Answer
$$171$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {\left( {3{x^2} + 4y} \right)}dxdy;\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \leqslant x \leqslant 3,\,\,\,\,\,\,\,1 \leqslant y \leqslant 4 \cr
& {\text{replacing the limits for the region }}R \cr
& \int_1^4 {\int_0^3 {\left( {3{x^2} + 4y} \right)dx} dy} \cr
& = \int_1^4 {\left[ {\int_0^3 {\left( {3{x^2} + 4y} \right)dx} } \right]} dy \cr
& {\text{solve the inner integral }} \cr
& \int_0^3 {\left( {3{x^2} + 4y} \right)dx} = \left[ {{x^3} + 4xy} \right]_0^3 \cr
& {\text{evaluating the limits in the variable }}x \cr
& = \left[ {{{\left( 3 \right)}^3} + 4\left( 3 \right)y} \right] - \left[ {{{\left( 0 \right)}^3} + 4\left( 0 \right)y} \right] \cr
& {\text{simplifying}} \cr
& = 27 + 12y \cr
& \cr
& \int_1^4 {\left[ {\int_0^3 {\left( {3{x^2} + 4y} \right)dx} } \right]} dy = \int_1^4 {\left( {27 + 12y} \right)} dy \cr
& {\text{integrating}} \cr
& = \left( {27y + 6{y^2}} \right)_1^4 \cr
& {\text{evaluate}} \cr
& = \left( {27\left( 4 \right) + 6{{\left( 4 \right)}^2}} \right) - \left( {27\left( 1 \right) + 6{{\left( 1 \right)}^2}} \right) \cr
& {\text{simplifying}} \cr
& = 204 - 33 \cr
& = 171 \cr} $$
