The Second Mean Value Theorem for Integrals says that for $f (x)$ and $g(x)$ continuous on $[a, b]$ and $g(x)\ge 0$ $$\int_a^bf(x)g(x)\,dx=f(a)\int_a^cg(x)\,dx+f(b)\int_c^bg(x)\,dx$$

I have a difficult time understanding what this means, as opposed to the first mean value theorem for integrals, which is easy to conceptualize. Is there a graphical or 'in words' interpretation of this theorem that I may use to understand it better?

  • $\begingroup$ en.wikipedia.org/wiki/Mean_value_theorem Check out the requirements of the second mean value theorem there. I think you need a further requirement on $f$, e.g. that it is increasing. $\endgroup$
    – muaddib
    Jun 24, 2015 at 21:58

1 Answer 1


I mentioned in a comment that you need more requirements on $f$ than just that is continuous. To give you a verbal explanation of the theorem I will assume it is non-decreasing. Then you can look at it as follows:

Since $f$ is non decreasing, $f(a)$ must be the minimum of $f$ over the interval, and $f(b)$ must be the maximum. Now it must be true that: $$\int_a^b f(x)g(x)dx \geq f(a) \int_a^b g(x) dx$$ and $$\int_a^b f(x)g(x)dx \leq f(b) \int_a^b g(x) dx$$

Now consider the function $F$ of $c$ given by $$F(c) = f(a)\int_a^c g(x)dx + f(b)\int_c^b g(x) dx$$ This function must satisfy $F(b) \leq \int_a^b f(x)g(x)dx$ and also $F(a) \geq \int_a^b f(x)g(x)dx$. Since it is continuous there must be a $c^*$ where equality holds. (By the intermediate value theorem).

So to put it in words. If you integrate a function $g$ from $a$ to $b$ and weight it by an increasing function $f$, then the weighted integral must be greater than the integral of $g$ times $f$'s min and less than the integral times $f$'s max. So there must be a point in between where $f$s min times some of $g$'s integral plus $f$'s max times the rest of $g$'s integral equals the total weighted integral.

  • $\begingroup$ Thank you, that clears it up. $\endgroup$ Jun 25, 2015 at 2:30
  • $\begingroup$ @Sky Happy to help. $\endgroup$
    – muaddib
    Jun 25, 2015 at 2:31
  • $\begingroup$ Is there a graphical interpretation when the nonnegativity condition for $g$ is dropped? $\endgroup$ Oct 16, 2020 at 15:26

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.