Answer
$$\left( {{e^3} - 1} \right)\left( {{e^2} - 1} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_0^3 {{e^{x + y}}} dxdy} \cr
& {\text{Recall that }}{e^a}{e^b} = {e^{a + b}} \cr
& \int_0^2 {\left[ {\int_0^3 {{e^x}{e^y}} dx} \right]dy} \cr
& \int_0^2 {{e^y}\left[ {\int_0^3 {{e^x}} dx} \right]dy} \cr
& {\text{Evaluate the inner integral}} \cr
& \int_0^3 {{e^x}} dx = \left[ {{e^x}} \right]_{x = 0}^{x = 3} \cr
& = {e^3} - {e^0} \cr
& = {e^3} - 1 \cr
& {\text{Therefore,}} \cr
& \int_0^2 {{e^y}\left[ {\int_0^3 {{e^x}} dx} \right]dy} = \int_0^2 {{e^y}\left( {{e^3} - 1} \right)dy} \cr
& = \left( {{e^3} - 1} \right)\int_0^2 {{e^y}dy} \cr
& {\text{Integrating}} \cr
& = \left( {{e^3} - 1} \right)\left[ {{e^y}} \right]_0^2 \cr
& = \left( {{e^3} - 1} \right)\left[ {{e^2} - {e^0}} \right] \cr
& = \left( {{e^3} - 1} \right)\left( {{e^2} - 1} \right) \cr} $$
