Prove that $\int_a^b f(x)dx=\int_{\alpha}^{\beta}f(\varphi(y))\varphi'(y)dy$

Question

Let $f:[a,b]\to\mathbb R$ continuous and $\varphi:[c,d]\longrightarrow [a,b]$ a surjective function in $\mathcal C^1([c,d])$. I denote $\alpha,\beta\in [c,d]$ such that $a=\varphi(\alpha)$ and $b=\varphi(\beta)$. I want to prove that

$$\int_a^b f(x)\mathrm dx=\int_{\alpha}^{\beta}f(\varphi(y))\varphi'(y)\mathrm dy$$

This is my proof:

Let $x=\varphi(y)$. Then $\mathrm dx=\varphi'(y)\mathrm dy$, and thus $$f(x)\mathrm dx=f(\varphi(y))\varphi'(y)\mathrm dy.$$

When $x$ traverses $[a,b]$, $\varphi(y)$ traverses $[\alpha,\beta]$, therefore $$\int_a^b f(x)\mathrm d x=\int_\alpha^\beta f(\varphi(y))\varphi'(y)\mathrm dy.$$

It looks a little bit ad-hoc, do you think it's correct ?

Answer 1

Here is how you make @Joe Aaron's comment more precise.

Let $F(x)$ be an antiderivative for $f$ (you could construct one if you needed that) and compute $\frac{d}{dx} F(\phi(x))$ and note that $F \circ \phi$ is an antiderivative for $f(\phi(x))\phi'(x)$. From there you just use the fundamental theorem of calculus.