## Calculus with Applications (10th Edition)

$$8\ln \left( 2 \right) + 4$$
\eqalign{ & \int_2^4 {\int_3^5 {\left( {\frac{x}{y} + \frac{y}{3}} \right)} } dxdy \cr & = \int_2^4 {\left[ {\int_3^5 {\left( {\frac{x}{y} + \frac{y}{3}} \right)} dx} \right]} dy \cr & {\text{solve the inner integral}} \cr & \int_3^5 {\left( {\frac{x}{y} + \frac{y}{3}} \right)} dx = \left( {\frac{{{x^2}}}{{2y}} + \frac{{xy}}{3}} \right)_3^5 \cr & = \left( {\frac{{{x^2}}}{{2y}} + \frac{{xy}}{3}} \right)_3^5 \cr & {\text{evaluate limits for }}x \cr & = \left( {\frac{{{{\left( 5 \right)}^2}}}{{2y}} + \frac{{\left( 5 \right)y}}{3}} \right) - \left( {\frac{{{{\left( 3 \right)}^2}}}{{2y}} + \frac{{\left( 3 \right)y}}{3}} \right) \cr & = \left( {\frac{{25}}{{2y}} + \frac{{5y}}{3}} \right) - \left( {\frac{9}{{2y}} + y} \right) \cr & = \frac{8}{y} + \frac{2}{3}y \cr & {\text{then}} \cr & \int_2^4 {\int_3^5 {\left( {\frac{x}{y} + \frac{y}{3}} \right)} } dxdy = \int_2^4 {\left( {\frac{8}{y} + \frac{2}{3}y} \right)} dy \cr & {\text{integrating}} \cr & \left[ {8\ln \left| y \right| + \frac{{{y^2}}}{3}} \right]_2^4 \cr & = \left( {8\ln \left| 4 \right| + \frac{{{4^2}}}{3}} \right) - \left( {8\ln \left| 2 \right| + \frac{{{2^2}}}{3}} \right) \cr & = 8\ln 4 + \frac{{16}}{3} - 8\ln 3 - \frac{4}{3} \cr & = 8\ln \left( {\frac{4}{2}} \right) + 4 \cr & = 8\ln \left( 2 \right) + 4 \cr}