Answer
$$88$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {\left( {{x^2} + 4{y^3}} \right)}dydx;\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \leqslant x \leqslant 2,\,\,\,\,\,\,\,0 \leqslant y \leqslant 3 \cr
& {\text{replacing the limits for the region }}R \cr
& \int_1^2 {\int_0^3 {\left( {{x^2} + 4{y^3}} \right)dy} dx} \cr
& = \int_1^2 {\left[ {\int_0^3 {\left( {{x^2} + 4{y^3}} \right)dy} } \right]dx} \cr
& {\text{solve the inner integral treat }}x{\text{ as a constant}} \cr
& = \int_0^3 {\left( {{x^2} + 4{y^3}} \right)dy} \cr
& {\text{integrate using the power rule}} \cr
& = \left[ {{x^2}y + {y^4}} \right]_0^3 \cr
& {\text{evaluating the limits in the variable }}y \cr
& = \left[ {{x^2}\left( 3 \right) + {{\left( 3 \right)}^4}} \right] - \left[ {{x^2}\left( 0 \right) + {{\left( 0 \right)}^4}} \right] \cr
& = 3{x^2} + 81 \cr
& {\text{then}} \cr
& \cr
& \int_1^2 {\left[ {\int_0^3 {\left( {{x^2} + 4{y^3}} \right)dy} } \right]dx} = \int_1^2 {\left( {3{x^2} + 81} \right)dx} \cr
& {\text{integrating}} \cr
& = \left( {{x^3} + 81x} \right)_1^2 \cr
& = \left( {{{\left( 2 \right)}^3} + 81\left( 2 \right)} \right) - \left( {{{\left( 1 \right)}^3} + 81\left( 1 \right)} \right) \cr
& {\text{simplifying}} \cr
& = 170 - 82 \cr
& = 88 \cr} $$
