Chapter 13 - Multiple Integration - 13.2 Double Integrals over General Regions - 13.2 Exercises - Page 983: 83

\[\frac{9}{8}\]

\[\begin{align} & \text{Let the region }R=\left\{ \left( x,y \right):\frac{1}{x}\le y\le \frac{5}{2}-x,\text{ }\frac{1}{2}\le x\le 2\text{ } \right\} \\ & \iint\limits_{R}{\left( x+y \right)}dA=\int_{1/2}^{2}{\int_{1/x}^{5/2-x}{\left( x+y \right)dydx}} \\ & =\int_{1/2}^{2}{\left[ xy+\frac{1}{2}{{y}^{2}} \right]_{1/x}^{5/2-x}}dx \\ & =\int_{1/2}^{2}{\left[ xy+\frac{1}{2}{{y}^{2}} \right]_{1/x}^{5/2-x}}dx \\ & =\int_{1/2}^{2}{\left[ x\left( \frac{5}{2}-x \right)+\frac{1}{2}{{\left( \frac{5}{2}-x \right)}^{2}}-x\left( \frac{1}{x} \right)-\frac{1}{2}{{\left( \frac{1}{x} \right)}^{2}} \right]}dx \\ & =\int_{1/2}^{2}{\left[ \frac{5}{2}x-{{x}^{2}}+\frac{1}{2}\left( \frac{25}{4}-5x+{{x}^{2}} \right)-1-\frac{1}{2{{x}^{2}}} \right]}dx \\ & =\int_{1/2}^{2}{\left( \frac{5}{2}x-{{x}^{2}}+\frac{25}{8}-\frac{5}{2}x+\frac{1}{2}{{x}^{2}}-1-\frac{1}{2{{x}^{2}}} \right)}dx \\ & =\int_{1/2}^{2}{\left( \frac{17}{8}-\frac{1}{2}{{x}^{2}}-\frac{1}{2{{x}^{2}}} \right)}dx \\ & \text{Integrating} \\ & =\left[ \frac{17}{8}x-\frac{1}{6}{{x}^{3}}+\frac{1}{2x} \right]_{1/2}^{2} \\ & =\left[ \frac{17}{8}\left( 2 \right)-\frac{1}{6}{{\left( 2 \right)}^{3}}+\frac{1}{2\left( 2 \right)} \right]-\left[ \frac{17}{8}\left( \frac{1}{2} \right)-\frac{1}{6}{{\left( \frac{1}{2} \right)}^{3}}+\frac{1}{2\left( 1/2 \right)} \right] \\ & =\frac{19}{6}-\frac{49}{24} \\ & =\frac{9}{8} \\ \end{align}\]