SOLUTION 22: Compute the area of the region enclosed by the graphs of the equations $y=2x, y= \displaystyle{ \frac{1}{2} }x-4, y=0,$ and $y=2$ . Now see the given graph of the enclosed region.

Using horizontal cross-sections to describe this region, we get that $$0 \le y \le 2 \ \ and \ \ \displaystyle{ \frac{1}{2}y } \le x \le 2y+8 \ ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{0}^{2} (Right \ - \ Left) \ dy }$$ $$= \displaystyle { \int_{0}^{2} \Big( (2y+8) - \frac{1}{2}y \Big) \ dy }$$ $$\displaystyle { = \Big( y^2+8y - \frac{y^2}{4} \Big) \Big\vert_{0}^{2} }$$ $$\displaystyle { = \Big( (2)^2 + 8(2) - \frac{(2)^2}{4} \Big) - \Big( (0)^2+8(0) - \frac{(0)^2}{4} \Big) }$$ $$= 19$$