Find the area of bounded by f(x)=2x^2
and g(x)=x^4-2x^2
It is easy to solve for their intersections.
I got x=0 , x= ± 2
Doing a rough sketch will show that there are 2 symmetrical regions, so let's do the one from 0 to 2
the height of the region is 2x^2 - (x^4 - 2x^2)
= 4x^2 - x^4
so the area of that region is the integral of the above from 0 to 2
= (4/3)x^3 - (1/5)x^5 │ from 0 to 2
= 32/3 - 32/5
= 64/15
the region on the left side of the y-axis is equal to that, so that the total area is
128/15
(check my arithmetic, I'm only on my first coffee)