Answer
$\displaystyle \frac{104}{15}$
Work Step by Step
Expressing the $y$'s as functions of $x$,
$y=f(x)=4-4x^{2},$
$y=g(x)=x^{4}-1.$
Graphing the functions, we find the intersection points at $x=-1$ and $x=1$.
Over the interval $x\in[-1,1],\ f(x)$ is above $g(x)$, so
$A=\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx=\int_{-1}^{1}[4-4x^{2}-(x^{4}-1)]dx$.
$=\displaystyle \int_{-1}^{1}[5-4x^{2}-x^{4}]dx$.
$=\displaystyle \left[5x-\frac{4x^{3}}{3}-\frac{x^{5}}{5} \right]_{-1}^{1}$
$=5(1+1)-\displaystyle \frac{4}{3}(1+1)-\frac{1}{5}(1+1)$
$=10-\displaystyle \frac{8}{3}-\frac{2}{5}$
$=\displaystyle \frac{104}{15}$
