## Calculus with Applications (10th Edition)

$$A = 2\ln 2 - \ln 3 + \frac{1}{4}$$
\eqalign{ & x = 2,\,\,\,\,\,x = 4,\,\,\,\,\,y = \frac{{x - 1}}{4},\,\,\,\,\,\,y = \frac{1}{{x - 1}} \cr & {\text{Find the points of intersection of the graphs}} \cr & \frac{{x - 1}}{4} = \frac{1}{{x - 1}} \cr & {\text{cross product}} \cr & {\left( {x - 1} \right)^2} = 4 \cr & x - 1 = \pm 2 \cr & x = 3,\,\,\,x = - 1 \cr & {\text{Then with the given points and the critical value }}x = 3{\text{ we have the intervals}} \cr & \left( {2,3} \right){\text{ and }}\left( {3,4} \right) \cr & {\text{For the interval }}\left( {2,3} \right){\text{ we have that }}\frac{1}{{x - 1}} > \frac{{x - 1}}{4}{\text{ }} \cr & {\text{and }} \cr & {\text{For the interval }}\left( {3,4} \right){\text{ we have that }}\frac{{x - 1}}{4} > \frac{1}{{x - 1}}{\text{ }} \cr & \cr & {\text{The area is given by}}: \cr & A = \int_2^3 {\left( {\frac{1}{{x - 1}} - \frac{{x - 1}}{4}} \right)} dx + \int_3^4 {\left( {\frac{{x - 1}}{4} - \frac{1}{{x - 1}}} \right)} dx \cr & {\text{Integrating}} \cr & A = \left( {\ln \left| {x - 1} \right| - \frac{{{{\left( {x - 1} \right)}^2}}}{8}} \right)_2^3 + \left( {\frac{{{{\left( {x - 1} \right)}^2}}}{8} - \ln \left| {x - 1} \right|} \right)_3^4 \cr & A = \left( {\ln \left| {3 - 1} \right| - \frac{{{{\left( {3 - 1} \right)}^2}}}{8}} \right) - \left( {\ln \left| {2 - 1} \right| - \frac{{{{\left( {2 - 1} \right)}^2}}}{8}} \right) + \left( {\frac{{{{\left( {4 - 1} \right)}^2}}}{8} - \ln \left| {4 - 1} \right|} \right) \cr & - \left( {\frac{{{{\left( {3 - 1} \right)}^2}}}{8} - \ln \left| {3 - 1} \right|} \right) \cr & \cr & A = \left( {\ln 2 - \frac{1}{2}} \right) - \left( { - \frac{1}{8}} \right) + \left( {\frac{9}{8} - \ln 3} \right) - \left( {\frac{1}{2} - \ln \left| 2 \right|} \right) \cr & A = \ln 2 - \frac{1}{2} + \frac{1}{8} + \frac{9}{8} - \ln 3 - \frac{1}{2} + \ln 2 \cr & A = 2\ln 2 - \ln 3 + \frac{1}{4} \cr}