Final Answer

${\int }_0^4{\int }_{\sqrt{y}}^2{e}^{}{x}^3\hspace{0.17em}dx\hspace{0.17em}dy=\frac{1}{3}(e^8-1)\backslash int\_\{0\}^\{4\}\; \backslash int\_\{\backslash sqrt\{y\}\}^\{2\}\; e^\{x^3\}\; \backslash ,\; dx\; \backslash ,\; dy\; =\; \backslash frac\{1\}\{3\}\; (e^8\; –\; 1)$