Area Bounded by Two Curves

The two curve represented by y = f (x), y = g (x), where f(x) ≥ g(x) in [a, b] as shown in  following Fig. 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. To take elementary area in the form of vertical strips has height f(x) - g(x) and width dx so that the elementary area 

dA = [f(x) – g(x)] dx, and the total area A can be taken as
A = `int _a^b [f(x) - g(x)]` dx , and the total area A can be taken as

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] 
= `int _a^b f(x) dx -  int _a^b g(x) dx = int _a^b[f(x) - g(x)] dx` , 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 following  Fig. then the area of the regions bounded by curves can be written as 

Total Area = Area of the region ACBDA + Area of the region BPRQB 
=`int _a^c [f(x) - g(x)] dx `+` int _c^b [g(x) - f(x)] dx`

Video link : https://youtu.be/tVKcJLMbgDo

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.Equations of above mentioned curves will be in their standard forms only as the cases in other forms.
For example : 
Find the area of the region bounded by the curve y = x2 and the line y = 4.
⇒ 

Since the given curve represented by the equation y = x2 is a parabola symmetrical about y-axis only, therefore, in above  Fig.
The required area of the region AOBA is given by  `2int _0^4 xdy` = 
`2("area of the region BONB bounded by curve, y - axis and the lines"
y =0 and y = 4)`
`= 2 int _0^4 sqrt y  dy = 2 xx 2/3 [y^(3/2)]_0^4 = 4/3 xx  8 = 32/3 `
Here,we have taken horizontal stripes as indicating in the above Fig.

Alternatively, we may consider the vertical strips like PQ as shown in the following Fig.

in fig the area of the region AOBA.  To this end, we solve the equations x2 = y and y = 4 which gives x = –2 and x = 2. Thus, the region AOBA may be stated as the region bounded by the curve y = x2, y = 4 and the ordinates x = –2 and x = 2. 
Therefore, the area of the region AOBA
=`int _-2^2 y dx`
[y=(y-coordinate of Q) - (y-coordinate of P) = 4 - `x^2`] 
= `2 int _0^2(4 - x^2) dx`
= `2[4x - x^3/3]_0^2 = 2[4 xx  2 - 8/3 ] = 32 / 3` 

Remark: From the above examples, it is inferred that we can consider either vertical strips or horizontal strips for calculating the area of the region.

Video link : https://youtu.be/RvjY8_QPBfo