Answer
The area of the region is $9/2$.
Work Step by Step
$x=y^2$ and $x=y+2$
1) Draft the graph:
The graph is drafted below. We would use integration with respect to $y$ here.
2) Find the limits of integration:
To find the limits of integration, we find intersection points with respect to $y$.
$$y^2=y+2$$ $$y^2-y-2=0$$ $$(y+1)(y-2)=0$$ $$y=-1\hspace{1cm}\text{or}\hspace{1cm}y=2$$
So the upper limit is $2$ and the lower limit is $-1$.
3) Find the area:
The region from $y=-1$ to $y=2$ is bounded on the right by $x=y+2$ and on the left by $x=y^2$. Therefore, the area of the region is $$A=\int^2_{-1}(y+2-y^2)dy$$ $$A=\frac{y^2}{2}+2y-\frac{y^3}{3}\Big]^2_{-1}$$ $$A=\Big(2+4-\frac{8}{3}\Big)-\Big(\frac{1}{2}-2+\frac{1}{3}\Big)$$ $$A=\frac{10}{3}-(-\frac{7}{6})=\frac{9}{2}$$
