What is the area bounded by y=x^2, y=(x-2)^2 and y=0, in the interval [0,2]?

Question

I know that I have to use integrals and I know that the defined integral of the top minus the defined integral of the bottom equals the area, but the area is bound by three functions...

Answer

The question is a little tricky because I failed to see the answer right away. After graphing everything on the paper, you can see y=x^2 and y=(x-2)^2 is symmetric with respect to x=1. So, we can get {integral from 0-1 of x^2} + {intergral from 1-2 of (x-2)^2}. Or, just {integral from 0-1 of x^2} * 2 = 2/3

What is the area bounded by y=x^2, y=(x-2)^2 and y=0, in the interval [0,2]?

Answers (1)