3

Let $f,g:[a,b] \to \mathbb{R}$ continuous, $g(x)>0 \forall x \in [a,b]$. Then there is $c \in (a,b)$ such that $\int_{a}^{b} f(x)g(x)dx = f(c)\int_{a}^{b} g(x)$

My attempt:

By mean value of integration, as $fg$ is continuous on $[a,b]$, there is $c \in (a,b)$ such that $\int_{a}^{b}f(x)g(x)dx=f(c)g(c)(b-a)$

Also there is $d \in (a,b)$ such that $\int_{a}^{b}g(x)dx=g(d)(b-a)$

If I could make $c=d$, then it would be done, because then $f(c)g(c)(b-a) = f(c)\int_{a}^{b} g(x)$. But I think I can't just choose them to be equal. Also I am not using $g(x)>0$, so definitely there is something wrong with my attempt.

Thanks.

  • 2
    Your thinking is correct, but try to use the mean value theorem with a different function. (Hint: it is written in the claim.) – Guy Sep 26 '18 at 20:23

2 Answers2

4

$f$ is continuous at $[a,b]$ . so it is bounded. Put $f[a,b])=[m,M]$.

For $x\in[a,b]$,

$$m\le f(x)\le M$$

thus

$$mg(x)\le g(x)f(x)\le Mg(x)$$ since $g(x)>0.$

by integrating

$$m\int_a^bg\le \int_a^bfg\le M\int_a^bg$$

but $\int_a^bg>0$, by division

$$m\le \frac{\int_a^bfg}{\int_a^bg}\le M$$

$$\implies \frac{\int_a^bfg}{\int_a^bg}\in f([a,b])$$ Done.

hamam_Abdallah
  • 62,951
3

Consider the constant

$$A:= \frac{\int_a^b f(x)g(x)dx}{\int_a^b g(x)dx}$$

I claim that $\min f \leq A \leq \max f$. This is because $\min f \cdot \int_a^b g(x)dx \leq \int_a^b f(x)g(x)dx \leq \max f \cdot \int_a^b g(x)dx$.

Since $f$ is continuous you can conclude that there exists $c$ such that $f(c)=A$.

Yanko
  • 13,758