Answer
$$e - 2$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {2{x^3}{e^{{x^2}y}}}dxdy;\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \leqslant x \leqslant 1,\,\,\,\,\,\,\,0 \leqslant y \leqslant 1 \cr
& or \cr
& \iint\limits_R {2{x^3}{e^{{x^2}y}}}dydx \cr
& \cr
& {\text{replacing the limits for the region }}R \cr
& = \int_0^1 {\int_0^1 {2{x^3}{e^{{x^2}y}}dy} dx} \cr
& = \int_0^1 {\left[ {\int_0^1 {2{x^3}{e^{{x^2}y}}dy} } \right]dx} \cr
& {\text{solve the inner integral by treating }}x{\text{ as a constant}} \cr
& = 2x\int_0^1 {{x^2}{e^{{x^2}y}}dy} \cr
& {\text{integrate by using the rule }}\int {{e^u}} du = {e^u} + C \cr
& = 2x\left[ {{e^{{x^2}y}}} \right]_0^1 \cr
& {\text{evaluating the limits in the variable }}y \cr
& = 2x\left[ {{e^{{x^2}\left( 1 \right)}} - {e^{{x^2}\left( 0 \right)}}} \right] \cr
& {\text{simplifying}} \cr
& = 2x\left( {{e^{{x^2}}} - 1} \right) \cr
& = 2x{e^{{x^2}}} - 2x \cr
& \cr
& \int_0^1 {\left[ {\int_0^1 {2{x^3}{e^{{x^2}y}}dy} } \right]dx} = \int_0^1 {\left( {2x{e^{{x^2}}} - 2x} \right)dx} \cr
& {\text{integrating}} \cr
& = \left[ {{e^{{x^2}}} - {x^2}} \right]_0^1 \cr
& {\text{evaluate}} \cr
& = \left( {{e^{{1^2}}} - {1^2}} \right) - \left( {{e^{{0^2}}} - {0^2}} \right) \cr
& {\text{simplifying}} \cr
& = e - 1 - 1 \cr
& = e - 2 \cr} $$
