Answer
\[\overline{f}=2e-4\]
Work Step by Step
\[\begin{align}
& \text{From the graph we can define the region }R\text{ as:} \\
& R=\left\{ \left( x,y \right):y-1\le y\le 1-y,\text{ 0}\le y\le 1\text{ } \right\} \\
& \text{The area of the region is } \\
& A=2\int_{0}^{1}{\left( -x+1 \right)}dx \\
& A=1 \\
& \text{The average of }f(x,y)\text{ on the region }R\text{ is} \\
& \overline{f}=\frac{1}{A}\iint_{R}{f\left( x,y \right)}dydx \\
& \overline{f}=\frac{1}{1}\int_{0}^{1}{\int_{y-1}^{1-y}{{{e}^{y}}}dx}dy \\
& \text{Integrate} \\
& \overline{f}=\int_{0}^{1}{\left[ x{{e}^{y}} \right]_{y-1}^{1-y}}dy \\
& \overline{f}=\int_{0}^{1}{\left[ \left( 1-y \right){{e}^{y}}-\left( y-1 \right){{e}^{y}} \right]}dy \\
& \overline{f}=\int_{0}^{1}{\left( {{e}^{y}}-y{{e}^{y}}-y{{e}^{y}}+{{e}^{y}} \right)}dy \\
& \overline{f}=\int_{0}^{1}{\left( 2{{e}^{y}}-2y{{e}^{y}} \right)}dy \\
& \text{Integrating } \\
& \overline{f}=\left[ 2{{e}^{y}}-2y{{e}^{y}}+2{{e}^{y}} \right]_{0}^{1} \\
& \overline{f}=\left[ 4{{e}^{y}}-2y{{e}^{y}} \right]_{0}^{1} \\
& \overline{f}=\left( 4e-2e \right)-\left( 4{{e}^{0}}-0 \right) \\
& \overline{f}=2e-4 \\
\end{align}\]
