Answer
$\displaystyle \frac{32}{3}$
Work Step by Step
Graph both functions in the same window (see below).
On the interval $x\in[-1,3]$, the graph of $y= 2x-x^{2}$ is above the graph of $y=-3.$
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,
$A=\displaystyle \int_{-1}^{3}[2x-x^{2}-(-3)]dx=\int_{-1}^{3}[2x-x^{2}+3]dx$
$=2[\displaystyle \frac{x^{2}}{2}]_{-1}^{3}-[\frac{x^{3}}{3}]_{-1}^{3}+3[x]_{-1}^{3}$
$=(9-1)-\displaystyle \frac{1}{3}(27+1)+3(3+1)$
$=8-\displaystyle \frac{28}{3}+12$
$=20-\displaystyle \frac{28}{3}$
$=\displaystyle \frac{32}{3}$
