Answer
$$A = 2\ln 2 + \frac{3}{2} - \ln 6$$
Work Step by Step
$$\eqalign{
& x = 1,\,\,\,\,\,x = 6,\,\,\,\,\,y = \frac{1}{x},\,\,\,\,\,\,y = \frac{1}{2} \cr
& {\text{Find the points of intersection of the graphs}} \cr
& \frac{1}{x} = \frac{1}{2} \cr
& x = 2 \cr
& {\text{Then with the given points and the critical value }}x = 2{\text{ we have the intervals}} \cr
& \left( {1,2} \right){\text{ and }}\left( {2,6} \right) \cr
& {\text{For the interval }}\left( {1,2} \right){\text{ we have that }}\frac{1}{x} \geqslant \frac{1}{2}{\text{ and for the interval }}\left( {2,6} \right) \cr
& \frac{1}{2} \geqslant \frac{1}{x} \cr
& {\text{The area is given by}}: \cr
& A = \int_1^2 {\left( {\frac{1}{x} - \frac{1}{2}} \right)} dx + \int_2^6 {\left( {\frac{1}{2} - \frac{1}{x}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left( {\ln \left| x \right| - \frac{1}{2}x} \right)_1^2 + \left( {\frac{1}{2}x - \ln \left| x \right|} \right)_2^6 \cr
& A = \left( {\ln \left| 2 \right| - \frac{1}{2}\left( 2 \right)} \right) - \left( {\ln \left| 1 \right| - \frac{1}{2}\left( 1 \right)} \right) + \left( {\frac{1}{2}\left( 6 \right) - \ln \left| 6 \right|} \right) - \left( {\frac{1}{2}\left( 2 \right) - \ln \left| 2 \right|} \right) \cr
& A = \ln 2 - 1 + \frac{1}{2} + 3 - \ln 6 - 1 + \ln 2 \cr
& A = 2\ln 2 + \frac{3}{2} - \ln 6 \cr} $$
