Answer
$$\frac{{38}}{3}$$
Work Step by Step
$$\eqalign{
& \int_1^3 {\int_1^2 {\left( {{y^2} + y} \right)} } dxdy \cr
& = \int_1^3 {\left[ {\int_1^2 {\left( {{y^2} + y} \right)} dx} \right]} dy \cr
& {\text{solve the inner integral}}{\text{, treat }}y{\text{ as a constant}} \cr
& \int_1^2 {\left( {{y^2} + y} \right)} dx \cr
& \left( {{y^2} + y} \right)\int_1^2 {dx} \cr
& = \left( {{y^2} + y} \right)\left[ x \right]_1^2 \cr
& {\text{evaluating the limits for }}x \cr
& = \left( {{y^2} + y} \right)\left( {2 - 1} \right) \cr
& = \left( {{y^2} + y} \right) \cr
& \cr
& \int_1^3 {\left[ {\int_1^2 {\left( {{y^2} + y} \right)} dx} \right]} dy = \int_1^3 {\left( {{y^2} + y} \right)} dy \cr
& = \int_1^3 {\left( {{y^2} + 3} \right)} dy \cr
& {\text{integrating}} \cr
& = \left( {\frac{{{y^3}}}{3} + \frac{{{y^2}}}{2}} \right)_1^3 \cr
& = \left( {\frac{{{{\left( 3 \right)}^3}}}{3} + \frac{{{{\left( 3 \right)}^2}}}{2}} \right) - \left( {\frac{{{{\left( 1 \right)}^3}}}{3} + \frac{{{{\left( 1 \right)}^2}}}{2}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{{27}}{2} - \frac{5}{6} \cr
& = \frac{{38}}{3} \cr} $$
