Chapter 13 - Multiple Integration - 13.2 Double Integrals over General Regions - 13.2 Exercises - Page 981: 27

\[2\]

\[\begin{align} & \iint_{R}{xy}dA \\ & \text{The region }R\text{ is represented in the graph shown below} \\ & \text{Then,} \\ & R=\left\{ \left( x,y \right):2x+1\le y\le -2x+5,\text{ }0\le x\le 1 \right\} \\ & \iint_{R}{xy}dA=\int_{0}^{1}{\int_{2x+1}^{-2x+5}{xy}}dydx \\ & \text{Integrate} \\ & =\int_{0}^{1}{\left[ \frac{x{{y}^{2}}}{2} \right]_{2x+1}^{-2x+5}}dx \\ & =\int_{0}^{1}{\left[ \frac{x{{\left( -2x+5 \right)}^{2}}}{2}-\frac{x{{\left( 2x+1 \right)}^{2}}}{2} \right]}dx \\ & =\int_{0}^{1}{\left[ \frac{x\left( 4{{x}^{2}}-20x+25 \right)}{2}-\frac{x\left( 4{{x}^{2}}+4x+1 \right)}{2} \right]}dx \\ & =\int_{0}^{1}{\left( \frac{4{{x}^{3}}-20{{x}^{2}}+25x}{2}-\frac{4{{x}^{3}}+4{{x}^{2}}+x}{2} \right)}dx \\ & =\int_{0}^{1}{\left( 2{{x}^{3}}-10{{x}^{2}}+\frac{25}{2}x-2{{x}^{3}}-2{{x}^{2}}-\frac{1}{2}x \right)}dx \\ & =\int_{0}^{1}{\left( -12{{x}^{2}}+12x \right)}dx \\ & =\left[ -4{{x}^{3}}+6{{x}^{2}} \right]_{0}^{1} \\ & =\left[ -4{{\left( 1 \right)}^{3}}+6{{\left( 1 \right)}^{2}} \right]-\left[ -4{{\left( 0 \right)}^{3}}+6{{\left( 0 \right)}^{2}} \right] \\ & =2 \\ \end{align}\]