Answer
$$\frac{{63}}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^3 {\int_{ - 2}^1 {\left( {2x + 3y} \right)} } dxdy \cr
& = \int_0^3 {\left[ {\int_{ - 2}^1 {\left( {2x + 3y} \right)} dx} \right]} dy \cr
& {\text{solve the inner integral}}{\text{, treat }}y{\text{ as a constant}} \cr
& \int_{ - 2}^1 {\left( {2x + 3y} \right)} dx = \left[ {{x^2} + 3xy} \right]_{ - 2}^1 \cr
& {\text{evaluating the limits for }}x \cr
& = \left[ {{{\left( 1 \right)}^2} + 3\left( 1 \right)y} \right] - \left[ {{{\left( { - 2} \right)}^2} + 3\left( { - 2} \right)y} \right] \cr
& = \left( {1 + 3y} \right) - \left( {4 - 6y} \right) \cr
& = 1 + 3y - 4 + 6y \cr
& = 9y - 3 \cr
& \cr
& = \int_0^3 {\left[ {\int_{ - 2}^1 {\left( {2x + 3y} \right)} dx} \right]} dy \cr
& = \int_0^3 {\left( {9y - 3} \right)} dy \cr
& {\text{integrating}} \cr
& = \left( {\frac{{9{y^2}}}{2} - 3y} \right)_0^3 \cr
& = \left( {\frac{{9{{\left( 3 \right)}^2}}}{2} - 3\left( 3 \right)} \right) - \left( {\frac{{9{{\left( 0 \right)}^2}}}{2} - 3\left( 0 \right)} \right) \cr
& {\text{simplifying}} \cr
& = \frac{{81}}{2} - 9 \cr
& = \frac{{63}}{2} \cr} $$
