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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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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