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