Answer
$$\int_{0}^{\ln 3} \int_{0}^{\ln 2} e^{x+y} d y d x=2$$
Work Step by Step
Given $$\int_{0}^{\ln 3} \int_{0}^{\ln 2} e^{x+y} d y d x$$
So, we get
\begin{aligned} I& = \int_{0}^{\ln 3} \int_{0}^{\ln 2} e^{x} \cdot e^{y} d y d x \\&= \int_{0}^{\ln 3} e^{x} d x \int_{0}^{\ln 2} e^{y} d y \\
& = \left[e^{x}\right]_{0}^{\ln 3}\left[e^{y}\right]_{0}^{\ln 2} \\&= \left[e^{\ln 3}-e^{0}\right]\left[e^{\ln 2}-e^{0}\right] \\
&=[3-1][2-1]=2 \end{aligned}
