Answer
The area of the region is $8/3$.
Work Step by Step
$y=x^2$ and $y=-x^2+4x$
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.
$$-x^2+4x=x^2$$ $$2x^2-4x=0$$ $$x(x-2)=0$$ $$x=0\hspace{1cm}\text{or}\hspace{1cm}x=2$$
So, the upper limit is $2$ and the lower one is $0$.
3) Find the area:
Looking at the draft, we see that the region from $x=0$ to $x=2$ is bounded above by $y=-x^2+4x$ and below by $y=x^2$. So, according to definition, the area of the region is
$$A=\int^2_{0}[-x^2+4x-x^2]dx=\int^2_{0}(-2x^2+4x)dx$$ $$A=-\frac{2x^3}{3}+2x^2\Big]^2_{0}$$ $$A=-\frac{2\times2^3}{3}+2\times2^2-0$$ $$A=-\frac{16}{3}+8=\frac{8}{3}$$
