Answer
$$\frac{{9 - {e^2}}}{2}$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {{e^{x + 2y}}}dA;\,\,\,\,\,\,\,R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant \ln 2,\,\,\,\,1 \leqslant y \leqslant \ln 3} \right\} \cr
& {\text{Convert to an iterated integral substituting the region }}R \cr
& = \int_1^{\ln 3} {\int_0^{\ln 2} {{e^{x + 2y}}} } dxdy \cr
& = \int_1^{\ln 3} {\left[ {\int_0^{\ln 2} {{e^{x + 2y}}} dx} \right]} dy \cr
& {\text{solve the inner integral}}{\text{, treat }}y{\text{ as a constant}} \cr
& = \int_0^{\ln 2} {{e^{x + 2y}}} dx \cr
& {\text{integrating}} \cr
& = \left( {{e^{x + 2y}}} \right)_0^{\ln 2} \cr
& {\text{evaluating the limits for the variable }}x \cr
& = {e^{\ln 2 + 2y}} - {e^{0 + 2y}} \cr
& {\text{simplifying}} \cr
& = {e^{\ln 2 + 2y}} - {e^{2y}} \cr
& \cr
& \int_1^{\ln 3} {\left[ {\int_0^{\ln 2} {{e^{x + 2y}}} dx} \right]} dy = \int_1^{\ln 3} {\left( {{e^{\ln 2 + 2y}} - {e^{2y}}} \right)} dy \cr
& {\text{integrating}} \cr
& = \left( {\frac{1}{2}{e^{\ln 2 + 2y}} - \frac{1}{2}{e^{2y}}} \right)_1^{\ln 3} \cr
& {\text{evaluate}} \cr
& = \frac{1}{2}\left( {{e^{\ln 2 + 2\ln 3}} - {e^{2\ln 3}}} \right) - \frac{1}{2}\left( {{e^{\ln 2 + 2\left( 1 \right)}} - {e^{2\left( 1 \right)}}} \right) \cr
& = \frac{1}{2}\left( {18 - 9} \right) - \frac{1}{2}\left( {2{e^2} - {e^2}} \right) \cr
& = \frac{9}{2} - \frac{1}{2}{e^2} \cr
& = \frac{{9 - {e^2}}}{2} \cr} $$
