Find the area bounded by y=2x^2 and y^2=4x

Answer 1

First you need the intersection
(2x^2)^2 = 4x
4x^4 - 4x = 0
x(x^3 - 1) = 0
x=0 or x = 1

see:
http://www.wolframalpha.com/input/?i=y%3D2x%5E2+,+y%5E2%3D4x

effective height = ?(4x) - 2x^2
= 2 x^(1/2) - 2x^2

area = ?(2 x^(1/2) - 2x^2) dx from x = 0 to 1
= [ (4/3)x^(3/2) - (2/3)x^3] from 0 to 1
= (4/3 - 2/3) - 0
= 2/3

Answer 2

The given curves are:
y=2x^2 and y=2x^(1/2)
The desired area lies between x=0 and x=1
The differential area is:
[2x^(1/2) - 2x^2]dx
The integral has limits:
x = 0, x = 1.
The result of integration is:
Area = 2/3.
QED