Answer
$${\left( {e - 1} \right)^2}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {e^{x + y}} \cr
& {\text{From the image of the rectangle shown below we obtain}} \cr
& {\text{the region }}R \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant x \leqslant 1,{\text{ }}x \leqslant y \leqslant 1} \right\} \cr
& {\text{The area }}A{\text{ of the region is:}} \cr
& A = \frac{1}{2}\left( 1 \right)\left( 1 \right) \cr
& A = \frac{1}{2} \cr
& {\text{The average Value of a Function Over a Region }}R{\text{ is}} \cr
& {\text{Average value}} = \frac{1}{A}\iint\limits_R {f\left( {x,y} \right)}dA,{\text{ then}} \cr
& {\text{Average value}} = \frac{1}{{1/2}}\int_0^1 {\int_x^1 {{e^{x + y}}} dydx} \cr
& = 2\int_0^1 {\left[ {\int_x^1 {{e^{x + y}}} dy} \right]dx} \cr
& {\text{Integrate with respect to }}y \cr
& = 2\int_0^1 {\left[ {{e^{x + y}}} \right]_x^1dx} \cr
& = 2\int_0^1 {\left( {{e^{x + 1}} - {e^{2x}}} \right)dx} \cr
& {\text{Integrate}} \cr
& = 2\left[ {{e^{x + 1}} - \frac{1}{2}{e^{2x}}} \right]_0^1 \cr
& = 2\left[ {{e^{1 + 1}} - \frac{1}{2}{e^2}} \right] - 2\left[ {{e^{0 + 1}} - \frac{1}{2}{e^0}} \right] \cr
& = 2{e^2} - {e^2} - 2\left( {e - \frac{1}{2}} \right) \cr
& = {e^2} - 2e + 1 \cr
& = {\left( {e - 1} \right)^2} \cr} $$