Answer
$\displaystyle \frac{32}{3}$
Work Step by Step
Graph both functions in the same window (see below).
On the interval $x\in[-2,2]$, the graph of $y= 2$ is above the graph of $y=x^{2}-2.$
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,
$A=\displaystyle \int_{-2}^{2}[2-(x^{2}-2)]dx=\int_{-2}^{2}(4-x^{2}]dx=$
$=4[x]_{-2}^{2}-[\displaystyle \frac{x^{3}}{3}]_{-2}^{2}$
$=4(2+2)-\displaystyle \frac{1}{3}(8+8)$
$=16-\displaystyle \frac{16}{3}$
$=\displaystyle \frac{48-16}{3}$
$=\displaystyle \frac{32}{3}$
