Answer
$$8$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {\int_0^2 {\left( {2x + 3y} \right)} dxdy} \cr
& \int_{ - 1}^1 {\left[ {\int_0^2 {\left( {2x + 3y} \right)} dx} \right]dy} \cr
& {\text{Evaluate inner integral}} \cr
& \int_0^2 {\left( {2x + 3y} \right)} dx = \left[ {{x^2} + 3xy} \right]_{x = 0}^{x = 2} \cr
& = \left[ {\left( {{{\left( 2 \right)}^2} + 3\left( 2 \right)y} \right) - \left( {{{\left( 0 \right)}^2} + 3\left( 0 \right)y} \right)} \right] \cr
& = 4 + 6y \cr
& {\text{Therefore,}} \cr
& \int_{ - 1}^1 {\int_0^2 {\left( {2x + 3y} \right)} dxdy} = \int_{ - 1}^1 {\left( {4 + 6y} \right)dy} \cr
& {\text{Integrating}} \cr
& {\text{ = }}\left[ {4y + 3{y^2}} \right]_{ - 1}^1 \cr
& = \left[ {4\left( 1 \right) + 3{{\left( 1 \right)}^2}} \right] - \left[ {4\left( { - 1} \right) + 3{{\left( { - 1} \right)}^2}} \right] \cr
& = \left[ {4 + 3} \right] - \left[ { - 4 + 3} \right] \cr
& = 8 \cr} $$
