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The statement of the rule is the following: For a continuous function $f:I\rightarrow\mathbb{R}$ on a real interval $I$, and a continuously differentiable function $\phi:[a,b]\rightarrow I$, it holds that \begin{equation}\int_a^bf(\varphi(t))\cdot\varphi'(t)dt=\int_{\varphi(a)}^{\varphi(b)}f(x)dx\tag{1}\end{equation}

My question is how one is supposed to actually use this equation, or how to make it intuitively agree with what one is actually doing, mechanically, when integrating by substitution. This has always bugged me, and I give the precise difficulties in my understanding below.

I first learned integration by substitution 'procedurally', and was confronted with the actual theorem only later. To demonstrate what I mean by procedurally, consider this example integral, disregarding whether integration by substitution is useful for solving it, as that is not the purpose of my question.

$$\int_1^2 t\exp(\frac{t}{2})dt$$

Naively, I could try to do the following: $$x:=\varphi(t)=\frac{t}{2} \implies t=2x$$ $$\frac{dx}{dt}=\frac{1}{2} \implies \text{'}dt=2dx\text{'}$$ and since I 'replace $x$ by $\frac{t}{2}$' the limits of integration become $\frac{1}{2}=0.5$ and $\frac{2}{2}=1$. Substituting $x$ and $dx$ in my integral accordingly, I arrive at

$$\int_{0.5}^1 2x \exp(x) 2dx$$

Now I see that the function $f$ which 'fits' my chosen $\varphi$ 'in hindsight' is $f(x)=4x \exp(x)$, and indeed $f(\varphi(t))=2t\exp(\frac{t}{2})$ and $\varphi'(t)=\frac{1}{2}$, which allows me to write my original integral in the form $$\int_1^2 2t\exp(\frac{t}{2})\frac{1}{2}dt$$ which agrees with the left-hand-side of (1) and thus justifies what I did after the fact, since $f$ is continuous and $\varphi$ is continuously differentiable on $[1,2]$.

However, assume I do not know the above procedure, and instead only equation (1). How could I apply it, not yet knowing $f$? If I were to just chose $\varphi(t)=\frac{t}{2}$ as above, I would be left with this partially substituted integral:

$$\int_1^2t\exp(\varphi(t))dt$$ and proceed how? To go strictly by the formula, I would still need to find $f$, whose form isn't immediate from the integral (or at least it isn't obvious to me that it would always be reasonably apparent, in particular for more complicated integrals).

You might object that I could start on the other side of equation (1), defining $f(x)=x\exp(\frac{x}{2})$, choose $\varphi(t)=2t$, and immediately obtain $$f(\varphi(t))\varphi'(t)=2t\exp(t)2$$ However, this leads to a problem with the limits of the integral: The rule for integration by substitution does not require $\varphi$ to have an inverse. And even if it did, then to make the equation agree with what is actually done it would be more sensible to write $$\int_a^b f(x)dx = \int_{\varphi^{-1}(a)}^{\varphi^{-1}(b)}f(\varphi(t))\varphi'(t)dt$$ because this is what I'm actually doing.

How to amend one's mental model so equation (1) and the actual procedure agree with each other?

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    $\begingroup$ In problem solving it is often the case that for a transformation to fit a problem perfectly you either need to know that it will, by experience, or know how to fix what you have such that it fits. There is nothing to amend, but you coming to terms with the fact that that is the case in the use of integration by substitution. $\endgroup$
    – plop
    Commented Jun 3, 2021 at 2:12
  • $\begingroup$ +1 Agree with @Plop: experience. $\endgroup$ Commented Jun 3, 2021 at 17:21

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