Answer
$$\bar f = 2$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( {x,y} \right) = 4 - x - y;{\text{ }}R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant 2,{\text{ 0}} \leqslant y \leqslant {\text{2}}} \right\} \cr
& {\text{The average value of an integrable function }}f{\text{ over a region }}R \cr
& {\text{is:}} \cr
& \bar f = \frac{1}{{{\text{area of }}R}}\iint\limits_R {f\left( {x,y} \right)}dA \cr
& {\text{The area of the region }}R{\text{ is:}} \cr
& {\text{Area of }}R = \int_0^2 {\int_0^2 {dx} dy} \cr
& {\text{Area of }}R = \left( 2 \right)\left( 2 \right) \cr
& {\text{Area of }}R = 4 \cr
& {\text{Therefore}}{\text{, }} \cr
& \bar f = \frac{1}{4}\iint\limits_R {f\left( {x,y} \right)}dA \cr
& \bar f = \frac{1}{4}\int_0^2 {\int_0^2 {\left( {4 - x - y} \right)dx} dy} \cr
& {\text{Integrate}} \cr
& \bar f = \frac{1}{4}\int_0^2 {\left[ {4x - \frac{1}{2}{x^2} - xy} \right]_0^2dy} \cr
& \bar f = \frac{1}{4}\int_0^2 {\left[ {4\left( 2 \right) - \frac{1}{2}{{\left( 2 \right)}^2} - 2y} \right]dy} \cr
& \bar f = \frac{1}{4}\int_0^2 {\left( {6 - 2y} \right)dy} \cr
& \bar f = \frac{1}{4}\left[ {6y - {y^2}} \right]_0^2 \cr
& \bar f = \frac{1}{4}\left[ {6\left( 2 \right) - {{\left( 2 \right)}^2}} \right] - \frac{1}{4}\left[ {6\left( 0 \right) - {{\left( 0 \right)}^2}} \right] \cr
& \bar f = 2 \cr} $$
