Answer
$$\frac{{15}}{2}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 2xy \cr
& {\text{From the image of the rectangle shown below we obtain}} \cr
& {\text{the region }}R \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant x \leqslant 5,{\text{ }}0 \leqslant y \leqslant 3} \right\} \cr
& {\text{The area }}A{\text{ of the region is:}} \cr
& A = \left( {5 - 0} \right)\left( {3 - 0} \right) \cr
& A = 15 \cr
& {\text{The average Value of a Function Over a Region }}R{\text{ is}} \cr
& {\text{Average value}} = \frac{1}{A}\iint\limits_R {f\left( {x,y} \right)}dA,{\text{ then}} \cr
& {\text{Average value}} = \frac{1}{{15}}\int_0^5 {\int_0^3 {2xy} dydx} \cr
& = \frac{1}{{15}}\int_0^5 {\left[ {\int_0^3 {2xy} dy} \right]dx} \cr
& {\text{Integrate with respect to }}y \cr
& = \frac{1}{{15}}\int_0^5 {\left[ {x{y^2}} \right]_0^3dx} \cr
& = \frac{1}{{15}}\int_0^5 {\left[ {x{{\left( 3 \right)}^2} - x{{\left( 0 \right)}^2}} \right]dx} . \cr
& = \frac{3}{5}\int_0^5 {xdx} \cr
& {\text{Integrate}} \cr
& = \frac{3}{{10}}\left[ {{x^2}} \right]_0^5 \cr
& = \frac{3}{{10}}\left[ {{{\left( 5 \right)}^2} - {{\left( 0 \right)}^2}} \right] \cr
& = \frac{{15}}{2} \cr} $$
