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This question already has an answer here:

Problem:

$f(x)$ is a continuous function, and it is periodic with period $T$. For any $a<b$, prove that $$\lim_{n\to\infty}\int_a^bf(nx)dx=\frac{b-a}{T}\int_0^Tf(x)dx$$

I tried substituting $nx=t$, but it gave me $\frac{1}{n}\int_{na}^{nb}f(t)dt$, and I don't know what to do. Can anyone give me hints to solve this? Or is there another way to solve this problem?

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marked as duplicate by Guy Fsone, José Carlos Santos, Did, user99914, hardmath Jan 17 '18 at 2:44

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  • $\begingroup$ Roughly speaking, there are about $n(b-a)/T$ periods between $na$ and $nb$. $\endgroup$ – Joey Zou Jun 26 '16 at 16:14
  • $\begingroup$ @JoeyZou Looking at the problem gives me that intuitively but I don't understand why there would be $n(b-a)/T$ periods between those two numbers... $\endgroup$ – zxcvber Jun 26 '16 at 16:15
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Your start is exactly right. Now note that the integral $\int_{na}^{nb}f(t)dt$ covers some number of full periods of the periodic function (about $n(b-a)/T$ of them, give or take one), plus some extra stuff (whose absolute area can't be more than $TB$, if $B$ is a bound on the absolute value of $f$ -- which must exist, since $f$ is continuous). Can you prove that the "extra stuff" and the "give or take one" don't contribute in the limit as $n\rightarrow\infty$?

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  • $\begingroup$ Sorry, I think your hint is great but I still don't know how I should proceed. $\endgroup$ – zxcvber Jun 26 '16 at 16:20
  • $\begingroup$ May I ask for a full answer?? $\endgroup$ – zxcvber Jun 26 '16 at 16:25

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