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 $\ln A^B = B \ln A$.) $$= 16 \ln 2 - 8 \ln 2 - 3$$ $$= 8 \ln 2 - 3$$