Answer
$${e^{16}} - 17$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {6{x^5}{e^{{x^3}y}}}dA;\,\,\,\,\,\,\,R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant 2,\,\,\,\,0 \leqslant y \leqslant 2} \right\} \cr
& {\text{Convert to an iterated integral substituting the region }}R \cr
& = \int_0^2 {\int_0^2 {6{x^5}{e^{{x^3}y}}} } dydx \cr
& = \int_0^2 {\left[ {\int_0^2 {6{x^5}{e^{{x^3}y}}} dy} \right]} dx \cr
& {\text{solve the inner integral}}{\text{, treat }}x{\text{ as a constant}} \cr
& = 6{x^5}\int_0^2 {{e^{{x^3}y}}} dy \cr
& {\text{integrating}} \cr
& = 6{x^5}\left( {\frac{{{e^{{x^3}y}}}}{{{x^3}}}} \right)_0^2 \cr
& {\text{evaluating the limits for the variable }}y \cr
& = 6{x^2}\left( {{e^{{x^3}y}}} \right)_0^2 \cr
& = 6{x^2}\left( {{e^{{x^3}\left( 2 \right)}} - {e^{{x^3}\left( 0 \right)}}} \right) \cr
& = 6{x^2}\left( {{e^{2{x^3}}} - 1} \right) \cr
& = 6{x^2}{e^{2{x^3}}} - 6{x^2} \cr
& \int_0^2 {\left[ {\int_0^2 {6{x^5}{e^{{x^3}y}}} dy} \right]} dx = \int_0^2 {\left( {6{x^2}{e^{2{x^3}}} - 6{x^2}} \right)} dx \cr
& {\text{integrating}}{\text{, }} \cr
& = \left( {{e^{2{x^3}}} - 2{x^3}} \right)_0^2 \cr
& {\text{evaluate}} \cr
& = \left( {{e^{2{{\left( 2 \right)}^3}}} - 2{{\left( 2 \right)}^3}} \right) - \left( {{e^{2{{\left( 0 \right)}^3}}} - 2{{\left( 0 \right)}^3}} \right) \cr
& = {e^{16}} - 16 - 1 \cr
& = {e^{16}} - 17 \cr} $$
