Integration by substitution: follow-up

This is a follow-up question to the one I ask here. Gono’s answer indicates that the following is true:

Suppose that $f:I \to \Bbb R$ is continuous and $\varphi: [a,b] \to I$ is continuously differentiable.
Then it holds $$
\int _{{a}}^{{b}}f(\varphi (t))\cdot \varphi ‘(t)\,{\mathrm {d}}t=\int _{{\varphi (a)}}^{{\varphi (b)}}f(x)\,{\mathrm {d}}x
This holds regardless of whether $\varphi$ is injective

I had been under the impression that this was only guaranteed to be valid if $\varphi$ is injective, since injectivity is one of the conditions given in a multivariate setting. In fact, I thought that I had a counterexample in my question. Note, however, that with the example I gave, the $f(\varphi(t))\varphi'(t)$ resulting from my substitution failed to coincide with the integrand $t^4$ over $[-1,1]$.

So, here’s my question: is the above statement true? Is there a corresponding statement that holds in the multivariate generalization regardless of whether $\varphi$ is injective?

Solutions Collecting From Web of "Integration by substitution: follow-up"

\begin{align*} F(u) &= \int _{{a}}^{{u}}f(\varphi (t))\cdot \varphi ‘(t)\,{\mathrm {d}}t \\ G(u) &=\int _{{\varphi (a)}}^{{\varphi (u)}}f(x)\,{\mathrm {d}}x\end{align*}

Then $F$ and $G$ are functions with the same derivative everywhere (namely $F'(u)= f(\varphi (u))\cdot \varphi ‘(u)$, and with the same value (zero) at $u= a$ Thus they have the same value at $b$. In short, this is really just an application of two parts of the FTC, plus the chain rule.

For the multivariate setting, the question of orientation becomes really critical, and I don’t know an easy way to say the right multi-variable statement. It’s something like “if φφ maps the boundary of the domain in an orientation-preserving way, and [probably some other stuff here], then the integrals are the same.” But I’ve never seen it written out. We have the advantage in 1-dimensional integrals that reversing the limits means something. 🙂