Answer
$\displaystyle \frac{22}{15}$
Work Step by Step
On the interval $x\in[0,1]$, the graph of $y=x^{2}$ is above he graph of $y=-2x^{4}$.
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,
$A=\displaystyle \int_{-1}^{1}[x^{2}-(-2x^{4})]dy=\int_{-1}^{1}(x^{2}+2x^{4})dx$
$=[\displaystyle \frac{x^{3}}{3}]_{-1}^{1}+2[\frac{x^{5}}{5}]_{-1}^{1}$
$=[\displaystyle \frac{1}{3}-(-\frac{1}{3})]-2[\frac{1}{5}-(-\frac{1}{5})]$
$=\displaystyle \frac{2}{3}+\frac{4}{5}$
$=\displaystyle \frac{2(5)+4(3)}{15}$
$=\displaystyle \frac{22}{15}$
