Application of Integrals for Class 12
From the previous classes, we are already aware of the various method of integration in Maths. In the class 12 maths chapter 2, students will learn a specific application of integrals to find the area under simple curves, area between lines and arcs of standard forms of different curves such as circles, parabolas and ellipses. We shall also deal with finding the area bounded by these curves along with the combination of lines. In this article, let us look at the various application of integrals Class 12 Maths Chapter 8.
Application of Integrals Class 12 Concepts
Chapter 8 – Application of Integrals class 12 CBSE covers the following concepts:
 Introduction
 Areas and Simple curves
 The area of a region bounded by a curve and a line
 The area between the two curves
From the previous classes, we are already aware of the various method of integration in maths. In this article, let us look at the various application of integrals class 12 Maths.
One of the major application of integrals is in determining the area under the curves.
Consider a function y = f(x), then the area is given as
[latex]A = \int_{a}^{b} dA = \int_{a}^{b}y.dx = \int_{a}^{b}f(x)dx[/latex]
Consider the two curves having equation of f(x) and g(x), the area between the region a,b of the two curves is given as
dA = f(x) – g(x)]dx, and the total area A can be taken as
[latex]A = \int_{a}^{b}[f(x) – g(x)]dx[/latex]Application of Integrals Examples
Example 1:
Determine the area enclosed by the circle x^{2} + y^{2 }= a^{2}
Solution:
Given, circle equation is x^{2} + y^{2 }= a^{2}
From the given figure, we can say that the whole area enclosed by the given circle is as
= 4(Area of the region AOBA bounded by the curve, coordinates x=0 and x=a, and the xaxis)
As the circle is symmetric about both xaxis and yaxis, the equation can be written as
= 4_{ 0}∫^{a }y dx (By taking the vertical strips) ….(1)
From the given circle equation, y can be written as
y = ±√(a^{2}x^{2})
As the region, AOBA lies in the first quadrant of the circle, we can take y as positive, so the value of y becomes √(a^{2}x^{2})
Now, substitute y = √(a^{2}x^{2}) in equation (1), we get
= 4_{ 0}∫^{a }√(a^{2}x^{2}) dx
Integrate the above function, we get
= 4 [(x/2)√(a^{2}x^{2}) +(a^{2}/2)sin^{1}(x/a)]_{0} ^{a}
Now, substitute the upper and lower limit, we get
= 4[{(a/2)(0)+(a^{2}/2)sin^{1 }1}{0}]
= 4(a^{2}/2)(π/2)
= πa^{2}.
Hence, the area enclosed by the circle x^{2} +y^{2} =a^{2 } is πa^{2}.
Let’s have a look at the example to understand how to find the area of the region bounded by a curve and a line.
Example 2:
Find the area of the region bounded between the line x = 2 and the parabola y^{2} = 8x.
Solution:
Given equation of parabola is y^{2} = 8x.
Equation of line is x = 2.
Here, y^{2} = 8x as a right handed parabola having its vertex at the origin and x = 2 is the line which is parallel to yaxis at x = 2 units distance
Similarly,
y^{2} = 8x has only even power of y and is symmetrical about xaxis.
So, the required area = Area of OAC + Area of OAB
= 2 (Area of OAB)
= 2 ∫_{0}^{2} y dx
Substituting the value of y, i.e. y2 = 8x and y = √(8x) = 2 √2 √x, we get;
= 2 ∫_{0}^{2} (2 √2 √x) dx
= 4√2 ∫_{0}^{2} (√x) dx
= 4√2 [x^{3/2}/ (3/2)]_{0}^{2}
By applying the limits,
= 4√2 {[2^{3/2}/ (3/2)] – 0}
= (8√2/3) × 2√2
= (16 × √2 × √2)
= 32/3
Go through the example given below to learn how to find the area between two curves.
Example 3:
Determine the area which lies above the xaxis and included between the circle and parabola, where the circle equation is given as x^{2}+y^{2} = 8x, and parabola equation is y^{2} = 4x.
Solution:
The circle equation x^{2}+y^{2} = 8x can be written as (x4)^{2}+y^{2}=16. Hence, the centre of the circle is (4, 0), and the radius is 4 units. The intersection of the circle with the parabola y^{2} = 4x is as follows:
Now, substitute y^{2} = 4x in the given circle equation,
x^{2}+4x = 8x
x^{2}– 4x = 0
On solving the above equation, we get
x=0 and x=4
Therefore, the point of intersection of the circle and the parabola above the xaxis is obtained as O(0,0) and P(4,4).
Hence, from the above figure, the area of the region OPQCO included between these two curves above the xaxis is written as
= Area of OCPO + Area of PCQP
= _{0}∫^{4} y dx + _{4}∫^{8} y dx
= 2 _{0}∫^{4} √x dx + _{4}∫^{8} √[4^{2}– (x4)^{2}]dx
Now take x4 = t, then the above equation is written in the form
= 2 _{0}∫^{4} √x dx + _{0}∫^{4 } √[4^{2}– t^{2}]dx …. (1)
Now, integrate the functions.
2 _{0}∫^{4} √x dx = (2)(⅔) (x^{3}/2)_{0}^{4}
2 _{0}∫^{4} √x dx = 32/3 …..(2)
_{0}∫^{4} √[4^{2}– t^{2}]dx = [(t/2)(√[4^{2}t^{2}] + (½)(4^{2})(sin^{1}(t/4)]_{0}^{4}
_{0}∫^{4} √[4^{2}– t^{2}]dx = 4π …..(3)
Now, substitute (2) and (3) in (1), we get
= (32/3) + 4π
= (4/3) (8+3π)
Therefore, the area of the region that lies above the xaxis, and included between the circle and parabola is (4/3) (8+3π).
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Algebra Formulas for Class 12  Differential Equations for Class 12 
Determinants for Class 12  Application of Derivatives for Class 12 