Answer
$${e^2}$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\int_0^1 {xy{e^{x + y}}} dxdy} \cr
& {\text{Recall that }}{e^{a + b}} = {e^a}{e^b} \cr
& \int_1^2 {\int_0^1 {xy{e^x}{e^y}} dxdy} \cr
& \int_1^2 {\left[ {\int_0^1 {xy{e^x}{e^y}} dx} \right]dy} \cr
& \int_1^2 {y{e^y}\left[ {\int_0^1 {x{e^x}} dx} \right]dy} \cr
& {\text{Evaluate inner integral}} \cr
& \int_0^1 {x{e^x}} dx = \left[ {x{e^x} - {e^x}} \right]_{x = 0}^{x = 1} \cr
& = \left( {{e^1} - {e^1}} \right) - \left( {0{e^0} - {e^0}} \right) \cr
& = 1 \cr
& {\text{Therefore,}} \cr
& \int_1^2 {y{e^y}\left[ {\int_0^1 {x{e^x}} dx} \right]dy} = \int_1^2 {y{e^y}\left( 1 \right)dy} \cr
& \int_1^2 {y{e^y}dy} \cr
& {\text{Integrating}} \cr
& {\text{ = }}\left[ {y{e^y} - {e^y}} \right]_1^2 \cr
& = \left( {2{e^2} - {e^2}} \right) - \left( {{e^1} - {e^1}} \right) \cr
& = {e^2} \cr} $$
