Answer
$\displaystyle \frac{4}{3}-\frac{4}{\pi}$
Work Step by Step
Graph the functions.
Over the interval $x\in[-1,1]$,
the graph of $f(x)=1-x^{2}$ is above the graph of $g(x)=\displaystyle \cos(\frac{\pi x}{2})$.
$A=\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx=\int_{-1}^{1}[1-x^{2}-\cos(\frac{\pi x}{2})]dx$.
$=\displaystyle \left[x -\frac{x^{3}}{3} \right]_{-1}^{1}-\int_{-1}^{1}\cos(\frac{\pi x}{2})dx\quad \left[\begin{array}{ll}
u= \frac{\pi x}{2} & du=\frac{\pi}{2}dx\\
x=-1\Rightarrow u=-\pi/2 & \\
x=1\Rightarrow u=\pi/2 &
\end{array}\right]$
$=(1+1)-\displaystyle \frac{1}{3}(1+1)-\int_{-\pi/2}^{\pi/2}\cos u(\frac{2du}{\pi})d$
$=\displaystyle \frac{4}{3}-\frac{2}{\pi}\left[\sin u \right]_{-\pi/2}^{\pi/2}$
$=\displaystyle \frac{4}{3}-\frac{2}{\pi}(1+1)$
=$\displaystyle \frac{4}{3}-\frac{4}{\pi}$
