3
$\begingroup$

I was thinking about integrals where we should be careful with the domain because we perform a trivial but not 1-1 change of variables. For example, in $$\int_{-1}^2\dfrac{4x^3dx}{1+x^4},$$ the classic change of variables $u=x^4$ is not injective; it can be evaluated simply by using the FTC, whose application here doesn't even see the fact that the $x^4$ is not 1-1. This fact actually freaked me out a bit: most of the examples that I know of for change of variable can be replaced by a direct (albeit uglier) application of the FTC, completely bypassing the bijectivity and other issues associated with change of variable.

So, I am wondering whether there really is a gap in my reasoning, that is, whether $$\int_a^b f'(g(x))g'(x)dx=f(g(b))-f(g(a))$$ actually always holds (assuming of course continuous differentiability), no matter how many oscillations $f$ and $g$ have, no matter how many pre-images every $g(x)$ has, etc.

In the above example, I would actually like to perform a change of variable $u=g(x),$ then—just then—apply the FTC; but that would require injectivity of $g,$ etc.

$\endgroup$
2
  • 2
    $\begingroup$ See also Definite integral with non-injective u-substitution. There’s no need to get into the discussion of “change variables” $u=g(x)$ either; this follows immediately by the chain rule (in reverse) and FTC. A theorem is a theorem (true statement); the reason people get all sorts of confusions is because they don’t apply it correctly/try to use it in situations where its hypotheses aren’t satisfied. $\endgroup$
    – peek-a-boo
    Feb 1, 2023 at 21:19
  • $\begingroup$ It helps to actually read carefully the statement and proof (if interested) of the theorem related to change of variables. A simpler version assumes continuity of $f$ as well as continuity of $g'$ and is proved easily using FTC. It is not necessary that $g$ is a bijection. $\endgroup$
    – Paramanand Singh
    Feb 2, 2023 at 4:53

1 Answer 1

2
$\begingroup$

I am wondering whether $$\int_a^b f'(g(x))g'(x)dx=f(g(b))-f(g(a))$$ actually always holds (assuming of course continuous differentiability),

In a word: yes.

In fact, it is sufficient that $g'$ is integrable (this is implied by $g$ being continuously differentiable). Full statement of the theorem.

but that would require injectivity of $g,$

No such inherent requirement, except precisely when the new variable (say, $u$) is an implicit function of the starting variable (say, $x$). In this case (say, with the substitution $x=\phi(u)$), the new integration limit $g(b)$ unambiguously equals $$u_b=\phi^{-1}(b)$$ due to the substitution being invertible.

Occasionally, like here, the substitution is substitutable into the given integrand only piecewise (as a consequence, every interval of application of the change-of-variable theorem has an injective substitution); even here, it is never unnecessary to separately verify that substitutions are injective or monotonic.

(That linked answer also contains a trivially valid non-injective-substitution example.)

$$\int_{-1}^2\dfrac{4x^3dx}{1+x^4}.$$ can be evaluated simply by using the FTC, whose application here doesn't even see the fact that the $x^4$ is not 1-1. This fact actually freaked me out a bit: most of the examples that I know of for change of variable can be replaced by a direct (albeit uglier) application of the FTC, completely bypassing the bijectivity and other issues associated with change of variable.

Your observation is pertinent, since change of integration variable is really just reversing the chain rule together with applying the FTC. Indeed, none of these applications really inherently require injectivity. It sounds like you will appreciate this integration example, where an algebra mistake is mistaken as integration variable change being inconsistent (apparently breaking down without an injective substitution, yet apparently never asking for an injective substitution), and where directly applying the FTC to the integral with that algebra mistake transplanted in similarly makes the FTC appear defective.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .