Answer
$4$
Work Step by Step
Graph both functions in the same window (see below).
On the interval $x\in[0,2]$, the graph of $y= 7-2x^{2}$ is above the graph of $y=x^{2}+4.$
The area between 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,
$A=\displaystyle \int_{-1}^{1}[7-2x^{2}-(x^{2}+4)]dx=\int_{-1}^{1}(3-3x^{2})dx$
$=3[x]_{-1}^{1}-3[\displaystyle \frac{x^{3}}{3}]_{-1}^{1}$
$=3(1+1)-(1+1)$
$=6-2$
$=4$
