SOLUTION 2: Compute the area of the region enclosed by the graphs of the equations $y=x^2$ and $y=x+2$ . Begin by finding the points of intersection of the two graphs. From $y=x^2$ and $y=x+2$ we get that \vskp $$x^2=x+2 \ \ \longrightarrow$$ $$x^2-x-2=0 \ \ \longrightarrow$$ $$(x-2)(x+1)=0 \ \ \ \ \longrightarrow$$ $$x=2 \ \ or \ \ x=-1$$ Now see the given graph of the enclosed region.

Using vertical cross-sections to describe this region, we get that $$-1 \le x \le 2 \ \ and \ \ x^2 \le y \le x+2 ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{-1}^{2} (Top \ - \ Bottom) \ dx }$$ $$= \displaystyle{ \int_{-1}^{2} ((x+2)-x^2) \ dx }$$ $$= \displaystyle{ \Big( {x^2 \over 2}+2x-{x^3 \over 3} \Big) \Big\vert_{-1}^{2} }$$ $$= \displaystyle{ \Big( {2^2 \over 2}+2(2)-{2^3 \over 3} \Big) - \Big( {(-1)^2 \over 2}+2(-1)-{(-1)^3 \over 3} \Big) }$$ $$= \displaystyle{ \Big(2+4-{8 \over 3}\Big) - \Big({1 \over 2}-2+{1 \over 3}\Big) }$$ $$= \displaystyle{ \Big(6-{8 \over 3}\Big) - \Big({3 \over 6}-{12 \over 6}+{2 \over 6}\Big) }$$ $$= \displaystyle{ \Big({36 \over 6}-{16 \over 6}\Big) - \Big({-7 \over 6}\Big) }$$ $$= \displaystyle{ {20 \over 6}+{7 \over 6} }$$ $$= \displaystyle{ {27 \over 6} }$$ $$= \displaystyle{ {9 \over 2} }$$