Answer
$$\frac{{14}}{3}$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {\left( {{x^2} + xy} \right)}dA;\,\,\,\,\,\,\,R = \left\{ {\left( {x,y} \right):1 \leqslant x \leqslant 2,\,\,\,\, - 1 \leqslant y \leqslant 1} \right\} \cr
& {\text{Convert to an iterated integral substituting the region }}R \cr
& = \int_1^2 {\int_{ - 1}^1 {\left( {{x^2} + xy} \right)} } dydx \cr
& = \int_1^2 {\left[ {\int_{ - 1}^1 {\left( {{x^2} + xy} \right)} dy} \right]} dx \cr
& {\text{solve the inner integral}}{\text{, treat }}x{\text{ as a constant}} \cr
& \int_{ - 1}^1 {\left( {{x^2} + xy} \right)} dy \cr
& = \left[ {{x^2}y + \frac{{x{y^2}}}{2}} \right]_{ - 1}^1 \cr
& {\text{evaluating the limits for the variable }}y \cr
& = \left[ {{x^2}\left( 1 \right) + \frac{{x{{\left( 1 \right)}^2}}}{2}} \right] - \left[ {{x^2}\left( { - 1} \right) + \frac{{x{{\left( { - 1} \right)}^2}}}{2}} \right] \cr
& {\text{simplifying}} \cr
& = \left( {{x^2} + \frac{x}{2}} \right) - \left( { - {x^2} + \frac{x}{2}} \right) \cr
& = {x^2} + \frac{x}{2} + {x^2} - \frac{x}{2} \cr
& 2{x^2} \cr
& = \int_1^2 {\left[ {\int_{ - 1}^1 {\left( {{x^2} + xy} \right)} dy} \right]} dx = \int_1^2 {2{x^2}} dx \cr
& {\text{integrating}} \cr
& = \left( {\frac{{2{x^3}}}{3}} \right)_1^2 \cr
& = \frac{2}{3}\left( {{x^3}} \right)_1^2 \cr
& {\text{evaluate}} \cr
& = \frac{2}{3}\left( {{{\left( 2 \right)}^3} - {{\left( 1 \right)}^3}} \right) \cr
& = \frac{2}{3}\left( 7 \right) \cr
& = \frac{{14}}{3} \cr} $$
