2
$\begingroup$

I had a question about Taylor's theorem proof in here:

and the key point in the answer given to me was this:

there's $c\in(a,b)$ such that

$$\int_a^b f(x)g(x)dx=f(c)\int_a^b g(x)dx$$

Can anyone help me identify this rule? Thank you! =)

$\endgroup$
3
$\begingroup$

Mean value theorem for integration. In case you want to know how this name came about: taking $g=1$, we see that there is $c$ with $$f(c)=\frac{1}{b-a}\int_a^b f \ dx,$$ the right hand side of which is called the integral mean of $f$ over $[a,b]$. (These integrals are fairly important, e.g. because it appears in the Lebesgue differentation theorem; in fact so important that there is even a standard notation. Well it should be without the absolute value bars, but somehow no code seems to work here.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for the additional info, appreciate it! =) I am very interested =) $\endgroup$ – jjepsuomi Jul 15 '14 at 19:14
  • 1
    $\begingroup$ @jjepsuomi: no problem. I thought I should add some information apart from the name itself. $\endgroup$ – user164074 Jul 15 '14 at 19:16
  • $\begingroup$ @user164074 and a good idea it was ;D thank you! =) $\endgroup$ – jjepsuomi Jul 15 '14 at 19:17

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.