Answer
$$2$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = x \cr
& {\text{From the image of the rectangle shown below we obtain}} \cr
& {\text{the region }}R \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant x \leqslant 4,{\text{ }}0 \leqslant y \leqslant 2} \right\} \cr
& {\text{The area }}A{\text{ of the region is:}} \cr
& A = \left( {4 - 0} \right)\left( {2 - 0} \right) \cr
& A = 8 \cr
& {\text{The average Value of a Function Over a Region }}R{\text{ is}} \cr
& {\text{Average value}} = \frac{1}{A}\iint\limits_R {f\left( {x,y} \right)}dA,{\text{ then}} \cr
& {\text{Average value}} = \frac{1}{8}\int_0^4 {\int_0^2 x dydx} \cr
& = \frac{1}{8}\int_0^4 {\left[ {\int_0^2 x dy} \right]dx} \cr
& {\text{Integrate with respect to }}y \cr
& = \frac{1}{8}\int_0^4 {\left[ {xy} \right]_0^2dx} \cr
& = \frac{1}{8}\int_0^4 {2xdx} \cr
& {\text{Integrate}} \cr
& = \frac{1}{8}\left[ {{x^2}} \right]_0^4 \cr
& = \frac{{{4^2}}}{8} \cr
& = 2 \cr} $$
