Answer
$$\int_0^{\ln{4}}\int_0^{\ln{3}}e^{x+y}dydx=6$$
Work Step by Step
We will start with the integral with respect to y:
$\int_0^{\ln{4}}\int_0^{\ln{3}}e^{x+y}dydx=\int_0^{\ln{4}}\int_0^{\ln{3}}e^xe^ydydx=\int_0^{\ln{4}}\left(e^xe^y\right)\bigg\vert_0^{\ln{3}}dx=\int_0^{\ln{4}}\left(e^xe^{\ln{3}}-e^xe^0\right)dx$
$=\int_0^{\ln{4}}\left(2e^x\right)dx=\left(2e^x\right)\bigg\vert_0^{\ln{4}}=2e^{\ln{4}}-2e^0=8-2=6$
