Answer
$A = 4\ln 15 - 7$
Work Step by Step
$$\eqalign{
& y = 4 - 2\cosh x,{\text{ }}y = \frac{1}{2}\sinh x \cr
& {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr
& \cr
& {\text{Find the intersection points }} \cr
& y = y \cr
& 4 - 2\cosh x = \frac{1}{2}\sinh x \cr
& {\text{Using the definition of the Hyperbolic Functions}} \cr
& 4 - 2\left( {\frac{{{e^x} + {e^{ - x}}}}{2}} \right) = \frac{1}{2}\left( {\frac{{{e^x} - {e^{ - x}}}}{2}} \right) \cr
& 4 - \left( {{e^x} + {e^{ - x}}} \right) = \frac{{{e^x} - {e^{ - x}}}}{4} \cr
& 16 - 4{e^x} - 4{e^{ - x}} = {e^x} - {e^{ - x}} \cr
& 16 - 5{e^x} - 3{e^{ - x}} = 0 \cr
& {\text{Multiplying the equation by }}{e^x} \cr
& 16{e^x} - 5{e^{2x}} - 3 = 0 \cr
& 5{e^{2x}} - 16{e^x} + 3 = 0 \cr
& {\text{Factoring}} \cr
& \left( {{e^x} - 3} \right)\left( {5{e^x} - 1} \right) = 0 \cr
& {\text{Zero - factor property}} \cr
& {e^x} - 3 = 0,{\text{ }}5{e^x} - 1 = 0 \cr
& {e^x} - 3 = 0,{\text{ }}5{e^x} = 1 \cr
& {e^x} = 3,{\text{ }}{e^x} = \frac{1}{5} \cr
& x = \ln 3,{\text{ }}x = \ln \left( {\frac{1}{5}} \right) \cr
& {\text{We have the interval }}\left[ {\ln \left( {\frac{1}{5}} \right),\ln 3} \right] \cr
& \cr
& {\text{We can find the area integrating with respect to }}x \cr
& A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx{\text{ }}\left( {\bf{1}} \right){\text{ }}\left( {{\text{see page 439}}} \right) \cr
& {\text{From the graph }} \cr
& 4 - 2\cosh x \geqslant \frac{1}{2}\sinh x{\text{ for the interval }}\left[ {\ln \left( {\frac{1}{5}} \right),\ln 3} \right] \cr
& {\text{Therefore}} \cr
& A = \int_{\ln \left( {1/5} \right)}^{\ln 3} {\left( {4 - 2\cosh x - \frac{1}{2}\sinh x} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {4x - 2\sinh x - \frac{1}{2}\cosh x} \right]_{\ln \left( {1/5} \right)}^{\ln 3} \cr
& {\text{Evaluate the limits}} \cr
& A = \left[ {4\left( {\ln 3} \right) - 2\sinh \left( {\ln 3} \right) - \frac{1}{2}\cosh \left( {\ln 3} \right)} \right] \cr
& - \left[ {4\ln \left( {\frac{1}{5}} \right) - 2\sinh \left( {\ln \frac{1}{5}} \right) - \frac{1}{2}\cosh \ln \left( {\ln \frac{1}{5}} \right)} \right] \cr
& {\text{Simplifying}} \cr
& = \left[ {4\left( {\ln 3} \right) - \frac{8}{3} - \frac{5}{6}} \right] - \left[ {4\ln \left( {\frac{1}{5}} \right) + \frac{{24}}{5} - \frac{{13}}{{10}}} \right] \cr
& A = 4\left( {\ln 3} \right) - \frac{7}{2} - 4\ln \left( {\frac{1}{5}} \right) - \frac{7}{2} \cr
& A = 4\left( {\ln 3 - \ln \frac{1}{5}} \right) - \frac{{14}}{2} \cr
& A = 4\ln 15 - 7 \cr} $$
