# Evaluating the Average value of f(x)

Determine the average value of $f(x)$ over the interval from $x=a$ to $x=b$, where $f(x)=\frac{1}{x}$, $a=\frac{1}{10}$, and $b=10$

Can someone please explain the simplifying in this problem step-by-step?

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This is not a complete problem. $f(x)$ must be defined as something, likely $\frac{1}{x}$. Additionally, $a$ and $b$ must have values for this to come out to an answer. I can assume they are $10$ and $\frac{1}{10}$, though. – Emrakul Nov 15 '13 at 0:05
Determine the average value of f(x) over the interval from x=a, where f(x)=1/x, a=1/10, and b=10 – Will Nov 15 '13 at 0:06
Mind adding that to your question? Also, your values for $a$ and $b$ conflict with the solution given. – Emrakul Nov 15 '13 at 0:06
$\ln$ is an antiderivative of $1/x$. – The Chaz 2.0 Nov 15 '13 at 0:10

Note that $$\ln\left(\frac{1}{x}\right) = - \ln{x}$$

In particular, we have that

$$\ln{a} - \ln\left(\frac{1}{a}\right) = 2 \ln{a}$$

Hence, we multiply $10$ by $2$ to get $20$.

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can u use the problems numbers – Will Nov 15 '13 at 0:31
@Sam I did. You have a $10$ that becomes a $20$, which is what I said in my answer. – user61527 Nov 15 '13 at 0:32
....ln(12)-ln(12)^-1 = ln(12/12^-1) ...is this correct? – Will Nov 15 '13 at 0:34
$\ln{12} - ln(12^{-1}) = \ln{12} - (-ln{12}) =\ln{12} + \ln{12} = 2 \ln{12}$ – Steve ODonnell Nov 15 '13 at 1:12
thank you@SteveODonnell – Will Nov 15 '13 at 1:26

Your values for $a$ and $b$ differ from the solution you've given, however, here is an answer anyway. Some steps are missing which should help clarify.

$$\frac{1}{b-a}\int_a^bf(x)dx$$

This turns into:

$$\frac{1}{12-\frac{1}{12}}\int_\frac{1}{12}^{12}\frac{1}{x}dx$$

We know that $\int\frac{1}{x}dx=\ln\left|x\right|$.

$$\left.\frac{12}{143}\ln|x|\right]_\frac{1}{12}^{12}$$

We substitue $12$ and $\frac{1}{12}$:

$$\frac{12}{143}\left(\ln(12)-\ln(12^{-1})\right)=\frac{24}{143}\left(\ln 12\right)$$

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how did you get ln(12)? ( this is what i need help on) – Will Nov 15 '13 at 0:16
@Sam Please review the fundamentals of integration; $12$ comes from the integral's limits, substituted into the integrand $\ln|x|$. – Emrakul Nov 15 '13 at 0:18
i know that but i am not understanding the ln12 in the answer. how did you use the properties of logarithms to get that? – Will Nov 15 '13 at 0:20
@Sam Substitute $12$ for $x$ in $\ln|x|$. – Emrakul Nov 15 '13 at 0:32
Emrakul also used $\ln(12^{-1}) = -\ln(12)$. – apt1002 Nov 15 '13 at 0:38