Question regarding integration by substitution.

Question

The theorem on integration by substitution says that $$\int_{\phi(a)}^{\phi(b)}f(x)dx=\int_{a}^{b}f(\phi(t))\phi'(t)dt$$ provided that $\phi$ has an integrable derivative. My question is, shouldn't $\phi$ be monotonic on $[a,b]$? I have this doubt as I am unable to prove this using Riemann Sums. Can someone tell how this works?

The proof on Wikipedia assumes that $$F(\phi(b))-F(\phi(a))=\int_{\phi(a)}^{\phi(b)}f(x)dx$$ but for this to happen, $\phi(a)<\phi(b)$, and $\phi$ should be increasing on $[a,b]$, isn't it?

You might need Riemann Stieltje sums – dunno Jun 18 '17 at 19:34 — dunno, Jun 18 '17 at 19:34

hamam_Abdallah · Answer 1 · 2017-06-18T19:35:20.610

1

You do not need that $\phi $ is bijective.

$\phi $ continuously differentiable at $[a,b] $
$f $ continuous at $\phi ([a,b]) $.

For the proof, consider

$$g (x)=\int_{\phi (a)}^{\phi (x)} f(t)dt-\int_a^x f (\phi (t))\phi'(t)dt $$

and you show by FTC , that $g'(x)=0$. thus $$g (b)=g (a)=0$$

edited Jun 18 '17 at 19:35

answered Jun 18 '17 at 19:04

hamam_Abdallah

62,951

then how would you prove the theorem using Riemann sums? – john Jun 18 '17 at 19:05
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@john You use FTC. – hamam_Abdallah Jun 18 '17 at 19:34

Question regarding integration by substitution.

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