Why does $u$-substitution require the variable to be injective?

Question

I read this answer but I didn't understand it. I expect a simple yet satisfying answer

Why does $u$-substitution require the variable to be injective? What's the reason for that? I didn't understand.

It doesn't require any injectivity assumptions; see the answer by Christian Blatter — user580918, Aug 30 '20 at 06:08
@user580918 I believe it is a case of "piecewise injective function". — Niraj Raut, Aug 30 '20 at 06:26
@Ryan But why does it require injective assumptions if the interval is not split up? Why does it? Why should it be injective in the first place? — Niraj Raut, Aug 30 '20 at 08:24
Please Include some details of what you do not understand specifically. Quote the relevant parts. — quid, Aug 30 '20 at 13:16

ryang · Accepted Answer · 2020-10-17T13:41:32.243

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EDIT:I consolidated both answers here: https://math.stackexchange.com/a/2518470/21813

edited Oct 17 '20 at 13:41

answered Aug 30 '20 at 12:56

ryang

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score 1 · Answer 2 · answered Aug 30 '20 at 13:00

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The general formula $$\int_{g(a)}^{g(b)}f(u)\mathrm{d}u=\int_a^bf(g(x))g'(x)\mathrm{d}x$$

is valid even if $g$ is not injective. But most time, we want to write something like $$\int_{a}^{b}f(u)\mathrm{d}u=\int_{g^{-1}(a)}^{g^{-1}(b)} f(g(x))g'(x)\mathrm{d}x$$

which of course require some inversibility of $g$.

answered Aug 30 '20 at 13:00

TheSilverDoe

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"...which of course requires some inversibility of $g$" — How does this prove that it needs to be injective? – Niraj Raut Aug 30 '20 at 17:27
"$g$ is injective" means that $g$ is a bijection on its image, i.e. that $g$ is invertible. – TheSilverDoe Aug 30 '20 at 22:06

Why does $u$-substitution require the variable to be injective?

2 Answers2