# State of the Mean Value theorem for inetgrals

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 ?

• $c\in (a,b)$. $\ \ \$ – Surb Apr 21 at 11:35
• 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 – Calvin Khor Apr 21 at 12:18
• 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. – Paramanand Singh Apr 22 at 6:03