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Lets say that $f(x) = (10 - x)\ln x$. Over the domain: $1 ≤ x ≤ 10$. How can I find the average value of $y$ over this domain and what is that value?

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I suppose you call $f(x)$ as your $y$. Then one (interpreted) way would be to calculate the area in that range and divide it by the width. To find the area, integrate the function over the domain. Width will be the length of domain. –  dineshdileep Nov 21 '12 at 14:15
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To find the average $\bar f$ of a function $f \colon [a,b] \to \mathbb R$ calculate $$ \bar f = \frac 1{b-a} \int_a^b f(x)\, dx $$ In your case, we have \begin{align*} \bar f &= \frac 1{10-1} \int_1^{10} (10 - x)\log x \, dx \\ &= \frac 1{10 - 1} \left[10x(\log x - 1) - \frac 14x^2(2\log x - 1)\right]_1^{10}\\ &= \frac 19 \left(100(\log 10 - 1) - 25(2\log 10 - 1) - 10(-1) + \frac 14(-1)\right)\\ &= \frac 19 \left(50\log 10 - 65 - \frac 14\right) \end{align*}

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The mean of an integrable function $\,f(x)\,$ over an interval $\,[a,b]\,$ is given by

$$\overline f:=\frac{1}{b-a}\int_a^bf(x)\,dx$$

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Antonio: There is an $(x)$ too much here. –  Christian Blatter Nov 21 '12 at 15:32
    
@ChristianBlatter, sure. Thanks –  DonAntonio Nov 21 '12 at 16:36
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