Now evaluate the iterated integral.

$\frac{1}{5}{\displaystyle \underset{0}{\overset{4}{}}}{\displaystyle \underset{0}{\overset{2}{}}}{x}^{2}ydydx$

First integrate with respect to $y.$

$\frac{1}{5}{\displaystyle \underset{0}{\overset{4}{}}}{\displaystyle \underset{0}{\overset{2}{}}}{x}^{2}ydydx=\frac{1}{5}{\displaystyle \underset{0}{\overset{4}{}}}[\frac{x^2}{2}{]}_0^{2}dx$

$=\frac{1}{5}{\displaystyle \underset{0}{\overset{4}{}}}[2x^2]dx$

$=\frac{2}{5}{\displaystyle \underset{0}{\overset{4}{}}}[x^2]dx$

Next, integrate with respect to $x.($Round your final answer to one decimal place.$)$

$\frac{2}{5}{\displaystyle \underset{0}{\overset{4}{}}}[x^2]dx=\frac{2}{5}[\frac{x^3}{3}{]}_0^4$

$=[\frac{15}{15}]$

$=$