Answer
$${e^4} - 13$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {x{e^y}dA} \cr
& {\text{Let the region }}R :y = 4 - x,{\text{ }}y = 0,{\text{ }}x = 0 \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant y \leqslant 4 - x,{\text{ }}0 \leqslant x \leqslant 4} \right\} \cr
& {\text{Therefore,}} \cr
& \iint\limits_R {x{e^y}dA} = \int_0^4 {\int_0^{4 - x} {x{e^y}dy} dx} \cr
& = \int_0^4 {x\left[ {\int_0^{4 - x} {{e^y}dy} } \right]dx} \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^4 {x\left[ {{e^y}} \right]_0^{4 - x}dx} \cr
& = \int_0^4 {x\left( {{e^{4 - x}} - {e^0}} \right)dx} \cr
& = \int_0^4 {\left( {x{e^{4 - x}} - x} \right)dx} \cr
& {\text{Integrating}} \cr
& = \left[ { - x{e^{4 - x}} - {e^{4 - x}} - \frac{1}{2}{x^2}} \right]_0^4 \cr
& = \left[ { - \left( 4 \right){e^{4 - 4}} - {e^{4 - 4}} - \frac{1}{2}{{\left( 4 \right)}^2}} \right] - \left[ { - \left( 0 \right){e^{4 - 0}} - {e^{4 - 0}} - \frac{1}{2}{{\left( 0 \right)}^2}} \right] \cr
& = \left( { - 4 - 1 - 8} \right) - \left[ {0 - {e^4} - 0} \right] \cr
& = {e^4} - 13 \cr} $$
