Answer
$$A = \frac{1}{e} + e - 2$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{x} - \frac{1}{e};{\text{ }}\left[ {1,{e^2}} \right] \cr
& {\text{First check if the graph crosses the }}x - {\text{axis in the given interval}} \cr
& \frac{1}{x} - \frac{1}{e} = 0 \cr
& {\text{solve for }}x \cr
& \frac{1}{x} = \frac{1}{e} \cr
& x = e \cr
& {\text{the graph crosses the }}x{\text{ - axis at }}x = e{\text{ and the given interval }}\left[ {1,{e^2}} \right],{\text{ }} \cr
& f\left( 1 \right) = \frac{1}{1} - \frac{1}{e} > 0,{\text{ }}f\left( x \right){\text{ is positive for }}\left( {1,e} \right) \cr
& {\text{then}} \cr
& {\text{the definite integral of the total area will be}}: \cr
& A = \int_1^e {\left( {\frac{1}{x} - \frac{1}{e}} \right)} dx + \left| {\int_e^{{e^2}} {\left( {\frac{1}{x} - \frac{1}{e}} \right)} dx} \right| \cr
& A = \left( {\ln \left| x \right| - \frac{x}{e}} \right)_1^e + \left| {\left( {\ln \left| x \right| - \frac{x}{e}} \right)_e^{{e^2}}} \right| \cr
& A = \left( {\ln \left| e \right| - \frac{e}{e}} \right) - \left( {\ln \left| 1 \right| - \frac{1}{e}} \right) + \left| {\left( {\ln \left| {{e^2}} \right| - \frac{{{e^2}}}{e}} \right) - \left( {\ln \left| e \right| - \frac{e}{e}} \right)} \right| \cr
& {\text{simplifying}} \cr
& A = \left( {1 - 1} \right) - \left( {0 - \frac{1}{e}} \right) + \left| {2 - e - 1 + 1} \right| \cr
& A = \frac{1}{e} + e - 2 \cr} $$
