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Prove that $$\int_a^b f(x)\,dx = c\int_{a/c}^{b/c} f(cy)\,dy$$

I've tried to use subtitution rule, but I'm afraid I can't do this if f is not continuous.


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you can use substitution. It is legal here. – Sabyasachi Mar 15 '14 at 14:36
can you explain please why it's legal here? – Shirly Geffen Mar 15 '14 at 14:39
The substitution $y=x/c$ is differentiable. – Unwisdom Mar 15 '14 at 14:41
@Unwisdom has explained it. – Sabyasachi Mar 15 '14 at 14:42

Substituting $x=g\left(y\right)$ is allowed here under certain conditions on function $g$ (so not conditions on the integrand $f$). Here you do that with function $g\left(y\right)=cy$ which is okay if $c\neq0$.

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This really answers the question. The asker forgot to mention that $c \neq 0$, and you brought this up. Yes. This is certainly true. – NasuSama Mar 15 '14 at 14:45

Let $x=cy$. Therefore, $dx=c dy$, and the limits become the desired ones: $a/c$ and $b/c$. Therefore, the integral is:

$$\int_a^b f(x)\,dx = c\int_{a/c}^{b/c} f(cy)\,dy$$

Which is what we wanted.

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does not answer the question. – Sabyasachi Mar 15 '14 at 14:43
@Sabyasachi. Why ? This is correctly proven as long as $c$ is not zero. – Claude Leibovici Mar 15 '14 at 14:50
@ClaudeLeibovici the OP did consider the substitution. (S)he isn't sure if it is correct. That explanation is required. – Sabyasachi Mar 15 '14 at 15:03
The OP's confusion regarding the requirement on the continuity of $f$(which is wrong) should also have been cleared up. – Sabyasachi Mar 15 '14 at 15:04

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