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
$$ AREA = \displaystyle{ \int_{0}^{2} (Right \ - \ Left) \ dy } $$
$$ = \displaystyle { \int_{0}^{2} \Big( (2y+8) - \frac{1}{2}y \Big) \ dy } $$
$$ \displaystyle { = \Big( y^2+8y - \frac{y^2}{4} \Big) \Big\vert_{0}^{2} } $$
$$ \displaystyle { = \Big( (2)^2 + 8(2) - \frac{(2)^2}{4} \Big) - \Big( (0)^2+8(0) - \frac{(0)^2}{4} \Big) } $$
$$ = 19 $$
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