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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.

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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

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

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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
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