SOLUTION 24: Compute the area of the region enclosed by the graphs of the equations $ y=2x, y=\displaystyle{ \frac{1}{2} }x, $
and $ y=4 $ . Begin by finding the points of intersection of the the three graphs. From $ y = 2x $ and
$ y = \displaystyle{ \frac{1}{2} }x $ we get that
$$ 2x = \displaystyle{ \frac{1}{2} }x \ \ \longrightarrow $$
$$ 4x = x \ \ \longrightarrow $$
$$ 3x = 0 \ \ \longrightarrow \ \ x=0 $$
From $ y = 4 $ and $ y = \displaystyle{ \frac{1}{2} }x $ we get that
$$ 4 = \displaystyle{ \frac{1}{2} }x \ \ \longrightarrow \ \ x=8 $$
From $ y = 4 $ and $ y = 2x $ we get that
$$ 4 = 2x \ \ \longrightarrow \ \ x=2 $$
Now see the given graph of the enclosed region.
a.) Using vertical cross-sections to describe this region, which is made up of two smaller regions, we get that
$$ 0 \le x \le 2 \ \ and \ \ \displaystyle{ \frac{1}{2} }x \le y \le 2x \ , $$
in addition to
$$ 2 \le x \le 8 \ \ and \ \ \displaystyle{ \frac{1}{2} }x \le y \le 4 \ , $$
so that the area of this region is
$$ AREA = \displaystyle{ \int_{0}^{2} (Top \ - \ Bottom) \ dx + \int_{2}^{8} (Top \ - \ Bottom) \ dx } $$
$$ = \displaystyle{ \int_{0}^{2} \Big( 2x - \frac{1}{2}x \Big) \ dx + \int_{2}^{8} \Big( 4 - \frac{1}{2}x \Big) \ dx } $$
$$ = \displaystyle{ \int_{0}^{2} \frac{3}{2}x \ dx + \int_{2}^{8} \Big( 4 - \frac{1}{2}x \Big) \ dx } $$
$$ \displaystyle { = \Big( \frac{3x^2}{4} \Big) \Big\vert_{0}^{2} + \Big( 4x - \frac{x^2}{4} \Big) \Big\vert_{2}^{8} } $$
$$ \displaystyle { = \Big( \frac{3(2)^2}{4} - \frac{3(0)^2}{4} \Big)
+ \Big( \Big( 4(8) - \frac{(8)^2}{4} \Big) - \Big( 4(2) - \frac{(2)^2}{4} \Big) \Big) } $$
$$ = (3) + (16) - (7) $$
$$ = 12 $$
b.) Using horizontal cross-sections to describe this region, we get that
$$ 0 \le y \le 4 \ \ and \ \ \displaystyle{ \frac{1}{2}y } \le x \le 2y \ , $$
so that the area of this region is
$$ AREA = \displaystyle{ \int_{0}^{4} ( Right \ - \ Left ) \ dy } $$
$$ AREA = \displaystyle{ \int_{0}^{4} \Big( 2y - \frac{1}{2}y \Big) \ dx } $$
$$ \displaystyle { = \Big( y^2 - \frac{y^2}{4} \Big) \Big\vert_{0}^{4} } $$
$$ \displaystyle { = \Big( (4)^2 - \frac{(4)^2}{4} \Big) - \Big( (0)^2 - \frac{(0)^2}{4} \Big) } $$
$$ \displaystyle { = (12) - (0) } $$
$$ = 12 $$
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