Answer
$$7$$
Work Step by Step
$$\eqalign{
& \int_0^{\ln 2} {\int_0^1 {6x{e^{3y}}} } dxdy \cr
& = \int_0^{\ln 2} {\left[ {\int_0^1 {6x{e^{3y}}} dx} \right]} dy \cr
& {\text{solve the inner integral}}{\text{, treat }}y{\text{ as a constant}} \cr
& = \int_0^1 {6x{e^{3y}}} dx \cr
& = 6{e^{3y}}\int_0^1 x dx \cr
& = 6{e^{3y}}\left[ {\frac{{{x^2}}}{2}} \right]_0^1 \cr
& = 3{e^{3y}}\left[ {{x^2}} \right]_0^1 \cr
& {\text{evaluating the limits for }}x \cr
& = 3{e^{3y}}\left[ {{1^2} - {0^2}} \right] \cr
& = 3{e^{3y}} \cr
& \cr
& \int_0^{\ln 2} {\left[ {\int_0^1 {6x{e^{3y}}} dx} \right]} dy = \int_0^{\ln 2} {3{e^{3y}}} dy \cr
& {\text{integrating}} \cr
& = \left( {{e^{3y}}} \right)_0^{\ln 2} \cr
& = {e^{3\ln 2}} - {e^0} \cr
& {\text{simplifying}} \cr
& = {e^{\ln 8}} - 1 \cr
& = 8 - 1 \cr
& = 7 \cr} $$
