SOLUTION 3: Compute the area of the region enclosed by the graphs of the equations $y=e^x$, $y=e^{-x}$, and $x=\ln 3$ . Begin by finding the points of intersection of the two graphs. From $y=e^{x}$ and $y=e^{-x}$ we get that $$e^{x} = e^{-x} \ \ \longrightarrow$$ $$e^{x} = 1/e^{x} \ \ \longrightarrow$$ $$e^{2x} = 1 \ \ \longrightarrow$$ $$2x = 0$$ $$x = 0$$ From $y=e^{x}$ and $x= \ln 3$ we get that (Recall that $e^{\ln A}=A$.) $$y=e^{\ln 3} \ \ \longrightarrow \ \ y=3$$ From $y=e^{-x}$ and $x= \ln 3$ we get that (Recall that $B \ln A= \ln A^B$.) $$y=e^{-\ln 3} = e^{\ln 3^{-1}} \ \ \longrightarrow \ \ y=3^{-1}= \frac{1}{3}$$ Now see the given graph of the enclosed region.

Using vertical cross-sections to describe this region, we get that $$0 \le x \le \ln 3 \ \ and \ \ e^{-x} \le y \le e^{x} ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{0}^{\ln 3} (Top \ - \ Bottom) \ dx }$$ $$= \displaystyle { \int_{0}^{\ln 3} (e^{x} - e^{-x}) \ dx }$$ $$= \displaystyle { \Big(e^{x} - (-e^{-x})\Big) \Big\vert_{0}^{\ln 3} }$$ $$= \displaystyle { \Big(e^{x} + e^{-x}\Big) \Big\vert_{0}^{\ln 3} }$$ $$= \displaystyle { \Big(e^{\ln 3} + e^{-\ln 3}\Big) - \Big(e^{0} + e^{-0}\Big) }$$ $$= \displaystyle { \Big(e^{\ln 3} + e^{\ln 3^{-1}}\Big) - \Big(1 + 1 \Big) }$$ $$= \displaystyle { \Big(3 + \frac{1}{3}\Big) - 2 }$$ $$= \displaystyle { \Big(\frac{10}{3}\Big) - 2 }$$ $$= \displaystyle { \frac{10}{3} - \frac{6}{3} }$$ $$= \displaystyle { \frac{4}{3} }$$