Answer
$$568$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {\left( {2x + 6y} \right)}dydx;\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \leqslant x \leqslant 4,\,\,\,\,\,\,\,2 \leqslant y \leqslant 3x \cr
& {\text{replacing the limits for the region }}R \cr
& \int_2^4 {\int_2^{3x} {\left( {2x + 6y} \right)dy} dx} \cr
& = \int_2^4 {\left[ {\int_2^{3x} {\left( {2x + 6y} \right)dy} } \right]dx} \cr
& {\text{solve the inner integral, treating }}x{\text{ as a constant}} \cr
& = \int_2^{3x} {\left( {2x + 6y} \right)dy} \cr
& {\text{integrate using the power rule}} \cr
& = \left[ {2xy + 3{y^2}} \right]_2^{3x} \cr
& {\text{evaluating the limits in the variable }}y \cr
& = \left[ {2x\left( {3x} \right) + 3{{\left( {3x} \right)}^2}} \right] - \left[ {2x\left( 2 \right) + 3{{\left( 2 \right)}^2}} \right] \cr
& \left( {6{x^2} + 27{x^2}} \right) - \left( {4x + 12} \right) \cr
& = 33{x^2} - 4x - 12 \cr
& {\text{then}} \cr
& \cr
& \int_2^4 {\left[ {\int_2^{3x} {\left( {2x + 6y} \right)dy} } \right]dx} = \int_2^4 {\left( {33{x^2} - 4x - 12} \right)dx} \cr
& {\text{integrating}} \cr
& = \left( {11{x^3} - 2{x^2} - 12x} \right)_2^4 \cr
& = \left( {11{{\left( 4 \right)}^3} - 2{{\left( 4 \right)}^2} - 12\left( 4 \right)} \right) - \left( {11{{\left( 2 \right)}^3} - 2{{\left( 2 \right)}^2} - 12\left( 2 \right)} \right) \cr
& {\text{simplifying}} \cr
& = 624 - 56 \cr
& = 568 \cr} $$
