#### Answer

The area of the region is $4$.

#### Work Step by Step

$y=7-2x^2$ and $y=x^2+4$
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 two curves.
$$7-2x^2=x^2+4$$ $$3x^2-3=0$$ $$3(x-1)(x+1)=0$$ $$x=1\hspace{1cm}\text{or}\hspace{1cm}x=-1$$
So, the upper limit is $1$ and the lower one is $-1$.
3) Find the area:
Looking at the draft, we see that the region from $x=-1$ to $x=1$ is bounded above by $y=7-2x^2$ and below by $y=x^2+4$. So, according to definition, the area of the region is
$$A=\int^1_{-1}[7-2x^2-(x^2+4)]dx=\int^1_{-1}(-3x^2+3)dx$$ $$A=-x^3+3x\Big]^1_{-1}$$ $$A=(-1+3)-(1-3)$$ $$A=2-(-2)=4$$