Answer
$$\frac{1}{9}{e^8} - 1$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_0^1 {{x^5}{y^2}{e^{{x^3}{y^3}}}dy} dx} \cr
& {\text{Rewrite}} \cr
& = \int_0^2 {\frac{1}{3}{x^2}\left[ {\int_0^1 {3{x^3}{y^2}{e^{{x^3}{y^3}}}} dy} \right]} dx \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^2 {\frac{1}{3}{x^2}\left[ {{e^{{x^3}{y^3}}}} \right]} _0^1dx \cr
& {\text{Evaluate the limits}} \cr
& = \int_0^2 {\frac{1}{3}{x^2}\left[ {{e^{{x^3}{{\left( 1 \right)}^3}}} - {e^{{x^3}{{\left( 0 \right)}^3}}}} \right]} dx \cr
& = \int_0^2 {\frac{1}{3}{x^2}\left[ {{e^{{x^3}}} - 1} \right]} dx \cr
& = \int_0^2 {\left[ {\frac{1}{3}{x^2}{e^{{x^3}}} - \frac{1}{3}{x^2}} \right]} dx \cr
& = \frac{1}{9}\int_0^2 {3{x^2}{e^{{x^3}}}} dx - \frac{1}{3}\int_0^2 {{x^2}} dx \cr
& {\text{Integrating}} \cr
& = \left[ {\frac{1}{9}{e^{{x^3}}} - \frac{1}{9}{x^3}} \right]_0^2 \cr
& = \left[ {\frac{1}{9}{e^{{{\left( 2 \right)}^3}}} - \frac{1}{9}{{\left( 2 \right)}^3}} \right] - \left[ {\frac{1}{9}{e^{{{\left( 0 \right)}^3}}} - \frac{1}{9}{{\left( 0 \right)}^3}} \right] \cr
& {\text{Simplifying}} \cr
& = \frac{1}{9}{e^8} - \frac{8}{9} - \frac{1}{9} \cr
& = \frac{1}{9}{e^8} - 1 \cr} $$
