let $f:[a,b] -> R $ be a continous function and $g : [a,b] - R$ a no negative integrable funtion , prove that there exists a $c$ in $(a, b)$ such that $$\int_a^b{f(x)g(x)dx}=f(c)\int_a^bg(x)dx$$ obviously I need help for the case in which g(x) is different from 0, I tried to found a lower and upper bound, but I just can't get it
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1See e.g. How do I prove this form of mean value theorem for integral? – Minus One-Twelfth Aug 30 '19 at 23:41
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let $\mu = dg$. then $\frac{1}{\int_a^b g(x)dx}\int_a^b f(x)g(x)dx = \frac{1}{\mu([a,b])}\int_a^b fd\mu$ is an average of $f$, so there's some point at which $f$ achieves it – mathworker21 Aug 30 '19 at 23:49
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@MinusOne-Twelfth my bad, I didn't find that one, but hey, I have a question, can you always just integrate an inequality and it will hold? – rorod8 Aug 31 '19 at 01:12
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Yes pretty much, have a read of this section of Wikipedia: https://en.wikipedia.org/wiki/Integral#Inequalities. – Minus One-Twelfth Aug 31 '19 at 01:16
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Let $c$ be the point in tthe interval where $f(c)$ is maximum. $\int_a^b f(x)g(x)dx \le \int_a^b f(c)g(x)dx=f(c)\int_a^b g(x)dx$, since $g(x)\ge 0$.
herb steinberg
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