#### Answer

The area of the region is $32/3$.

#### Work Step by Step

$y=2x-x^2$ and $y=-3$
1) Draft the graph
The graph is enclosed below. From the graph, we choose integration with respect to $x$ to employ here for calculation of the region area.
2) Find the limits of integration:
We can find the limits of integration by finding points of intersection between the curve and the line.
$$2x-x^2=-3$$ $$x^2-2x-3=0$$ $$x=-1\hspace{1cm}\text{or}\hspace{1cm}x=3$$
So, the upper limit is $3$ and the lower one is $-1$.
3) Find the area:
Looking at the draft, we see that the region is bounded above by $y=2x-x^2$ and below by $y=-3$. So, according to definition, the area of the region is
$$A=\int^3_{-1}[2x-x^2-(-3)]dx=\int^3_{-1}[2x-x^2+3]dx$$ $$A=\Big(x^2-\frac{x^3}{3}+3x\Big)\Big]^3_{-1}$$ $$A=(9-9+9)-(1-\frac{(-1)}{3}-3)$$ $$A=9-(-\frac{5}{3})=\frac{32}{3}$$