## Calculus with Applications (10th Edition)

$$3$$
\eqalign{ & \iint\limits_R {\left( {2x + 3y} \right)}dxdy;\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \leqslant y \leqslant 1,\,\,\,\,\,\,\,y \leqslant x \leqslant 2 - y \cr & {\text{replacing the limits for the region }}R \cr & \int_0^1 {\int_y^{2 - y} {\left( {2x + 3y} \right)dx} dy} \cr & = \int_0^1 {\left[ {\int_y^{2 - y} {\left( {2x + 3y} \right)dx} } \right]dy} \cr & {\text{solve the inner integral by treating }}y{\text{ as a constant}} \cr & \int_y^{2 - y} {\left( {2x + 3y} \right)dx} \cr & {\text{integrate using the power rule}} \cr & = \left[ {{x^2} + 3xy} \right]_y^{2 - y} \cr & {\text{evaluating the limits in the variable }}x \cr & = \left[ {{{\left( {2 - y} \right)}^2} + 3\left( {2 - y} \right)y} \right] - \left[ {{{\left( y \right)}^2} + 3\left( y \right)y} \right] \cr & = \left( {4 - 4y + {y^2} + 6y - 3{y^2}} \right) - \left( {{y^2} + 3{y^2}} \right) \cr & = 4 - 4y + {y^2} + 6y - 3{y^2} - {y^2} - 3{y^2} \cr & = 4 + 2y - 6{y^2} \cr & {\text{then}} \cr & \cr & \int_0^1 {\left[ {\int_y^{2 - y} {\left( {2x + 3y} \right)dx} } \right]dy} = \int_1^3 {\left( {4 + 2y - 6{y^2}} \right)dy} \cr & {\text{integrating}} \cr & = \left( {4y + {y^2} - 2{y^3}} \right)_0^1 \cr & = \left( {4\left( 1 \right) + {{\left( 1 \right)}^2} - 2{{\left( 1 \right)}^3}} \right) - \left( {4\left( 0 \right) + {{\left( 0 \right)}^2} - 2{{\left( 0 \right)}^3}} \right) \cr & {\text{simplifying}} \cr & = 4 + 1 - 2 - 0 \cr & = 3 \cr}