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I am getting confused by the Mean value theorem for integrals, differently stated in different sources. The idea is that for any continuous functions $f$ on ana interval $[a,b]$, there exists some $c$ such taht $$ \int_a^b f(x)dx =f(c)(b-a) $$ My question is: should $c$ be in $(a,b) $ or $[a,b]$ ?

In Wikipédia it is stated with $c$ in the closed interval, however in the proof, where they used the Intermediate value theorem, they took $c$ in the open inetrval.

Could someone clarify it for me ?

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    $\begingroup$ $c\in (a,b)$. $ \ \ \ $ $\endgroup$ – Surb Apr 21 at 11:35
  • $\begingroup$ Strictly speaking, they only get $\frac1{b-a}\int_a^b f \in [f(m),f(M)]$, which does not allow them to apply their stated version of the intermediate value theorem, which needs $\frac1{b-a}\int_a^bf \in (f(m),f(M))$. But if its is equal to $f(m)$ or $f(M)$, you can show that $f$ is actually a constant on $[a,b]$, so it doesn't matter $\endgroup$ – Calvin Khor Apr 21 at 12:18
  • $\begingroup$ Let $F(x) =\int_a^x f(t) \, dt$ then applying usual mean value theorem on $F$ shows that $c\in(a, b) $ works fine. $\endgroup$ – Paramanand Singh Apr 22 at 6:03

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