Answer
$$\frac{{64}}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^4 {\int_1^{{e^x}} {\frac{x}{y}} dydx} \cr
& = \int_0^4 {\left[ {\int_1^{{e^x}} {\frac{x}{y}} dy} \right]dx} \cr
& {\text{solve the inner integral by treating }}x{\text{ as a constant}} \cr
& = x\int_1^{{e^x}} {\frac{1}{y}} dy \cr
& {\text{integrate}} \cr
& = x\left[ {\ln \left| y \right|} \right]_1^{{e^x}} \cr
& {\text{evaluating the limits}} \cr
& = x\left( {\ln \left| {{e^x}} \right| - \ln \left| 1 \right|} \right) \cr
& = x\left( {x - 0} \right) \cr
& = {x^2} \cr
& \cr
& \int_0^4 {\left[ {\int_1^{{e^x}} {\frac{x}{y}} dy} \right]dx} = \int_0^4 {{x^2}dx} \cr
& {\text{then}} \cr
& = \left( {\frac{{{x^3}}}{3}} \right)_0^4 \cr
& {\text{evaluate}} \cr
& = \left( {\frac{{{4^3}}}{3}} \right) - \left( {\frac{{{0^3}}}{3}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{{64}}{3} - 0 \cr
& = \frac{{64}}{3} \cr} $$
