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\begin{center}
\textbf{Multiple Integration}

\textit{\textbf{Iteration of Double Integrals}}
\end{center}

\textbf{Question}


Calculate the given double integral by iteration in the region
defined by the given curves.

$\ds \int\!\!\!\int_D \ln x \,dA$

With $D$ being the region bounded by $2x+2y=5$ and $xy=1$ in the first
quadrant.


\textbf{Answer}

For intersection: $xy=1, 2x+2y=5$.

$\Rightarrow 2x^2 - 5x+2=0$, or $(2x-1)(x-2) = 0$. And so the
intersections are at $x=1/2$ and $x=2$.

\begin{eqnarray*}
I & = & \int\!\!\!\int_D \ln x \,dA = \int_{1/2}^2 \ln x
\,dx\int_{1/x}^{(5/2)-x} \,dy\\
& = & \int_{1/2}^2 \ln x \left ( \frac{5}{2} - x - frac{1}{x} \right )
\,dx\\
& = & \int_{1/2}^2 \ln x \left ( \frac{5}{2} - x \right ) \,dx -
\frac{1}{2} \left. (\ln x)^2 \right |_{1/2}^2
\end{eqnarray*}

$$U = \ln x \ \ \ \ \ dV= \left ( \frac{5}{2} - x \right ) dx$$
$$dU = \frac{dx}{x} \ \ \ \ \ V=\frac{5}{2}x - \frac{x^2}{2}$$

\begin{eqnarray*}
I & = & -\frac{1}{2} \left. \left ( (\ln 2)^2 - \left ( \ln
\frac{1}{2} \right )^2\right ) + \left ( frac{5}{2}x-\frac{x^2}{2}
\right ) \ln x \right |_{1/2}^2\\
& & -\int_{1/2}^2 \left ( \frac{5}{2} - \frac{x}{2} \right ) \,dx\\
& = & (5-2)\ln 2 - \left ( \frac{5}{4} - \frac{1}{8} \right ) ln
\frac{1}{2} - \frac{15}{4} + \frac{15}{16}\\
& = & \frac{33}{8} ln 2 - \frac{45}{16}
\end{eqnarray*}

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