Answer
$$e + \frac{1}{e} - 2$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_0^1 {{e^{x - y}}} dxdy} \cr
& {\text{Recall that }}{e^a}{e^b} = {e^{a + b}} \cr
& \int_0^1 {\int_0^1 {{e^x}{e^{ - y}}} dxdy} \cr
& \int_0^1 {{e^{ - y}}\left[ {\int_0^1 {{e^x}} dx} \right]dy} \cr
& {\text{Evaluate inner integral}} \cr
& \int_0^1 {{e^x}} dx = \left[ {{e^x}} \right]_{x = 0}^{x = 1} \cr
& = {e^1} - {e^0} \cr
& = e - 1 \cr
& {\text{Therefore,}} \cr
& \int_0^1 {{e^{ - y}}\left[ {\int_0^1 {{e^x}} dx} \right]dy} = \int_0^1 {{e^{ - y}}\left( {e - 1} \right)dy} \cr
& = \left( {e - 1} \right)\int_0^1 {{e^{ - y}}dy} \cr
& {\text{Integrating}} \cr
& = \left( {e - 1} \right)\left[ { - {e^{ - y}}} \right]_0^1 \cr
& = \left( {e - 1} \right)\left[ { - {e^{ - 1}} + {e^0}} \right] \cr
& = \left( {e - 1} \right)\left( { - \frac{1}{e} + 1} \right) \cr
& = - 1 + e + \frac{1}{e} - 1 \cr
& = e + \frac{1}{e} - 2 \cr} $$
