SOLUTION 8: Compute the area of the
region enclosed by the graphs of the equations $ y = \displaystyle \frac{8}{x} $,
$ y = 2x $ and $ y = 2 $ . Begin by finding the points of
intersection of the three graphs. From $ y = \displaystyle \frac{8}{x} $ and $ y
= 2x $ we get that
$$ \displaystyle \frac{8}{x} = 2x \ \ \longrightarrow $$
$$ 8 = 2x^{2} \ \ \longrightarrow $$
$$ 4 = x^{2} \ \ \longrightarrow \ \ x = 2 $$
From $ y = \displaystyle \frac{8}{x} $ and $ y
= 2 $ we get that
$$ \displaystyle \frac{8}{x} = 2 \ \ \longrightarrow \ \ x = 4 $$
From $ y = 2x $ and $ y
= 2 $ we get that
$$ \displaystyle 2x = 2 \ \ \longrightarrow \ \ x = 1 $$
Now see the given graph of the enclosed region.
Using vertical cross-sections to describe this region, which is made up of two smaller regions, we get that
$$ 1 \le x \le 2 \ \ and \ \ 2 \le y \le 2x $$
in addition to
$$ 2 \le x \le 4 \ \ and \ \ 2 \le y \le \displaystyle \frac{8}{x} ,$$
so that the area of this region is
$$ AREA = \displaystyle{ \int_{1}^{2} (Top \ - \ Bottom ) \ dy } + \displaystyle{ \int_{2}^{4} (Top \ - \ Bottom ) \ dy } $$
$$ = \displaystyle{ \int_{1}^{2} (2x - 2 ) \ dy } + \displaystyle{ \int_{2}^{4} \Big(\frac{8}{x} - 2 \Big) \ dy } $$
$$ = \displaystyle { \Big( x^2-2x \Big) \Big\vert_{1}^{2} } + \displaystyle { \Big( 8 \ln |x| - 2x\Big) \Big\vert_{2}^{4} } $$
$$ = { \Big( (2)^2-2(2) \Big) - \Big( (1)^2-2(1) \Big) }
+ { \Big( 8 \ln 4 - 2(4) \Big) - \Big( 8 \ln 2 - 2(2) \Big) } $$
$$ = (0)-(-1) + 8 \ln 2^2 - 8 - 8 \ln 2 + 4 $$
(Recall that $ \ln A^B = B \ln A $.)
$$ = 16 \ln 2 - 8 \ln 2 - 3 $$
$$ = 8 \ln 2 - 3 $$
Click HERE to return to the list of problems.