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Theorem: Let $f$ be a continuous function on $[a,b] \in \Bbb R$. Prove that there exists a $c \in [a,b]$ so that $$ f(c) = \frac{1}{b-a}\int_a^b f(x)\,dx $$

I cannot use derivatives or any other integral theorems, I am doing these proofs using Riemann Sums and continuity.

My Thoughts:

I am planning on approaching this problem using the following theorem:

Theorem: Let $f$ be a continuous function on the interval $[a,b]$. Then $$\left| \int_a^bf(x)\,dx \right| \ \le \ (b-a)\sup\limits_{[a,b]}|f(x)| $$

Or I could show that this $f(c)$ exists, but then I am unsure as to how I would prove that $c \in [a,b]$. Hints would be appreciated.

Edit: So I have realized that I have the following inequality for $m := \inf\limits_{[a,b]}f(x)(b-a)$, $M := (b-a)\sup\limits_{[a,b]}|f(x)|$: $$ m \le \frac{1}{b-a}\left| \int_a^bf(x)\,dx\right| \le M $$

So now how do I relate this to $f(c)$?

Edit 2: I changed the title as upon considering the existing MVT, I don't think this is the same thing.

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Observe that $(b-a)\min f(x)\leq\int_a^b f(x)dx\leq(b-a)\max f(x)$, then use the Intermediate Value Theorem. – enzotib Aug 18 '12 at 18:18
I see now. So the middle quantity is bounded by $m, M$ and $f$ must take the value of some $c$ so that $f(c) \in [m,M]$, and so this necessarily means that $c\in [a,b]$? – Moderat Aug 18 '12 at 18:22
Not the $m,M$ that you use in your question, that use the absolute value. – enzotib Aug 18 '12 at 18:23
This is the MVT expressed in integral form. – copper.hat Aug 18 '12 at 18:24
Also note that the problem is equivalent to finding $c$ so that $\int_a^b (f(x)-f(c)) dx = 0$. If $f(c) = \max_{t \in [a,b]} f(t)$, then the integral is $\leq 0$, similarly, for $f(c) = \min_{t \in [a,b]} f(t)$, the integral is $\geq 0$. – copper.hat Aug 18 '12 at 18:28
up vote 2 down vote accepted

Let $I=\frac1{b-a}\int\limits_a^bf$. Assume that the conclusion does not hold, then either $f(x)\lt I$ for every $x$ in $[a,b]$, or $f(x)\gt I$ for every $x$ in $[a,b]$ (otherwise, $f(x)\lt I\lt f(y)$ for some $x$ and $y$ in $[a,b]$ and, by the intermediate value theorem, $f(z)=I$ for some $z$ between $x$ and $y$).

Without loss of generality, the first case happens. Then $\int\limits_a^bf(x)\,\mathrm dx\lt\int\limits_a^bI\,\mathrm dx=(b-a)I=\int\limits_a^bf$, this is absurd.

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I like your answers, including this one. – Quinn Culver Aug 19 '12 at 3:40
@QuinnCulver $\langle$ Blushes $\rangle$... – Did Aug 19 '12 at 9:07

Define $F(y)=\int_{a}^{y}f(x)dx$ use the theorem fundamental calculus and mean value theorem

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$F(x)$ is a constant? – enzotib Aug 18 '12 at 18:29
You mean $\int_a^x f(t) dt$, of course. – copper.hat Aug 18 '12 at 18:30
Yeah, I made a type. – user29999 Aug 18 '12 at 18:30
You mean a type of typo., of course :-). – copper.hat Aug 18 '12 at 18:41
The mean value theorem can be used only if one is allowed to mention derivatives. – Michael Hardy Aug 18 '12 at 21:11

If you have the extreme value theorem that says that a continuous function on a closed bounded interval actually reaches its sup and its inf, then you have $f(x_0)=M$ and $f(x_1)=m$ for some $x_0,x_1\in[a,b]$. Then the intermediate value theorem implies $f(x_2)=\text{the desired value}$, since you say you've already shown that the desired value is intermediate.

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Let's consider $M=\max{f(x)}$, $m=\min{f(x)}$ and then we have that $$(b-a) m\leq f(c) ({b-a})= \int_a^b f(x)\,dx \leq (b-a) M$$ If f is continuous on a closed interval [a,b], and c is any number between f(a) and f(b) inclusive, then there is at least one number x in the closed interval such that f(x)=c. Hence, we are done.


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