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


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.


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 $c\in [a,b]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.