Answer
$\displaystyle \frac{128}{15}$
Work Step by Step
On the interval $[-2,2]$, the graph of $y=2x^{2}$ is above the graph of $y=x^{4}-2x^{2}$,
The area between the 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,
$A=\displaystyle \int_{-2}^{2}[2x^{2}-(x^{4}-2x^{2})]dx=\int_{-2}^{2}(4x^{2}-x^{4})dx$
$=4\displaystyle \int_{-2}^{2}x^{2}dx -\int_{-2}^{2}x^{4}dx$
$=4[\displaystyle \frac{x^{3}}{3}]_{-2}^{2}-[\frac{x^{5}}{5}]_{-2}^{2}$
$=4(\displaystyle \frac{2^{3}}{3}-\frac{(-2)^{3}}{3})-[\frac{32}{5}-(-\frac{32}{5})]$
$=\displaystyle \frac{64}{3}-\frac{64}{5}$
$=\displaystyle \frac{64(5)-64(3)}{15}$
= $\displaystyle \frac{128}{15}$
