Change of variable vs Fundamental Theorem of Calculus

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.

• 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. Feb 1, 2023 at 21:19
• 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. Feb 2, 2023 at 4:53

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.

$$\int_{-1}^2\dfrac{4x^3dx}{1+x^4}.$$ 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.