Application of Integrals
Introduction
Calculus is widely seen as functions, differentiation, and integration. It has a lot of applications in the fields of Engineering, Science, and Mathematics. We have already studied to find the area bounded by the curve y = f(x), the ordinates x = a, x = b and x-axis, while calculating definite integral as the limit of a sum.
In this chapter, we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses. We will learn how to use integrals to find out areas under different conditions.
Area under Simple Curves
Let we want to find the area bounded be the curve y = f(x), x-axis and the ordinates x = a and x = b.
From the figure, we assume that area under the curve is composed of large number of very thin vertical strips. Let us take an arbitrary strip of height y and width dx, then dA (area of the elementary strip) = y dx, where, y = f(x). This area is called the elementary area which is located at an arbitrary position within the region which is specified by some value of x between a and b.
Total area A of the region between x-axis, ordinates x = a, x = b and the curve y = f(x) is calculated by adding up the elementary areas of thin strips across the region PQRSP.
The area A under the curve f(x) bounded by x = a and x = b is given by
A = aꭍb dA = aꭍb y dx = aꭍb f(x) dx
The area A of the region bounded by the curve x = g (y), y-axis
and the lines y = c, y = d as shown in the figure, is given by
A = cꭍd x dy = cꭍd g(y) dy
Example: Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.
Solution:
The area of the region bounded by the curve, y2 = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.
So, area of ABCD = ʃ14 y dx
= ʃ14 √x dx
= [x3/2/(3/2)]14
= (2/3)[43/2 - 13/2]
= (2/3)[8 – 1]
= (2/3) * 7
= 14/3 units
Example: Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
Solution:
The area of the region bounded by the curve, x2 = 4y, y = 2, and y = 4, and the y-axis is the area ABCD.
So, area of ABCD = ʃ24 x dy
= ʃ24 2√y dy
= 2[y3/2/(3/2)]24
= (4/3)[43/2 - 23/2]
= (4/3)[8 – 2√2]
= (32 – 8√2)/3 units
The area of the region bounded by a curve and a line
In this section, we will find the area of the region bounded by a line and a circle, a line and a parabola, a line and an ellipse when their standard forms are given.
Example: Find the area of the region in the first quadrant enclosed by x-axis, line x = √3y and the circle x2 + y2 = 4
Solution:
The area of the region bounded by the circle x2 + y2 = 4, x = √3y, and the x-axis is the area OAB.
The point of intersection of the line and the circle in the first quadrant is (√3, 1).
Area OAB = Area ∆OCA + Area ACB
Area ∆OCA = (1/2) * OC * AC = (1/2) * √3 * 1 = √3/2
Area ACB = ʃ√32 y dx
= ʃ√32 √(4 – x2) dx
= [x√(4 – x2)/2 + (4/2) * sin-1 x/2]√32
= [2 * π/2 - (√3/2) * √(4 - 3) - 2sin-1 √3/2]
= [π - √3/2 – 2(π/3)]
= [π/3 - √3/2]
Therefore, area enclosed by x-axis, x = √3y and the circle x2 + y2 = 4 in the first quadrant
= √3/2 + [π/3 - √3/2] = π/3 units
Example: Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line x = a/√2
Solution:
The area of the smaller part of the circle, x2 + y2 = a2 cut off by the line x = a/√2 is the area ABCDA.
It can be observed that the area ABCD is symmetrical about x-axis.
So, Area ABCD = 2 * Area ABC
Now, Area of ABC = ʃa/√2a y dx
= ʃa/√2a √(a2 – x2) dx
= [x√( a2 – x2)/2 + (a2/2) * sin-1 x/a]a/√2a
= [a2/2 * π/2 - (a/2√2) * √( a2 - a2/2) - a2/2 * sin-1 1/√2]
= a2/2 * π/2 - (a/2√2) * (a/√2) - a2/2 * (π/4)
= a2π/4 - a2/4 - a2π/8
= (a2/4)(π - 1 - π/2)
= (a2/4)(π/2 - 1)
So, area of ABCD = 2[(a2/4)(π/2 - 1)] = (a2/2)(π/2 - 1)
Therefore, The area of the smaller part of the circle, x2 + y2 = a2 cut off by the line x = a/√2 is
(a2/2)(π/2 - 1) units.
Area between Two Curves
The most general class of problems in calculating the area of bounded regions is the one involving the regions between two curves. Let we are given two curves represented by y = f(x), y = g(x), where f(x) ≥ g(x) in [a, b] as shown in figure.
Here the points of intersection of these two curves are given by x = a and x = b obtained by taking common values of y from the given equation of two curves.
Elementary strip has height f(x) – g(x) and width dx so that the elementary area
dA = [f(x) – g(x)] dx
Total area A can be calculated as A = aꭍb [f(x) – g(x)]
Alternatively,
A = [area bounded by y = f (x), x-axis and the lines x = a, x = b]
– [area bounded by y = g (x), x-axis and the lines x = a, x = b]
A = aꭍb f(x) = aꭍb g(x) = aꭍb [f(x) – g(x)], where f(x) ≥ g(x) in [a, b]
If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], where a < c < b as shown in the figure, then the area of the regions bounded by curves can be calculated as
Total Area = Area of the region ACBDA + Area of the region BPRQB.
= aꭍc [f(x) – g(x)] + cꭍb [g(x) – f(x)]
Example:
Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y.
Solution:
The required area is represented by the shaded area OBCDO.
Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we get the point of intersection as B(√2, 1/2) and D(-√2, 1/2).
It can be observed that the required area is symmetrical about y-axis.
So, Area OBCDO = 2 * Area OBCO
Now, draw BM perpendicular to OA.
Therefore, the coordinates of M is (√2, 0).
Therefore, Area OBCO = Area OMBCO – Area OMBO
= ʃ0√2 {√(9 – 4x2)/4} dx - ʃ0√2 x2/4 dx
= (1/2)ʃ0√2 √(9 – 4x2) dx – (1/4)ʃ0√2 x2 dx
= (1/4)[x√(9 – 4x2) + (9/2) * sin-1 2x/3]0√2 - (1/4)[x3/3]0√2
= (1/4)[√2 * √(9 – 8) + (9/2) * sin-1 2√2/3] - (1/12)(√2)3
= (√2/4) + (9/8) * sin-1 2√2/3 - √2/6
= (√2/12) + (9/8) * sin-1 2√2/3
= (1/2)[√2/6 + (9/4) * sin-1 2√2/3]
Therefore, the required area OBCDO = 2 * (1/2)[√2/6 + (9/4) * sin-1 2√2/3]
= [√2/6 + (9/4) * sin-1 2√2/3] units
Example:
Find the area between the curves y = x and y = x2
Solution:
The required area is represented by the shaded area OBAO as
The points of intersection of the curves, y = x and y = x2 is A (1, 1).
Now, draw AC perpendicular to x-axis.
So, Area (OBAO) = Area (∆OCA) – Area (OCABO)
= ʃ01 x dx - ʃ01 x2 dx
= [x2/2]01 – [x3/3]01
= 1/2 - 1/3
= 1/6 units
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