integral mean value theorem

The Integral Mean Value Theorem.

If $f$ and $g$ are continuous real functions on an interval $\mathrm{[}a\mathrm{,}b\mathrm{]}$, and $g$ is additionally non-negative on $\mathrm{(}a\mathrm{,}b\mathrm{)}$, then there exists a $\zeta \mathrm{\in }\mathrm{(}a\mathrm{,}b\mathrm{)}$ such that

$${\int }_a^{b}f(x)g(x)dx=f(\zeta ){\int }_a^{b}g(x)dx.$$

Proof.

Since $f$ is continuous on a closed bounded set, $f$ is bounded and attains its bounds, say $f\left(x_0\right)\le f(x)\le f\left(x_1\right)$ for all $x\in [a,b]$. Thus, since $g$ is non-negative for all $x\in [a,b]$

$$f\left(x_0\right)g(x)\le f(x)g(x)\le f\left(x_1\right)g(x).$$

Integrating both sides gives

$$f\left(x_0\right){\int }_a^{b}g(x)dx\le {\int }_a^{b}f(x)g(x)dx\le f\left(x_1\right){\int }_a^{b}g(x)dx.$$

If ${\int }_a^{b}g(x)dx=0$, then $g(x)$ is identically zero, and the result follows trivially. Otherwise,

$$f\left(x_0\right)\le \frac{{\int }_a^{b}f(x)g(x)dx}{{\int }_a^{b}g(x)dx}\le f\left(x_1\right),$$

and the result follows from the intermediate value theorem. ∎

Title	integral mean value theorem
Canonical name	IntegralMeanValueTheorem
Date of creation	2013-03-22 17:15:56
Last modified on	2013-03-22 17:15:56
Owner	me_and (17092)
Last modified by	me_and (17092)
Numerical id	9
Author	me_and (17092)
Entry type	Theorem
Classification	msc 26A06
Related topic	EstimatingTheoremOfContourIntegral