Answer
\[\frac{2}{3}\]
Work Step by Step
\[\begin{align}
& \text{From the graph we can define the region }R\text{ as:} \\
& R=\left\{ \left( x,y \right):0\le y\le -x+2,\text{ }0\le x\le 2\text{ } \right\} \\
& \text{The area of the region is }A=\frac{1}{2}\left( 2 \right)\left( 2 \right)=2 \\
& \text{The average of }f(x,y)\text{ on the region }R\text{ is} \\
& \overline{f}=\frac{1}{A}\iint_{R}{f\left( x,y \right)}dydx \\
& \overline{f}=\frac{1}{2}\int_{0}^{2}{\int_{0}^{2-x}{y}dy}dx \\
& \text{Integrate} \\
& \overline{f}=\frac{1}{2}\int_{0}^{2}{\left[ \frac{1}{2}{{y}^{2}} \right]_{0}^{2-x}}dx \\
& \overline{f}=\frac{1}{4}\int_{0}^{2}{{{\left( 2-x \right)}^{2}}}dx \\
& \overline{f}=-\frac{1}{4}\left[ \frac{{{\left( 2-x \right)}^{3}}}{3} \right]_{0}^{2} \\
& \overline{f}=-\frac{1}{4}\left[ \frac{{{\left( 2-2 \right)}^{3}}}{3}-\frac{{{\left( 2-0 \right)}^{3}}}{3} \right] \\
& \overline{f}=\frac{2}{3} \\
\end{align}\]
