Answer
$$\frac{1}{6}\left( {{e^{14}} - {e^7} - {e^{10}} + {e^3}} \right)$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {{x^2}{e^{{x^3} + 2y}}}dxdy;\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \leqslant x \leqslant 2,\,\,\,\,\,\,\,1 \leqslant y \leqslant 3 \cr
& {\text{replacing the limits for the region }}R \cr
& \int_1^3 {\int_1^2 {{x^2}{e^{{x^3} + 2y}}dx} dy} \cr
& = \int_1^3 {\left[ {\int_1^2 {{x^2}{e^{{x^3} + 2y}}dx} } \right]dy} \cr
& {\text{solve the inner integral, treating }}y{\text{ as a constant}} \cr
& = \int_1^2 {{x^2}{e^{{x^3} + 2y}}dx} \cr
& = \frac{1}{3}\int_1^2 {{e^{{x^3} + 2y}}\left( {3{x^2}} \right)dx} \cr
& {\text{integrate by using the rule }}\int {{e^u}} du = {e^u} + C \cr
& = \frac{1}{3}\left[ {{e^{{x^3} + 2y}}} \right]_1^2 \cr
& {\text{evaluating the limits in the variable }}x \cr
& = \frac{1}{3}\left[ {{e^{{{\left( 2 \right)}^3} + 2y}} - {e^{{{\left( 1 \right)}^3} + 2y}}} \right] \cr
& {\text{simplifying}} \cr
& = \frac{1}{3}\left( {{e^{8 + 2y}} - {e^{1 + 2y}}} \right) \cr
& \cr
& \int_1^3 {\left[ {\int_1^2 {{x^2}{e^{{x^3} + 2y}}dx} } \right]dy} = \int_1^3 {\frac{1}{3}\left( {{e^{8 + 2y}} - {e^{1 + 2y}}} \right)dy} \cr
& = \frac{1}{{3\left( 2 \right)}}\int_1^3 {\left( {2{e^{8 + 2y}} - 2{e^{1 + 2y}}} \right)dy} \cr
& {\text{integrating}} \cr
& = \frac{1}{6}\left( {{e^{8 + 2y}} - {e^{1 + 2y}}} \right)_1^3 \cr
& {\text{evaluate}} \cr
& = \frac{1}{6}\left( {{e^{8 + 2\left( 3 \right)}} - {e^{1 + 2\left( 3 \right)}}} \right) - \frac{1}{6}\left( {{e^{8 + 2\left( 1 \right)}} - {e^{1 + 2\left( 1 \right)}}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{6}\left( {{e^{14}} - {e^7}} \right) - \frac{1}{2}\left( {{e^{10}} - {e^3}} \right) \cr
& = \frac{1}{6}\left( {{e^{14}} - {e^7} - {e^{10}} + {e^3}} \right) \cr} $$
