# Visual Calculus

## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.

#### SIGN UP FOR A FREE TRIAL

Calculus is a very important section in mathematics which analyzes the behavior of different functions and with the help of the concepts in calculus; we can evaluate the values of the functions in different cases. Visual Calculus is a method of solving complex functions by using a visual approach and we can get the solutions of the functions using simple techniques. In order to find the area covered under the graph, we can easily use the method of integration and find the area covered from one point to another point.

Example 1: Find the area covered by the curve y = x2 and the X-axis(y = 0) between x = 0 and x = 2.

To find the area covered under the curve, we find the integral of the functions.
∫xn dx= x(n+1)/ (n+1)
∫(x2 – 0)dx = x2+1/(2 + 1)= x3/3
First substitute x = 0 and x= 2 in the above answer.
When x= 0, ∫f(x) dx= 03/3= 0
When x=2, ∫f(x)dx= 23/3= 8/3
8/3 - 0= 8/3units2
Hence the area covered between x= 0 and x= 2 is 8/3units2

Example 2: Find the area covered by the curve y = x3 and the X-axis(y = 0) between x = 0 and x = 2.

To find the area covered under the curve, we find the integral of the functions.
∫xn dx= x(n+1)/ (n+1)
∫(x3 – 0)dx = x3+1/(3 + 1)= x4/4
First substitute x = 0 and x= 2 in the above answer.
When x= 0, ∫f(x) dx= 04/4= 0
When x=2, ∫f(x)dx= 24/4= 4
4 - 0= 4units2
Hence the area covered between x= 0 and x= 2 is 4units2