Answer
$\displaystyle \frac{e^{-6}+5}{2}$
Work Step by Step
$\displaystyle \int\int_{R}ye^{-xy}dA=\int_{0}^{3}\int_{0}^{2}ye^{-xy}dxdy=\int_{0}^{3}[\int_{0}^{2}ye^{-xy}dx]dy$
$\left[\begin{array}{l}
\text{... treat y as a constant in the inner integral} \\
\displaystyle\int_{0}^{2}ye^{-xy}dx=y\cdot(-\frac{1}{y}[e^{-xy}]_{x=0}^{x=2})\\ \\
=[-e^{-xy}]_{x=0}^{x=2}\\ \\
=-e^{-2y}+1
\end{array}\right]$
$=\displaystyle \int_{0}^{3}(-e^{-2y}+1)dy$
$=[\displaystyle \frac{1}{2}e^{-2y}+y]_{0}^{3}$
$=\displaystyle \frac{1}{2}e^{-6}+3-(\frac{1}{2}+0)$
$=\displaystyle \frac{e^{-6}+5}{2}$
