# integral mean value theorem

###### The Integral Mean Value Theorem.

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

 $\int_{a}^{b}\!{f(x)g(x)}\,\mathrm{d}{x}=f(\zeta)\int_{a}^{b}\!{g(x)}\,\mathrm{% d}{x}.$
###### Proof.

Since $f$ is continuous on a closed bounded set, $f$ is bounded and attains its bounds, say $f\left(x_{0}\right)\leq f(x)\leq f\left(x_{1}\right)$ for all $x\in[a,b]$. Thus, since $g$ is non-negative for all $x\in[a,b]$

 $f\left(x_{0}\right)g(x)\leq f(x)g(x)\leq f\left(x_{1}\right)g(x).$

Integrating both sides gives

 $f\left(x_{0}\right)\int_{a}^{b}\!{g(x)}\,\mathrm{d}{x}\leq\int_{a}^{b}\!{f(x)g% (x)}\,\mathrm{d}{x}\leq f\left(x_{1}\right)\int_{a}^{b}\!{g(x)}\,\mathrm{d}{x}.$

If $\int_{a}^{b}\!{g(x)}\,\mathrm{d}{x}=0$, then $g(x)$ is identically zero, and the result follows trivially. Otherwise,

 $f\left(x_{0}\right)\leq\frac{\int_{a}^{b}\!{f(x)g(x)}\,\mathrm{d}{x}}{\int_{a}% ^{b}\!{g(x)}\,\mathrm{d}{x}}\leq f\left(x_{1}\right),$

and the result follows from the intermediate value theorem. ∎

Title integral mean value theorem IntegralMeanValueTheorem 2013-03-22 17:15:56 2013-03-22 17:15:56 me_and (17092) me_and (17092) 9 me_and (17092) Theorem msc 26A06 EstimatingTheoremOfContourIntegral