Answer
$$A = {e^2} + 1$$
Work Step by Step
$$\eqalign{
& y = {e^2},{\text{ }}y = {e^x} \cr
& {e^2} > {e^x}{\text{ on the interval }}\left( {0,2} \right) \cr
& {\text{From the graph and using symmetry properties, we obtain }} \cr
& A = \int_0^2 {\left( {{e^2} - {e^x}} \right)} dx \cr
& {\text{Integrate and evaluate}} \cr
& A = \left[ {{e^2}x - {e^x}} \right]_0^2 \cr
& A = \left[ {{e^2}\left( 2 \right) - {e^2}} \right] - \left[ {{e^2}\left( 0 \right) - {e^0}} \right] \cr
& A = \left( {2{e^2} - {e^2}} \right) - \left( {0 - 1} \right) \cr
& A = {e^2} + 1 \cr} $$
