Answer
$$\frac{{49}}{{20}}$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\int_0^{\ln x} {{x^3}{e^y}} } dydx \cr
& = \int_1^2 {\left[ {\int_0^{\ln x} {{x^3}{e^y}} dy} \right]} dx \cr
& {\text{solve the inner integral}}{\text{, treat }}x{\text{ as a constant}} \cr
& \int_0^{\ln x} {{x^3}{e^y}} dy = {x^3}\int_0^{\ln x} {{e^y}} dy \cr
& = {x^3}\left[ {{e^y}} \right]_0^{\ln x} \cr
& {\text{evaluating the limits in the variable }}y \cr
& = {x^3}\left[ {{e^{\ln x}} - {e^0}} \right] \cr
& {\text{simplifying}} \cr
& = {x^3}\left( {x - 1} \right) \cr
& = {x^4} - {x^3} \cr
& \cr
& \int_1^2 {\left[ {\int_0^{\ln x} {{x^3}{e^y}} dy} \right]} dx = \int_1^2 {\left( {{x^4} - {x^3}} \right)} dx \cr
& {\text{integrating by using the power rule}} \cr
& = \left( {\frac{{{x^5}}}{5} - \frac{{{x^4}}}{4}} \right)_1^2 \cr
& {\text{evaluate}} \cr
& = \left( {\frac{{{2^5}}}{5} - \frac{{{2^4}}}{4}} \right) - \left( {\frac{{{1^5}}}{5} - \frac{{{1^4}}}{4}} \right) \cr
& = \frac{{12}}{5} - \left( { - \frac{1}{{20}}} \right) \cr
& = \frac{{12}}{5} + \frac{1}{{20}} \cr
& = \frac{{49}}{{20}} \cr} $$
