Answer
$\displaystyle \frac{16}{3}$
Work Step by Step
We treat the x-axis as the graph of $y=f_{1}(x)=0$,
and the given function as $y=f(x)=x\sqrt{4-x^{2}}$.
The area between the two graphs is over [a,b], where one graph is above the other; it is given as:
$\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx$.
On $[-2,0] \quad y=0$ is above $y=x\sqrt{4-x^{2}}$.
On $[0,2] \quad y=x\sqrt{4-x^{2}}$is above $ y=0$ .
so
$A=\displaystyle \int_{-2}^{0}[0-x\sqrt{4-x^{2}}]dx++\int_{0}^{2}[x\sqrt{4-x^{2}}-0]dx$
$A=-\displaystyle \int_{-2}^{0}x\sqrt{4-x^{2}}dx+\int_{0}^{2}x\sqrt{4-x^{2}}dx$
$A=-I_{1}+I_{2}$.
Applying (to both integrals):
$\quad \left[\begin{array}{ll}
u=4-x^{2} & du=g'(x)dx\\
& du=-2xdx \\
& xdx=\frac{du}{-2}
\end{array}\right]$,
The borders change to
integral $I_{1}:\quad x=-2 \ \rightarrow\ u=0$ and $\quad x=0 \ \rightarrow\ u=4$
integral $I_{2}:\quad x=0 \ \rightarrow\ u=4$ and $\quad x=2 \ \rightarrow\ u=0$
$I_{1}=$ (after substituing) $=\displaystyle \int_{0}^{4}u^{1/2}(-\frac{du}{2})$
$=-\displaystyle \int_{0}^{4}u^{1/2}du=$
$=-\displaystyle \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{4}$
$=-\displaystyle \frac{2}{3}(\frac{4^{3/2}}{2}-0)$
$=-\displaystyle \frac{2}{3}\cdot\frac{(2^{2})^{3/2}}{2}$
$=-\displaystyle \frac{1}{3}\cdot 8$
$=-\displaystyle \frac{8}{3}$
$I_{2}=$ (after substituing) $=-\displaystyle \int_{4}^{0}u^{1/2}du=$
Reversing the bounds, we find that
$I_{2}=-I_{1}=\displaystyle \frac{8}{3}$
$A=-(-\displaystyle \frac{8}{3})+\frac{8}{3}=\frac{16}{3}$
