SOLUTION 20: Compute the area of the region enclosed by the graphs of the equations $ \displaystyle y= \frac{2}{x}, y= \frac{1}{2}x,$
and $ y=2 $ . Begin by finding the points of intersection
of the three graphs. From $ \displaystyle y= \frac{2}{x} $ and $ y= \frac{1}{2}x $
we get that
$$ \displaystyle \frac{2}{x} = \frac{1}{2}x \ \ \longrightarrow $$
$$ x^2 = 4 \ \ \longrightarrow $$
$$ x=2 \ \ or x=-2 (No!) \ \ \longrightarrow \ \ y = \frac{2}{(2)}=1 $$
Now see the given graph of the enclosed region.
Using horizontal cross-sections to describe this region, we get that
$$ 1 \le y \le 2 \ \ and \ \ \frac{2}{x} \le x \le \frac{1}{2}x \ ,$$
so that the area of this region is
$$ AREA = \displaystyle{ \int_{1}^{2} (Right \ - \ Left) \ dy } $$
$$ = \displaystyle { \int_{1}^{2} \Big( 2y - \frac{2}{y} \Big) \ dy } $$
$$ = \displaystyle { ( y^2 - 2 \ln |y| ) \Big\vert_{1}^{2} } $$
$$ = \displaystyle { ( (2)^2 - 2 \ln 2 ) - ( (1)^2 - 2 \ln 1 ) } $$
$$ = 4 - 2 \ln 2 - 1 + 2(0) $$
$$ = 3 - 2 \ln 2 $$
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