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The mean-value theorem for integrals is:

$$\int_a^bf(x)dx=(b-a)f(c)$$

My calculus-book merely "observes" that $f(c)$ is also the average value of $f$ on $[a,b]$:

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

But why is this so? How can one prove this?

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    $\begingroup$ What would be the definition of 'average value of $f$'? The value $\frac{1}{b-a}\int_a^b f(t)\, \mathrm dt$ sounds like a logical one for me, but it's easy to see this is equal to $f(c )$ by dividing the first expression by $(b-a)$. $\endgroup$ – SvanN Sep 9 '17 at 10:59
  • $\begingroup$ The average value of a function is defined as $$ f_{\text{avg}}=\frac{1}{b-a}\int_a^bf(x)dx $$ $\endgroup$ – Sachchidanand Prasad Sep 9 '17 at 11:00
  • $\begingroup$ What you want, you want the proof of your first expression (Mean Value Theorem)? $\endgroup$ – Sachchidanand Prasad Sep 9 '17 at 11:01
  • $\begingroup$ See en.wikipedia.org/wiki/… $\endgroup$ – dromastyx Sep 9 '17 at 11:02
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If $f$ is Riemann-integrable, you know that the integral is a limit of Riemann sums, so that for example when $n$ is large $$ \int_a^b f(t) d t \approx \Delta x \big(f(a) + f(a + \Delta x) + f(a + 2 \Delta x)+ \cdots+ f(b-\Delta x)\big) $$ with $\Delta x = \frac{b-a}{n}$. You see that $$ \frac{1}{b-a}\int_a^b f(t) d t \approx \frac{1}{n} \big(f(a) + f(a + \Delta x) + f(a + 2 \Delta x)+ \cdots+ f(b-\Delta x)\big) $$ is actually close to the average of $n$ values of $f$ at regularly separated points.

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From the definition of the integral as limit of Riemann sums it becomes natural to call $${1\over b-a}\int_a^b f(x)\>dx$$ the average of $f$ on $[a,b]$. The MVT then says that there is a point $c\in[a,b]$ such that $f(c)$ is exactly equal to this average value.

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  • $\begingroup$ Why is it "natural"? It seems I am lacking some crucial insight here :) $\endgroup$ – GambitSquared Sep 9 '17 at 11:09
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Hint: Suppose $f(x)$ is continues on $[a,b]$ so it has maximum and minimum

assume $max f(x)=M ,min f(x)=n$ so $$\forall x \in [a,b] :n \leq f(x)\leq M$$ so avegrage of $f(x)$ is between max,min $$average=\overline {f} \to \\n \leq \overline {f}\leq M\\$$ $f(x) $ is continues ,so $F(x)=\int_a^x f(t)dt$ is continues too . So $ n(b-a)\leq F(x) \leq M(b-a)$ now $F(x)$ now use intermediate point theorem for $F(x)$ so ,must be $\exists c \in (a,b) \to F'(c)=\frac{F(b)-F(a)}{b-a}\to f(c)=\frac{F(b)-F(a)}{b-a}=\frac{\int_a^b f(x)dx-0}{b-a}=\overline {f}$

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  • $\begingroup$ I understand that $f(c)$ is between $n$ and $M$, but I don't see why $f(c)=\overline{f}$ $\endgroup$ – GambitSquared Sep 9 '17 at 11:11
  • $\begingroup$ @GambitSquared : I add more description . $\endgroup$ – Khosrotash Sep 9 '17 at 11:20

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