Answer
$$\frac{{1232}}{9}$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_0^4 {\left( {{x^2}{y^2} + 5x} \right)dx} dy} \cr
& = \int_0^2 {\left[ {\int_0^4 {\left( {{x^2}{y^2} + 5x} \right)dx} } \right]dy} \cr
& {\text{solve the inner integral}} \cr
& \int_0^4 {\left( {{x^2}{y^2} + 5x} \right)dx} = \left( {\frac{{{x^3}{y^2}}}{3} + \frac{{5{x^2}}}{2}} \right)_0^4 \cr
& {\text{evaluate limits for }}x \cr
& = \left( {\frac{{{{\left( 4 \right)}^3}{y^2}}}{3} + \frac{{5{{\left( 4 \right)}^2}}}{2}} \right) - \left( {\frac{{{{\left( 0 \right)}^3}{y^2}}}{3} + \frac{{5{{\left( 0 \right)}^2}}}{2}} \right) \cr
& = \frac{{64{y^2}}}{3} + 40 \cr
& {\text{then}} \cr
& \int_0^2 {\int_0^4 {\left( {{x^2}{y^2} + 5x} \right)dx} dy} = \int_0^2 {\left( {\frac{{64{y^2}}}{3} + 40} \right)} dy \cr
& {\text{integrating}} \cr
& \left[ {\frac{{64{y^3}}}{9} + 40y} \right]_0^2 \cr
& = \left( {\frac{{64{{\left( 2 \right)}^3}}}{9} + 40\left( 2 \right)} \right) - \left( {\frac{{64{{\left( 0 \right)}^3}}}{9} + 40\left( 0 \right)} \right) \cr
& = \frac{{512}}{9} + 80 \cr
& = \frac{{1232}}{9} \cr} $$
