#### Answer

The area of the region is $9/2$.

#### Work Step by Step

$y=x^2-2x$ and $y=x$
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.
$$x^2-2x=x$$ $$x^2-3x=0$$ $$x(x-3)=0$$ $$x=0\hspace{1cm}\text{or}\hspace{1cm}x=3$$
So, the upper limit is $3$ and the lower one is $0$.
3) Find the area:
Looking at the draft, we see that the region from $x=0$ to $x=3$ is bounded above by $y=x$ and below by $y=x^2-2x$. So, according to definition, the area of the region is
$$A=\int^3_{0}[x-(x^2-2x)]dx=\int^3_{0}(-x^2+3x)dx$$ $$A=-\frac{x^3}{3}+\frac{3x^2}{2}\Big]^3_{0}$$ $$A=-\frac{27}{3}+\frac{27}{2}-0$$ $$A=\frac{9}{2}$$