## Calculus with Applications (10th Edition)

$$395$$
\eqalign{ & \int_0^3 {\int_0^5 {\left( {2x + 6y + {y^2}} \right)dy} dx} \cr & = \int_0^3 {\left[ {\int_0^5 {\left( {2x + 6y + {y^2}} \right)dy} } \right]dx} \cr & {\text{solve the inner integral}} \cr & \int_0^5 {\left( {2x + 6y + {y^2}} \right)dy} = \left( {2xy + 3{y^2} + \frac{{{y^3}}}{3}} \right)_0^5 \cr & {\text{evaluate limits for }}y \cr & = \left( {2x\left( 5 \right) + 3{{\left( 5 \right)}^2} + \frac{{{{\left( 5 \right)}^3}}}{3}} \right) - \left( {2x\left( 0 \right) + 3{{\left( 0 \right)}^2} + \frac{{{{\left( 0 \right)}^3}}}{5}} \right) \cr & = 10x + 75 + \frac{{125}}{3} \cr & = 10x + \frac{{350}}{3} \cr & {\text{then}} \cr & \int_0^3 {\int_0^5 {\left( {2x + 6y + {y^2}} \right)dy} dx} = \int_0^3 {\left( {10x + \frac{{350}}{3}} \right)} dy \cr & {\text{integrating}} \cr & \left[ {5{x^2} + \frac{{350}}{3}x} \right]_0^3 \cr & = \left( {5{{\left( 3 \right)}^2} + \frac{{350}}{3}\left( 3 \right)} \right) - \left( {5{{\left( 0 \right)}^2} + \frac{{350}}{3}\left( 0 \right)} \right) \cr & = 45 + 350 \cr & = 395 \cr}