Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Let $f:[x_1,x_2]\rightarrow \mathbb {R}$ be continuous and $g:[t_1,t_2]\rightarrow[x_1,x_2]$ be continuous and monotonic. Suppose $g(t_1)=x_1$ and $g(t_2)=x_2$ (in the case $g$ is increasing). Is it true that: $$\dfrac {\int _{x_1}^{x_2}f(x)dx}{x_2-x_1}=\dfrac {\int _{t_1}^{t_2}f(g(t))dt}{t_2-t_1} \quad ?$$ For example, if $g(t)=kt$ it is easily seen to be true: $$\dfrac {\int _{x_1}^{x_2}f(x)dx}{x_2-x_1}=\dfrac {\int _{t_1}^{t_2}f(g(t))g'(t)dt}{x_2-x_1}=\dfrac {\int _{t_1}^{t_2}f(g(t))dt}{t_2-t_1}$$

Thanks for your help.

share|cite|improve this question
up vote 1 down vote accepted

No. Let $x=g(t)$, then $dx = g'(t) dt$. Then

$$\frac{1}{x_2-x_1}\int_{x_1}^{x_2} dx \: f(x) = \frac{1}{x_2-x_1}\int_{t_1}^{t_2} dt \: g'(t) f(g(t)) $$

In your case, $g(t)=k t$, this is OK, but in general, no. Example, let $g(t) = k t^2$; then

$$\frac{1}{x_2-x_1}\int_{x_1}^{x_2} dx \: f(x) = \frac{2}{t_2^2-t_1^2}\int_{t_1}^{t_2} dt \: t f(k t^2) $$

share|cite|improve this answer
Ok, but this doesn't prove that: $$\dfrac{1}{x_2-x_1}\int _{t_1}^{t_2}dtg'(t)f(g(t)) \neq \dfrac{1}{t_2-t_1}\int _{t_1}^{t_2}dtf(g(t)) $$ – pppqqq Feb 17 '13 at 11:59
Ok, thanks, i got the counterexample. To complete yours, one may take: $f(x)=x \quad g(t)=t^2 \quad t_i ^2 = x_i $. – pppqqq Feb 17 '13 at 12:26

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.