From the previous classes, we are already aware of the various method of integration in Maths.Â 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 12Â 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.Â 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
\(A = \int_{a}^{b} dA = \int_{a}^{b}y.dx = \int_{a}^{b}f(x)dx\)
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
\(A = \int_{a}^{b}[f(x) – g(x)]dx\)Application of Integrals Examples
Example 1:Â
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 asÂ
= 4_{ 0}âˆ«^{a }y dx (By taking the vertical strips) â€¦.(1)
From the given circle equation, y can be written as
y = Â±âˆš(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 âˆš(a^{2}x^{2})
Now, substitute y = âˆš(a^{2}x^{2}) in equation (1), we get
= 4_{ 0}âˆ«^{a }âˆš(a^{2}x^{2}) dxÂ
Integrate the above function, we get
= 4 [(x/2)âˆš(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)(Ï€/2)
= Ï€a^{2}.
Hence, the area enclosed by the circle x^{2} +y^{2} =a^{2 }Â is Ï€a^{2}.
Example 2:
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}âˆ«^{4} ydx + _{4}âˆ«^{8} y dx
= 2 _{0}âˆ«^{4} âˆšx dx + _{4}âˆ«^{8} Â âˆš[4^{2}– (x4)^{2}]dx
Now take x4 = t, then the above equation is written in the formÂ
= 2 _{0}âˆ«^{4} âˆšx dx + _{0}âˆ«^{4 }Â âˆš[4^{2}– t^{2}]dx â€¦. (1)
Now, integrate the functions.
2 _{0}âˆ«^{4} âˆšx dx = (2)(â…”) (x^{3}/2)_{0}^{4}
2 _{0}âˆ«^{4} Â âˆšx dx = 32/3Â â€¦..(2)
_{0}âˆ«^{4} Â âˆš[4^{2}– t^{2}]dx = [(t/2)(âˆš[4^{2}t^{2}] + (Â½)(4^{2})(sin^{1}(t/4)]_{0}^{4}
_{0}âˆ«^{4} âˆš[4^{2}– t^{2}]dxÂ = 4Ï€ â€¦..(3)
Now, substitute (2) and (3) in (1), we get
= (32/3) + Â 4Ï€Â
= (4/3) (8+3Ï€)Â
Therefore, the area of the region that lies above the xaxis, and included between the circle and parabola is (4/3) (8+3Ï€).
To learn more class 12 mathematics concepts, download BYJU’S – The Learning App and also watch engaging videos to learn with ease
Related Links 

Algebra Formulas for Class 12  Differential Equations for Class 12 
Determinants for Class 12  Application of Derivatives for Class 12 