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 $f(x)$ is integrable on $[a,b]$, with $f(x)\ge 0$ on $[a,b]$, and that $g(x)$ is continuous on $[a,b]$. Assuming that $f(x)g(x)$ is integrable on $[a,b]$, show that $\mathrm{\exists}c\in [a,b]$ so that

${\int}_{a}^{b}f(x)g(x)dx=g(c){\int}_{a}^{b}f(x)dx.$

Thank you in advance.

Suppose $f(x)$ is integrable on $[a,b]$, with $f(x)\ge 0$ on $[a,b]$, and that $g(x)$ is continuous on $[a,b]$. Assuming that $f(x)g(x)$ is integrable on $[a,b]$, show that $\mathrm{\exists}c\in [a,b]$ so that

${\int}_{a}^{b}f(x)g(x)dx=g(c){\int}_{a}^{b}f(x)dx.$

Thank you in advance.

Marley Stone

Beginner2022-09-18Added 13 answers

$g(x)$ takes its maximum and minimum values $M$ and $m$ on $[a,b]$. Say, $g({x}_{1})=M$ and ${g}_{(}{x}_{2})=m$. Then

$g({x}_{1}){\int}_{a}^{b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx=M{\int}_{a}^{b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx\ge {\int}_{a}^{b}f(x)g(x)\phantom{\rule{thinmathspace}{0ex}}dx\ge m{\int}_{a}^{b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx=g({x}_{2}){\int}_{a}^{b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx.$

Can you now see why there is a $c\in [a,b]$ with

${\int}_{a}^{b}f(x)g(x)\phantom{\rule{thinmathspace}{0ex}}dx=g(c){\int}_{a}^{b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx?$

$g({x}_{1}){\int}_{a}^{b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx=M{\int}_{a}^{b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx\ge {\int}_{a}^{b}f(x)g(x)\phantom{\rule{thinmathspace}{0ex}}dx\ge m{\int}_{a}^{b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx=g({x}_{2}){\int}_{a}^{b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx.$

Can you now see why there is a $c\in [a,b]$ with

${\int}_{a}^{b}f(x)g(x)\phantom{\rule{thinmathspace}{0ex}}dx=g(c){\int}_{a}^{b}f(x)\phantom{\rule{thinmathspace}{0ex}}dx?$

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