Ivan Buckley

2022-09-17

I received a question on a previous exam, but I had no clue how to go about doing it. I know I'm supposed to use the MVT, IVT and FTC, but I'm not sure where. The question is
Suppose $f\left(x\right)$ is integrable on $\left[a,b\right]$, with $f\left(x\right)\ge 0$ on $\left[a,b\right]$, and that $g\left(x\right)$ is continuous on $\left[a,b\right]$. Assuming that $f\left(x\right)g\left(x\right)$ is integrable on $\left[a,b\right]$, show that $\mathrm{\exists }c\in \left[a,b\right]$ so that
${\int }_{a}^{b}f\left(x\right)g\left(x\right)dx=g\left(c\right){\int }_{a}^{b}f\left(x\right)dx.$

Marley Stone

$g\left(x\right)$ takes its maximum and minimum values $M$ and $m$ on $\left[a,b\right]$. Say, $g\left({x}_{1}\right)=M$ and ${g}_{\left(}{x}_{2}\right)=m$. Then
$g\left({x}_{1}\right){\int }_{a}^{b}f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx=M{\int }_{a}^{b}f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx\ge {\int }_{a}^{b}f\left(x\right)g\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx\ge m{\int }_{a}^{b}f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx=g\left({x}_{2}\right){\int }_{a}^{b}f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx.$
Can you now see why there is a $c\in \left[a,b\right]$ with
${\int }_{a}^{b}f\left(x\right)g\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx=g\left(c\right){\int }_{a}^{b}f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx?$

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