Answer
$\displaystyle \frac{8}{3}$
Work Step by Step
Graph both functions in the same window (see below).
On the interval $x\in[0,2]$, the graph of $y= -x^{2}+4x$ is above the graph of $y=x^{2}.$
The area between two graphs over [a,b], where one graph is above the other, is given with $\displaystyle \int_{a}^{b}(y_{above}-y_{below})dt$.
Thus,
$A=\displaystyle \int_{0}^{2}[-x^{2}+4x-(x^{2})]dx=\int_{0}^{2}(4x-2x^{2})dx$
$=4[\displaystyle \frac{x^{2}}{2}]_{0}^{2}-2[\frac{x^{3}}{3}]_{0}^{2}$
$=2(4-0)-\displaystyle \frac{2}{3}(8-0)$
$=8-\displaystyle \frac{16}{3}$
$=\displaystyle \frac{24-16}{3}$
$=\displaystyle \frac{8}{3}$
