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Suppose that $f:[a,b]\rightarrow R$ is continuous and $g:[a,b]\rightarrow R$ is integrable and such that $g(x) \ge 0$ for all $x \in [a, b]$. Prove that there is a number $c$ in $[a,b]$ such that

$$ \int_{a}^{b}f(x)g(x)dx = f(c)\int_{a}^{b}g(x)dx $$

My proof:

Consider $F(x) =f(x)\int_{a}^{b}g(x)dx$, $F$ is continuous since it's a scalar multiplication of $f$. Moreover, $m\int_{a}^{b}g(x) \le \int_{a}^{b}f(x)g(x)dx\le M\int_{a}^{b}g(x)$, if $m\le f(x)\le M$ (by extremum value theorem).

The statement follows by an application of Intermediate value theorem on $F$.

I suspect that my proof is flawed since I didn't use $g(x)\ge 0$ in my proof (at least not explicitly). Could you please check my attempt and give me hints to why it's flawed? I prefer hints to complete solutions.

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  • $\begingroup$ Why is $m \int_a^b g(x)\,dx \leqslant \int_a^b f(x)g(x)\,dx$? $\endgroup$ Aug 18 '20 at 21:18
  • $\begingroup$ It's because $m \le f(x) \le M$ on the interval $[a,b]$ and if $ f(x) \le g(x)$, then $\int f(x) \le \int g(x)$. $\endgroup$ Aug 18 '20 at 21:22
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    $\begingroup$ So why is $m g(x) \leqslant f(x)g(x)$? $\endgroup$ Aug 18 '20 at 21:23
  • $\begingroup$ @DanielFischer Ahhh. Thank you! $\endgroup$ Aug 18 '20 at 21:23

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