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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](http://en.wikipedia.org/wiki/Intermediate_value_theorem).2012-08-18
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    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 $c\in [a,b]$?2012-08-18
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    Not the $m,M$ that you use in your question, that use the absolute value.2012-08-18
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    This is the MVT expressed in integral form.2012-08-18
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    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 $\geq 0$.2012-08-18

4 Answers 4

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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.2012-08-19
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    @QuinnCulver $\langle$ Blushes $\rangle$...2012-08-19
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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?2012-08-18
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    You mean $\int_a^x f(t) dt$, of course.2012-08-18
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    Yeah, I made a type.2012-08-18
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    You mean a type of typo., of course :-).2012-08-18
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    As you can see, I don't have a good English too.2012-08-18
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    The mean value theorem can be used only if one is allowed to mention derivatives.2012-08-18
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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.

Q.E.D.