# Proof of integral property

Suppose $f$ is continuous on $[a,b]$. Prove that there exists an $x$ so that:

$f(x)= \frac{1}{b-a} \int_{a}^{b} f(x)dx.$

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Use the mean value theorem with a suitable function (and the fundamental theorem of calculus). – Zircht Feb 11 '14 at 21:59

Think about proving there is an $x$ so that $$f(x) (b-a) = \int_a^b f(z) dz.$$ Right-hand side is an area over $[a,b]$ with variable height, and left-hand side is over the same interval but fixed height. Intuitively, you are looking for some sort of average value of $f$ on $[a,b]$ to make this work.

Certainly if you take the values $m,M$ where $f$ achieves minimum and maximum on $[a,b]$ then $$f(m) (b-a) \leq \int_a^b f(z)dz \leq f(M) (b-a)$$ Can you take it from here?

EDIT Now you have $$f(m) \leq \frac{1}{b-a} \int_a^b f(z)dz \leq f(M)$$ so $f$ achieves $f(m)$ and $f(M)$ on $[a,b]$ and by the Intermediate Value Theorem, if $f$ is continuous, it must achieve all values between $[f(m), f(M)]$ on $[a,b]$.

You can do the last step yourself.

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Hmm, I understand the intuition, but I'm not quite sure how to complete the proof. Would the last inequality guarantee the existence of the f(z) integral, which is the integral we need? – kiwifruit Feb 11 '14 at 22:08
@kiwifruit I edited to make another step – gt6989b Feb 12 '14 at 0:49

Define $F(x)=\int_a^x f(t) dt$ (it can be easily checked by definition and squeeze theorem, that $F$ is differentiable for continuous function $f$ and $F'=f$). Then, by mean-value theorem, we have $$F(b)-F(a)=F'(\xi)(b-a),$$ for some $\xi\in (a,b)$, which is same as $$\int_a^b f(t) dt-\int_a^a f(t) dt=(\int_a^x f(t) dt)'|_{\xi} (b-a)$$ or $$\int_a^b f(t) dt=f(\xi)(b-a).$$ By dividing, we get $$f(\xi)=\frac{\int_a^b f(t) dt}{b-a}.$$

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Why is $F'(\xi) = \left( \int_a^x f(t) dt \right)'$? who mentioned that such an $x$ exists? Or do you mean take $x= \xi$? – gt6989b Feb 12 '14 at 0:48
Yes, derivative of $F$ in point $x=\xi$. – alans Feb 12 '14 at 0:52
You might want to mention where the continuity of $f$ is used, since it is necessary for the result to hold and slightly hidden in this proof of the result. – Pete L. Clark Feb 12 '14 at 1:28

The result you are inquiring about is often called the Mean Value Theorem for Integrals (or sometimes in older texts the "Second Mean Value Theorem" or something like that; I find numbered theorem names confusing). It is a rather standard result: for instance it is the first result in the "Integral Miscellany" chapter of my honors calculus notes.

(The result there is stated in a slightly more general form: if also $f,g: [a,b] \rightarrow \mathbb{R}$ with $f$ continuous and $g$ Riemann-integrable and non-negative then there is $c \in (a,b)$ with $\int_a^b fg = f(c) \int_a^b g$.)

The other two answers to this question give each of the standard proofs of the Mean Value Theorem for Integrals: one may either appeal to the Extreme Value Theorem and the Intermediate Value Theorem and the fact that if $f \leq g$ on an interval $[a,b]$ then $\int_a^b f \leq \int_a^b g$ or use the Fundamental Theorem of Calculus and then the Mean Value Theorem. For some reason I prefer the first argument although I'm not really sure why: maybe because the Mean Value Theorem is proved using the Intermediate Value Theorem anyway, it seems more basic. (Also the hypothesis of continuity of $f$ is used in the first argument in perhaps a more perspicuous way.) On the other hand the second argument is a bit easier...Finally, the more general version of the last paragraph is proved using the former method; since it involves the integral of a discontinuous function, I don't see at the moment how to prove it using the Fundamental Theorem of Calculus.

It also true that in the setting of the Mean Value Theorem, when we assume moreover that $f'$ is continuous on $[a,b]$, by applying the Mean Value Theorem for integrals to $f'$ and applying the other part of the Fundamental Theorem of Calculus, one deduces the conclusion of the Mean Value Theorem (in this special case, which would probably be sufficient for all "real-life applications").

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