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Find the f(c) guaranteed by the Mean Value Theorem for Integration on the
function f(x)=ln(x)/x on the interval [1, 100].

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What have you tried Lias? –  Nameless Dec 9 '12 at 11:44
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I suggest you write out what the Mean Value Theorem says. It will have an integral in it. I suggest you evaluate that integral. I bet if you do all that you will be able to find $f(c)$. –  Gerry Myerson Dec 9 '12 at 11:46
    
i tried the theorem: [f(b)- f(a)]/[b-a] ... that's as far as i understood @Nameless –  lias Dec 9 '12 at 12:00
    
@lias That's the Mean Value Theorem for derivatives not the Mean Value Theorem for Integration. –  Nameless Dec 9 '12 at 12:03
    
ahhh ok, thanku –  lias Dec 9 '12 at 12:57
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1 Answer

up vote 1 down vote accepted

One has $$\int_1^{100}{\log x\over x}\ dx={1\over2}\bigl(\log x\bigr)^2\Biggr|_1^{100}={1\over2}\bigl(\log 100\bigr)^2\ .$$ In order to "find the $f(c)$ whose existence is guaranteed by the mean value theorem" we therefore have to solve the equation $$(100-1) f(c)={1\over2}\bigl(\log 100\bigr)^2$$ for $f(c)$. The result is $$f(c)={1\over198}\bigl(\log 100\bigr)^2\doteq 0.107\ .$$ Actually we have not used the MVT at all. We just have computed the average value of $f$ on the interval $[1,100]$. The essence of the MVT is that MVT guarantees the existence of a $c\in[1,100]$ such that $f(c)$ is equal to this average. As $f(1)=0$ and $f(10)\doteq0.23$, by the intermediate value theorem there has to be such a $c\in[1,10]$, even.

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