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.

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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 $.) $$ = 16 \ln 2 - 8 \ln 2 - 3 $$ $$ = 8 \ln 2 - 3 $$

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