Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $f$ and $g$ are continuous functions on $[a,b]$ and that $g(x)\ge 0$ for all $x\in[a,b]$. Prove that there exists $x$ in $[a,b]$ such that $$\int_a^b f(t)g(t)dt=f(x)\int_a^b g(t)dt$$

I think I need to do something with this theorem:

Intermediate Value Theorem for Integrals

If $f$ is a continuous function on $[a,b]$ then for at least one $x$ in $[a,b]$ we have $$f(x)=\frac{1}{b-a} \int_a^bf$$

share|cite|improve this question
up vote 4 down vote accepted

Let $m = \min \{f(x): x \in [a,b]\}$ and $M = \max \{f(x): x \in [a,b]\}$.

First let us assume that $\displaystyle \int_a^b g(x)dx > 0$. Then we have that $$m \leq \dfrac{\displaystyle \int_a^b f(x) g(x) dx}{\displaystyle \int_a^b g(x)dx} \leq M$$ Now use intermediate value theorem to get what you want.

If $\displaystyle \int_a^b g(x) dx = 0$ and since $g(x) \geq 0$ and is continuous, we have that $g(x) = 0$ on $[a,b]$. Hence, $$\displaystyle \int_a^b f(x) g(x) dx = \displaystyle \int_a^b g(x)dx =0$$ Hence, $$\displaystyle \int_a^b f(x) g(x) dx = f(t) \displaystyle \int_a^b g(x)dx $$ for any $t \in [a,b]$.

share|cite|improve this answer
Note that $g(x)\geq 0$, not $g(x)>0$. – Mario Carneiro Dec 15 '12 at 22:55
@MarioCarneiro Thanks. Have updated it. – user17762 Dec 15 '12 at 23:14
I don't understand how you get this: $$m \leq \dfrac{\displaystyle \int_a^b f(x) g(x) dx}{\displaystyle \int_a^b g(x)dx} \leq M$$ – Kasper Dec 15 '12 at 23:16
@Kasper since $m \leq f(x) \leq M$, we have $m g(x) \leq f(x) g(x) \leq M g(x)$ as $g(x) \geq 0$. Now integration preserves $ \leq $ and $\geq$. Hence, you get I have written. – user17762 Dec 15 '12 at 23:19
Since $g(x)\geq 0$, $f(x)g(x)\leq Mg(x)$, so $\int_a^bf(x)g(x)\,dx\leq\int_a^bMg(x)\,dx=M\int_a^bg(x)\,dx$. Thus $\dfrac{\int_a^bf(x)g(x)\,dx}{\int_a^bg(x)\,dx}\leq M$. A similar argument holds for $m$. (Edit: Ninja'd!) – Mario Carneiro Dec 15 '12 at 23:20

Define $$ \bar{f}=\frac{\int_a^bf(t)\,g(t)\,\mathrm{d}t}{\int_a^bg(t)\,\mathrm{d}t}\tag{1} $$ Then $$ \int_a^b\left(f(t)-\bar{f}\right)\,g(t)\,\mathrm{d}t=0\tag{2} $$ Suppose that $f(t_+)-\bar{f}\gt0$ and $f(t)-\bar{f}\ge0$ for all $t\in[a,b]$, then $f-\bar{f}$ is positive in some neighborhood of $t^+$ and therefore $\int_a^b\left(f(t)-\bar{f}\right)\,g(t)\,\mathrm{d}t\gt0$. Thus, there must be some $t_-$ where $f(t_-)-\bar{f}\lt0$.

Suppose that $f(t_-)-\bar{f}\lt0$ and $f(t)-\bar{f}\le0$ for all $t\in[a,b]$, then $f-\bar{f}$ is negative in some neighborhood of $t_-$ and therefore $\int_a^b\left(f(t)-\bar{f}\right)\,g(t)\,\mathrm{d}t\lt0$. Thus, there must be some $t_+$ where $f(t_+)-\bar{f}\gt0$.

Thus, if $f(t)-\bar{f}$ is not identically $0$ on $[a,b]$, we must have $t_+$ and $t_-$ where $f(t_+)-\bar{f}\gt0$ and $f(t_-)-\bar{f}\lt0$.

By the intermediate value theorem, there must be an $x$ between $t_+$ and $t_-$ such that $f(x)-\bar{f}=0$, which is the same as $$ \int_a^bf(t)\,g(t)\,\mathrm{d}t=\bar{f}\int_a^bg(t)\,\mathrm{d}t=f(x)\int_a^bg(t)\,\mathrm{d}t\tag{3} $$

share|cite|improve this answer

Let $h(x)$ be an antiderivative of $g(x)$, so that $h'(x)=g(x)$. Then using the substitution $$u=h(t)\Rightarrow du=g(t)\,dt$$ we get $$\int_{h^{-1}(a)}^{h^{-1}(b)}f(t)\,du=f(x)\int_{h^{-1}(a)}^{h^{-1}(b)}du=f(x)(h^{-1}(b)-h^{-1}(a))$$ which we can validate using the IVT for integrals.

share|cite|improve this answer
Unfortunately, this proof only works for $g(t)>0$ on $[a,b]$, or at best $g(t)=0$ at countably many points, since it relies on $h^{-1}(x)$, which may not exist if $h$ is not 1-1, since it may not be strictly increasing. Marvis' proof is more general. – Mario Carneiro Dec 15 '12 at 23:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.