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I'm trying to understand the proof for the mean value theorem of integrals. I was looking at the following method of proof where after using the extreme value theorem to obtain a maximum and minimum on a closed interval we obtain the following inequality,

$$f\left(m\right)\leq\frac{1}{b-a}\int_{a}^{b}f\left(x\right)dx\leq f\left(M\right)$$ Where $$M:=\max\{f\left(x\right)|x\in[a,b]\}$$

$$m:=\min\{f\left(x\right)|x\in[a,b]\}$$

At this point, the proofs I've seen apply the intermediate value theorem. My question is, to what function are we applying the intermediate value theorem? I.e. Are we arguing that $g(x)=\frac{1}{b-a}\int_{a}^{b}f\left(x\right)dx$ will obtain all values between $f(m)$ and $f(M)$? But how do we know that, as our inequality only states that our function is bounded by these two values.

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No.We are not arguing that $g(x)=\frac{1}{b-a}\int_{a}^{b}f\left(x\right)dx$, because expression $\frac{1}{b-a}\int_{a}^{b}f\left(x\right)dx$ is constant value(doesn't depends on $x$), which lies in $[f(m),f(M)]=R$.

Since, continuous function $f$ takes every value in it's range $R$( By IVT), we hace $f(c)=\frac{1}{b-a}\int_{a}^{b}f\left(x\right)dx$, for some $c\in [a,b]$.

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