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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?

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  • $\begingroup$ 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 '15 at 21:58
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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.

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  • $\begingroup$ Thank you, that clears it up. $\endgroup$ – mysatellite Jun 25 '15 at 2:30
  • $\begingroup$ @Sky Happy to help. $\endgroup$ – muaddib Jun 25 '15 at 2:31

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