Answer
$\displaystyle \frac{(e^{3}-1)^{2}}{3}$
Work Step by Step
$e^{x+3y}=e^{x}\cdot e^{3y}$
Let $g(x)=e^{x}, \quad h(y)=e^{3y}.$
Apply formula (5), $\displaystyle \iint_{R}g(x)\cdot h(y)dA=\int_{a}^{b}g(x)dx\cdot\int_{c}^{d}h(y)dy$
$\displaystyle \int_{0}^{3}e^{x}dx=[e^{x}]_{0}^{3}=e^{3}-1$
$\displaystyle \int_{0}^{1}e^{3y}dy=[\frac{e^{3y}}{3}]_{0}^{1}=\frac{e^{3}-1}{3}$
So,
$\displaystyle \int_{0}^{1}\int_{0}^{3}e^{x+3y}dxdy=\int_{0}^{3}e^{x}dx\int_{0}^{1}e^{3y}dy$
$= \displaystyle \frac{(e^{3}-1)^{2}}{3}$
