Answer
$$\ \iint \limits_{R} y e^{-x y} d A =\frac{1}{2e^{6}} + \frac{5}{2} $$
Work Step by Step
Given$$\ \iint \limits_{R} y e^{-x y} d A, \quad R=[0,2] \times[0,3]$$
So, we get \begin{aligned} I&= \int_{0}^{3} \int_{0}^{2} y e^{-x y} dx \ dy \\
&=-\int_{0}^{3} \left[ e^{-x y}\right]_{0}^{2} dy \\
&=-\int_{0}^{3} e^{-2 y} dy +\int_{0}^{3} dy \\
&=-\int_{0}^{3} e^{-2 y} dy + y|_{0}^{3} \\
&=\frac{1}{2} e^{-2 y}|_{0}^{3} +3 -0 \\
&=\frac{1}{2} e^{-6} - \frac{1}{2} + 3 \\
&=\frac{1}{2e^{6}} + \frac{5}{2} \\
\end{aligned}
