Chapter 8
APPLICATION OF INTEGRALS
8.1 Overview
This chapter deals with a specific application of integrals to find the area under simple
curves, area between lines and arcs of circles, parabolas and ellipses, and finding the
area bounded by the above said curves.
8.1.1 The area of the region bounded by the curve y = f (x), x-axis and the lines
x = a and x = b (b > a) is given by the formula:
b b
Area = ∫ ydx = ∫ f ( x) dx
a a
8.1.2 The area of the region bounded by the curve x = (y), y-axis and the lines
y = c, y = d is given by the formula:
d d
=
Area ∫ xdy = ∫ ( y) dy
c c
8.1.3 The area of the region enclosed between two curves y = f (x), y = g (x) and the
lines x = a, x = b is given by the formula.
b
Area = ∫ [ f ( x) – g ( x)] dx , where f (x) ≥ g (x) in [a, b]
a
8.1.4 If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], a < c < b, then
c b
Area = ∫ [ f ( x) – g ( x)] dx + ∫ ( g ( x) – f ( x)) dx
a c
8.2 Solved Examples
Short Answer (S.A.)
Example 1 Find the area of the curve
y = sin x between 0 and π.
Solution We have
Area OAB = = – cos x 0
= cos0 – cosπ = 2 sq units.
, APPLICATION OF INTEGRALS 171
Example 2 Find the area of the region bounded by the curve ay2 = x3, the y-axis and
the lines y = a and y = 2a.
Solution We have
2a 2a 1 2
Area BMNC = ∫
a
xdy = ∫ a 3 y 3 dy
a
1
2a
3a 3 53
= y
5 a
1
5 5
3a 3
= ( 2a ) – a 3
3
5
15 5
3
= a 3 a 3 ( 2)3 – 1
5
2
3 2 3
= a 2.2 – 1 sq units.
5
Example 3 Find the area of the region
bounded by the parabola y2 = 2x and the
straight line x – y = 4.
Solution The intersecting points of the given
curves are obtained by solving the equations
x – y = 4 and y2 = 2x for x and y.
We have y2 = 8 + 2y i.e., (y – 4) (y + 2) = 0
which gives y = 4, –2 and
x = 8, 2.
Thus, the points of intersection are (8,
4), (2, –2). Hence
4
1
∫ 4 + y – 2 y
2
Area = dy
–2
4
y2 1 3
= 4 y + – y = 18 sq units.
2 6 –2
Example 4 Find the area of the region
bounded by the parabolas y2 = 6x and
x2 = 6y.
APPLICATION OF INTEGRALS
8.1 Overview
This chapter deals with a specific application of integrals to find the area under simple
curves, area between lines and arcs of circles, parabolas and ellipses, and finding the
area bounded by the above said curves.
8.1.1 The area of the region bounded by the curve y = f (x), x-axis and the lines
x = a and x = b (b > a) is given by the formula:
b b
Area = ∫ ydx = ∫ f ( x) dx
a a
8.1.2 The area of the region bounded by the curve x = (y), y-axis and the lines
y = c, y = d is given by the formula:
d d
=
Area ∫ xdy = ∫ ( y) dy
c c
8.1.3 The area of the region enclosed between two curves y = f (x), y = g (x) and the
lines x = a, x = b is given by the formula.
b
Area = ∫ [ f ( x) – g ( x)] dx , where f (x) ≥ g (x) in [a, b]
a
8.1.4 If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], a < c < b, then
c b
Area = ∫ [ f ( x) – g ( x)] dx + ∫ ( g ( x) – f ( x)) dx
a c
8.2 Solved Examples
Short Answer (S.A.)
Example 1 Find the area of the curve
y = sin x between 0 and π.
Solution We have
Area OAB = = – cos x 0
= cos0 – cosπ = 2 sq units.
, APPLICATION OF INTEGRALS 171
Example 2 Find the area of the region bounded by the curve ay2 = x3, the y-axis and
the lines y = a and y = 2a.
Solution We have
2a 2a 1 2
Area BMNC = ∫
a
xdy = ∫ a 3 y 3 dy
a
1
2a
3a 3 53
= y
5 a
1
5 5
3a 3
= ( 2a ) – a 3
3
5
15 5
3
= a 3 a 3 ( 2)3 – 1
5
2
3 2 3
= a 2.2 – 1 sq units.
5
Example 3 Find the area of the region
bounded by the parabola y2 = 2x and the
straight line x – y = 4.
Solution The intersecting points of the given
curves are obtained by solving the equations
x – y = 4 and y2 = 2x for x and y.
We have y2 = 8 + 2y i.e., (y – 4) (y + 2) = 0
which gives y = 4, –2 and
x = 8, 2.
Thus, the points of intersection are (8,
4), (2, –2). Hence
4
1
∫ 4 + y – 2 y
2
Area = dy
–2
4
y2 1 3
= 4 y + – y = 18 sq units.
2 6 –2
Example 4 Find the area of the region
bounded by the parabolas y2 = 6x and
x2 = 6y.
