Answer
$\frac{8}{3}$
Work Step by Step
$y=x^2$, $y=4x-x^2$
Find their intersections:
$x^2=4x-x^2$
$2x^2-4x=0$
$x(2x-4)=0$
$x=0$ or $2x-4=0$
If $2x-4=0$, then $2x=4$, so $x=2$.
So the intersections are $x=0$ and $x=2$. Note that between 0 and 2 $4x-x^2>x^2$, so the integrand is $(4x-x^2)-x^2$.
The area is:
$\int_0^2((4x-x^2)-x^2)dx$
$=\int_0^2(4x-2x^2)dx$
$=(\frac{4x^2}{2}-\frac{2x^3}{3})|_0^2$
$=(\frac{4*2^2}{2}-\frac{2*2^3}{3})-(\frac{4*0^2}{2}-\frac{2*0^3}{3})$
$=(8-\frac{16}{3})-(0-0)$
$=\frac{8}{3}$