Answer
$$\frac{32}{3}$$
Work Step by Step
Since the region bounded by
$$ y = 4- x ^2,\ \ \ \ \ \ \ y = 0 $$
First, we find the intersection points
\begin{aligned}
4-x^2&=0\\
(2-x)(2+x)&=0
\end{aligned}
Then $$x=2,\ \ \ x= -2$$
Since
$$ 4-x^2\geq 0 \ \ \ \text{for}\ \ \ -2\leq x\leq 2 $$
Then area is given by
\begin{aligned} \text { Area }&=\int_{-2}^{2}\left(4-x^{2}\right) d x\\
&=\left[4 x-\frac{1}{3} x^{3}\right]_{-2}^{2}\\
&=\left(8-\frac{8}{3}\right)-\left(-8+\frac{8}{3}\right)\\
&=\frac{32}{3}\end{aligned}
