Try to generalize First Mean Value Theorem For Integrals

Question

1.

Suppose that $f(x)$ is (Riemann) integrable on $[a,b]$ and $F^{'}(x)=f(x)$ for all $x\in[a,b]$,then there is a number $\color{red}{\xi\in(a,b)}$ such that $$\int_{a}^{b}f(x)dx=f(\xi)(b-a)$$

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2.

I generalize the above statement which is ture to the following version:

Suppose that $f(x)$ and $g(x)$ are (Riemann)integrable on $[a,b]$ with $g(x)\geq 0$ for all $x\in[a,b]$ and $F$is a primitive function of $f$ on $[a,b]$ ,then there is a number $\color{red}{\xi\in(a,b)}$ such that $$\int_{a}^{b}f(x)g(x)dx=f(\xi)\int_{a}^{b}g(x)dx$$

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3.

I need to proof the generalisation is ture,or give some counterexamples to disprove it .

From the First mean value theorem for definite integrals,we have $$\int_{a}^{b}f(x)g(x)dx=\mu\int_{a}^{b}g(x)dx,\quad \inf_{x\in[a,b]}\{f(x)\}\leq \mu\leq \sup_{x\in[a,b]}\{f(x)\}.$$ If $$\inf_{x\in[a,b]}\{f(x)\}< \mu <\sup_{x\in[a,b]}\{f(x)\},$$ Darboux's theorem tells us there is a $\xi\in(a,b)$ such that $f(\xi)=\mu.$

If $$\mu=\inf_{x\in[a,b]}\{f(x)\}\quad \text{or}\quad \mu=\sup_{x\in[a,b]}\{f(x)\},$$

is there also a $\xi\in(a,b)$ such that $f(\xi)=\mu ?$

Answer 1

The first result is nothing but Langrange mean value theorem applied to anti-derivative $F$ of $f$. To get to your generalization we must assume the existence of anti-derivative of $fg$ and $g$ and apply Cauchy Mean Value Theorem on $F$ and $G$ where $F'=fg, G'=g$. I am not sure if the result holds without the existence of anti-derivatives involved.

Answer 2

In 3. we are given that $f$ and $g$ are Riemann integrable on $[a,b]$ and $g(x) \geqslant 0$ for all $x \in [a,b]$. We also are given that $f$ has a primitive $F$ implying that $f(x) = F'(x)$ for all $x \in [a,b]$.

Proof of the mean value theorem for integrals is, of course, straightforward under the stronger condition that $f$ is continuous. It is also true with this weaker condition that $f$ is a derivative. Let's first prove this carefully and then consider your final question where in one case $\mu = \inf f(x)$.

Since $f$ is Riemann integrable, it is bounded and there exist finite numbers $m = \inf_{x \in [a,b]}\, f(x)$ and $M = \sup_{x \in [a,b]}\, f(x)$. Since $mg(x) \leqslant f(x)g(x) \leqslant Mg(x)$ for all $x \in [a,b]$ we have

$$m\int_a^b g(x) \, dx \leqslant \int_a^b f(x) \, g(x) \, dx \leqslant M \int_a^b g(x) \, dx.$$

In the case where $\int_a^b g(x) \, dx = 0$ it is easy to show that we can choose any $\xi \in (a,b)$ and the theorem holds.

Otherwise take $\mu = \int_a^b f(x) \, g(x) \, dx / \int_a^b g(x) \, dx$. We know that $m \leqslant \mu \leqslant M$. If $m < \mu < M$, by the properties of infimum and supremum there exist $\alpha , \beta \in [a,b]$ such that

$$m < f(\alpha) < \mu < f(\beta) < M.$$

By Darboux's theorem $f$ (as a derivative) has the intermediate value property. Hence, there exists $\xi \in (\alpha,\beta) \subset [a,b]$ such that $f(\xi) = \mu$ and we are done.

Suppose however that $\mu = m$. Since $f(x) \geqslant m$ and $g(x) \geqslant 0$, we have

$$\int_a^b |f(x) - m| \, g(x) \, dx = \int_a^b (f(x) - m) \, g(x) \, dx = (\mu -m) \int_a^b g(x) \, dx = 0,$$

and it follows that $(f(x) - m) \, g(x) = 0$ almost everywhere. For this case where $\int_a^b g(x) \, dx > 0$ we have $g(x) > 0$ almost everywhere and there must be a point $\xi \in (a,b)$ such that $f(\xi) = m$.

The case where $\mu = M$ is handled in a similar way.