Answer
$$\frac{{99}}{8}$$
Work Step by Step
$$\eqalign{
& \int_0^3 {\int_1^2 {\left( {x{y^3} - x} \right)} } dydx \cr
& = \int_0^3 {\left[ {\int_1^2 {\left( {x{y^3} - x} \right)dy} } \right]} dx \cr
& {\text{solve the inner integral}} \cr
& \int_1^2 {\left( {x{y^3} - x} \right)dy} = \left[ {x\left( {\frac{{{y^4}}}{4}} \right) - x\left( y \right)} \right]_1^2 \cr
& {\text{then}} \cr
& = \left[ {\frac{{x{y^4}}}{4} - xy} \right]_1^2 \cr
& {\text{evaluating the limits in the variable }}y \cr
& = \left[ {\frac{{x{{\left( 2 \right)}^4}}}{4} - x\left( 2 \right)} \right] - \left[ {\frac{{x{{\left( 1 \right)}^4}}}{4} - x\left( 1 \right)} \right] \cr
& {\text{simplifying}} \cr
& = \left( {4x - 2x} \right) - \left( {\frac{1}{4}x - x} \right) \cr
& = 2x + \frac{3}{4}x \cr
& = \frac{{11x}}{4} \cr
& \cr
& \int_0^3 {\left[ {\int_1^2 {\left( {x{y^3} - x} \right)dy} } \right]} dx = \int_0^3 {\frac{{11x}}{4}} dx \cr
& {\text{integrating}} \cr
& = \frac{{11}}{4}\left( {\frac{{{x^2}}}{2}} \right)_0^3 \cr
& = \frac{{11}}{8}\left( {{x^2}} \right)_0^3 \cr
& = \frac{{11}}{8}\left( {9 - 0} \right) \cr
& = \frac{{99}}{8} \cr} $$
