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Let $$f:[a,b]\rightarrow \mathbb R$$ be continuous . Show that there exists $c\in [a,b]$ s.t. $${{1}\over{b-a}}\int_a^b f(x)dx=f(c)$$

Trivially it holds for constant functions . So necessarily consider non-constant functions . The above equation looks somewhat similar to the MVT equation :$${{\psi(b)-\psi(a)}\over {b-a}} =\psi'(c)$$
If we consider $$\psi(t)=\int_a^t f(x)dx$$ and prove that this function $\psi$ is continuous and differentiable then showing $\psi' (c)=f(c)$ would bring the result . This was my idea ,. I don't know though if any of it is possible at all .

Will this tecnique work $?$

Thanks for your help.

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  • $\begingroup$ en.wikipedia.org/wiki/Mean_value_theorem $\endgroup$
    – Svetoslav
    Oct 2, 2015 at 19:26
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    $\begingroup$ Yes, that is exactly what you have to do. $\endgroup$
    – Chappers
    Oct 2, 2015 at 19:29
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    $\begingroup$ $\psi(t)=\int_a^t f(x)dx$ is continuously differentiable if $f$ is. math.stackexchange.com/questions/752545/… Might help. $\endgroup$
    – Almentoe
    Oct 2, 2015 at 19:31
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    $\begingroup$ That is the statement of the First fundamental theorem of integral calculus. $\endgroup$
    – Bernard
    Oct 2, 2015 at 19:31
  • $\begingroup$ @Almentoe : Thanks . That is exactly what I needed . $\endgroup$
    – user118494
    Oct 2, 2015 at 19:36

1 Answer 1

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One need not invoke the MVT for differentiable functions (although that approach will work since $\int_0^xf(t)\,dt$ is differentiable for $f$ continuous). Here, we work from properties of the integral and continuous functions only.

Since $f$ is continuous on $[a,b]$ it attains both its minimum $m$ and its maximum $M$ there. Then, we know that

$$m(b-a)\le \int_a^bf(x)\,dx\le M(b-a)\implies m\le\frac{1}{b-a}\int_a^bf(x)\,dx\le M$$

Now, we also know that since $f$ is continuous on $[a,b]$, that it attains all values between $m$ and $M$. Thus, for any number $m\le y\le M$, there exists a number $c\in[a,b]$ such that $y=f(c)$. Letting $y=\frac{1}{b-a}\int_a^bf(x)\,dx$, there exists a number $c\in[a,b]$, such that

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

where upon solving for $\int_a^bf(x)\,dx$ yields

$$\bbox[5px,border:2px solid #C0A000]{\int_a^bf(x)\,dx=f(c)(b-a)}$$

for some $c\in [a,b]$.

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