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Question

# Find the average value of f(x, y) over the plane region R. f(x, y) = x R: rectangle with vertices (0, 0), (4, 0), (4, 2), (0, 2)

Solution

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Step 1
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If $f$ is integrable over the plane region $R$, then the average value of $f$ over $R$ is

$A_{vg} = \dfrac{1}{A}\int_{R}^{} \int_{}^{} f(x,y) \,dA$

where

$A$ is the area of region $R$

Now

\begin{align*} \dfrac{1}{A}\int_{R}^{} \int_{}^{} f(x,y) \,dA &= \dfrac{1}{4 \times 2}\int_{0}^{2} \int_{0}^{4} x \,dx \,dy \\ &= \dfrac{1}{8}\int_{0}^{2} \eval{\dfrac{x^2}{2}}_{0}^{4} \,dy \\ &= \dfrac{1}{8}\int_{0}^{2} \qty(\dfrac{4^2}{2} - \dfrac{0^2}{2} ) \,dy \\ &= \dfrac{1}{8}\int_{0}^{2} (8 - 0 ) \,dy \\ &= \int_{0}^{2} 1 \,dy \\ &= \eval{y}_{0}^{2} \\ &= 2-0 \\ &= 2 \end{align*}

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