Calculus (3rd Edition)

$$e^{2} +1.$$
First, we find the intersections between the two curves by putting $$e^2=e^x\Longrightarrow x=2.$$ Now, the area is given by \begin{align*} Area\ &=\int_0^2e^{2}-e^{x}dx \\ &=( xe^{2}-e^{x})_0^2\\ &=2e^{2} -e^{2} -(0-1)\\ &=e^{2} +1. \end{align*}