Choose the correct answer:
The area bounded by the curve y = x | x | , x-axis and the
ordinates x = – 1 and x = 1 is given by
1 2 4
(A) 0 (B) (C) (D) .
3 3 3
Sol. Equation of the curve is
y = x | x | = x(x) = x2 if x ≥ 0 ...(i)
and y = x | x | = x(– x)
= – x2 if x ≤ 0 ...(ii)
Eqn. (i) namely x2 = y (x ≥ 0) represents the arc of the upward
parabola in first quadrant and
Y x2= y
equation (ii) namely x2 = – y (x ≤ 0)
represents the arc of the downward A
parabola in the third quadrant.
We are to find the area bounded x=1
by the given curve, x-axis and the N
X′ X
ordinates x=–1 O M
x = – 1 and x = 1.
Required area = Area ONBO
+ Area OAMO B
2
x=–y Y′

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 Class 12 Chapter 8 - Applications of Integrals

0
= ∫ − 1 y dx (for y given by (ii))
1
+ ∫0 y dx (for y given by (i))
0 1
∫−1 − x ∫0
2
= dx + x 2 dx

0 1
 – x3   x3   −1 1 2
=  3  +  3  = 0 –   + – 0 =
 −1  0  3  3 3
∴ Option (C) is the correct answer.
18. Choose the correct answer:
The area of the circle x2 + y2 = 16 exterior to the parabola
y2 = 6x
4 4
(A) π – 3)
(4π (B) π + 3)
(4π
3 3
4 4
(C) (8ππ – 3) (D) π + 3 ).
(8π
3 3
Sol. Equation of circle is
x2 + y2 = 16 ...(i)
and that of parabola is
y2 = 6x ...(ii)
Now (i) is the circle with
centre at O(0, 0) and
radius 4.
∴ A ↔ (4, 0)
Also, this circle is
symme-trical about x-axis
(... on changing y to – y,
its equa tion remains
unaltered.)
Also Circle (i) is
symmetrical about y-axis.
Equation (ii) represents
a rightward parabola with vertex at origin O. It is also
symmetrical about x-axis.
To find the points of intersection of the two curves, let us
solve them for x and y.
Putting y2 = 6x from (ii) in (i),
x2 + 6x – 16 = 0 or (x + 8) (x – 2) = 0
⇒ x = – 8 or 2
When x = – 8, from (ii), y2 = – 48 < 0 so x = – 8 is not possible.
When x = 2, from (ii), y2 = 12 ⇒ y = ± 2 3
∴ The two points of intersection are B(2, 2 3 ) and B′ (2, – 2 3 ).
Required area (shaded) = Area of circle – area of circle
interior to the parabola

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 Class 12 Chapter 8 - Applications of Integrals

= π × 42 – area OBAB′O
(... area of circle = πr2, here r = 4)
= 16π – 2 × area OBACO ...(iii)
(... the two curves are symmetrical about x-axis.)
Now area OBACO = area OBCO + area BACB
= (area under arc OB of parabola and x-axis)
+ (area under arc BA of circle and x-axis)
2 4
= ∫0 6x dx + ∫2 16 − x2 dx
from (ii) from (i)
x 3/2  2 4
x 2 16 −1 x 
= 6 .   +  16 − x + sin 
 3 / 2  0 2 2 4 2
 x a2 x
∵ ∫ a2 − x 2 dx =
2
a2 − x 2 +
2
sin −1 
a 

2 1
6 (2 2 ) + 8 sin 1 – 12 – 8 sin– 1 2
– 1
=
3
8 π π
or area OBACO = + 8 . – 2 3 – 8 .
3 2 6
 π π 1
∵ sin 2 = 1 and sin 6 = 2 
 
8 1 1
= – 2 3 + 8π  − 
3 2 6
8−6 3 −1 2 8π
= + 8π  6  = +
3   3 3
Putting this value of area OBACO in (i),
 2 8π 
Required area = 16π – 2  + 
 3 3 
4 16π
= 16π – –
3 3
 1 4 32 π 4
= 16π  1 −  – = –
 3  3 3 3
32 π 4 3 4
=
– = (8π – 3 ) sq. units.
3 3 3
∴ Option (C) is the correct answer.
19. Choose the correct answer:
The area bounded by the y-axis, y = cos x and y = sin x when
π
0 ≤ x ≤ is
2
(A) 2( 2 – 11) (B) 2 – 1 (C) 2 + 1 (D) 2 .
Sol. We are to find the area bounded by y-axis, y = cos x, y = sin x
π
when 0 ≤ x ≤ .
2
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Class 12 Chapter 8 - Applications of Integrals

 π
Table of values for y = cos x  0 ≤ x ≤
 2 
π π π π
x 0
6 4 3 2
3 1 1
y 1 0
2 2 2
 π
Table of values for y = sin x  0 ≤ x ≤ 
 2
π π π π
x 0
6 4 3 2
1
3 1
y 0 1
22 2
From the two tables of values, we observe that graphs of
 π
y = sin x and y = cos x  0 ≤ x ≤  have a common point i.e.,
 2 
π 1 
intersect at the point B  4 , .
 2
Now required shaded area OAB
= Area OABM – Area OBM
= (Area bounded by the curve y= cos x, x-axis and the vertical
π
lines x = 0 to x = )
4
– (Area bounded by the curve y = sin x, x-axis and the vertical
π
lines x = 0 to x = )
4
π 4 π 4
∫0 cos x dx – ∫
0
sin x dx = (sin x)0π 4 – (–cos x)0π 4

π π
= sin – sin 0 + (cos – cos 0)
4 4

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Class 12 Chapter 8 - Applications of Integrals

1 1 2
=
2
– 0 +
2
– 1 =
2
– 1 = ( )
2 – 1 = sq units.

∴ Option (B) is the correct answer.

Remark. We were required to find area bounded by y-axis.
The second possible solution was:
1/ 2
Required area = ∫0 x dy where x = sin–1 y from y = sin x
1
+ ∫ 1/ 2
x dy where x = cos–1 y from y = cos x

∫ sin
−1
Since it is laborious to evaluate y dy

∫ sin ∫ cos y dy = ∫
−1 −1
= y . 1 dy and cos −1 y . 1 dy ,
so, we have chosen to the solution by first method.

MathonGo 46