## Calculus with Applications (10th Edition)

$$255$$
\eqalign{ & \int_{16}^{25} {\int_2^7 {\frac{{3 + 5y}}{{\sqrt x }}} } dydx \cr & = \int_{16}^{25} {\left[ {\int_2^7 {\frac{{3 + 5y}}{{\sqrt x }}} dy} \right]} dx \cr & {\text{solve the inner integral}} \cr & \int_2^7 {\frac{{3 + 5y}}{{\sqrt x }}} dy = \frac{1}{{\sqrt x }}\int_2^7 {\left( {3 + 5y} \right)} dy \cr & {\text{using the power rule }} \cr & = \frac{1}{{\sqrt x }}\left( {3y + \frac{{5{y^2}}}{2}} \right)_2^7 \cr & {\text{evaluating the limits in the variable }}y \cr & = \frac{1}{{\sqrt x }}\left( {3\left( 7 \right) + \frac{{5{{\left( 7 \right)}^2}}}{2}} \right) - \frac{1}{{\sqrt x }}\left( {3\left( 2 \right) + \frac{{5{{\left( 2 \right)}^2}}}{2}} \right) \cr & {\text{simplifying}} \cr & = \frac{1}{{\sqrt x }}\left( {\frac{{287}}{2}} \right) - \frac{1}{{\sqrt x }}\left( {16} \right) \cr & = \left( {\frac{{287}}{2} - 16} \right)\frac{1}{{\sqrt x }} \cr & = \frac{{255}}{{2\sqrt x }} \cr & {\text{then}} \cr & \int_{16}^{25} {\int_2^7 {\frac{{3 + 5y}}{{\sqrt x }}} } dydx = \int_{16}^{25} {\frac{{255}}{{2\sqrt x }}} dx \cr & = 255\int_{16}^{25} {\frac{1}{{2\sqrt x }}} dx \cr & = 255\left[ {\sqrt x } \right]_{16}^{25} \cr & = 255\left( {\sqrt {25} - \sqrt {16} } \right) \cr & = 255 \cr}