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In Michael Spivak's Calculus (p.368), he makes the following remark about integration by substitution:

The substitution formula is often written in the form $$\int f(u) \, du= \int f(g(x))g'(x) \, dx, \quad u=g(x) \, .$$This formula cannot be taken literally (after, all $\int f(u) \, du$ should mean a primitive of $f$ and $\int f(g(x))g'(x) \, dx$ should mean a primitive of $(f \circ g) \cdot g'$; these are certainly not equal). However, it may be regarded as a symbolic summary of the procedure which we have developed.

I don't understand what Spivak is referring to here. Say $F$ is a primitive of $f$. Then, \begin{align} \text{LHS} &= \int f(u) \, du = F(u)+C \\ \text{RHS} &= \int f(g(x))g'(x) = F(g(x))+C=F(u)+C \end{align} I'm also not quite on board with the view that the LHS is a primitive of $f$. If this were true, then wouldn't we simply write $$ \int f(u) \, du = F \, ? $$ I think if we were being precise then we would say the LHS is a primitive of $f$, evaluated at the point $u$. I'm probably misunderstanding something here, but I'm not what it exactly is.

I am aware that this question has been asked before, but I wasn't entirely satisfied with the answer, and I'm worried that I might have a deeper confusion about integration in general.

Joe
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  • I'm not sure why you say you're not on board with the view that $\int f(u),du$ should mean a primitive of $f$. That's just by definition. The occurrence of the letter $u$ is irrelevant. By definition when we use the $\int$ symbol without any bounds we're literally asking the question of finding a function whose derivative is the one we started with (i.e a primitive). I suggest you take a look at this previous answer or mine and links therein where I try to explain the substitution rule more carefully. – peek-a-boo Jan 10 '21 at 21:07
  • @peek-a-boo In that case, I wouldn't know how to parse the statement $$\int x^2 , dx = \frac{x^3}{3}+C , .$$Strictly speaking, the RHS isn't a function or even a family of functions—it's just a number. Can you elaborate please? Thanks for providing the link; I'll check it out. – Joe Jan 10 '21 at 21:18
  • I do explain this in that other link but just to answer it directly once again: the way to parse this is we consider the function $f:\Bbb{R}\to\Bbb{R}$ defined as $f(x)=x^2$, and we consider the set $P_f:={F:\Bbb{R}\to\Bbb{R},,|,, F'=f}$ (i.e the set of primitives of $f$). The claim being made is that for every $F\in P_f$, there exists a unique $C\in\Bbb{R}$ such that for all $x\in\Bbb{R}$, $F(x)=\frac{x^3}{3}+C$. In other words, we have that $P_f={x\mapsto \frac{x^3}{3}+C,|, C\in\Bbb{R}}$, so we have an explicit representation for the set of all primitives of our original $f$. – peek-a-boo Jan 10 '21 at 21:22
  • As you can see, this is ALOT of words, so we just summarize all of this into the simple expression $\int x^2,dx = \frac{x^3}{3}+C$ (where $C\in\Bbb{R}$). (By the way, you said the RHS is just a number... well that's also not true, because you haven't put a quantifier over $x$. So strictly speaking as a logical statement this is very bad... but of course most of the time everyone knows what is meant, and everyone is comfortable with sacrificing logical precision because this is what we're most comfortable with) – peek-a-boo Jan 10 '21 at 21:23
  • @peek-a-boo Okay, thanks a lot. Sorry, I hadn't looked at the link yet. – Joe Jan 10 '21 at 21:26
  • @peek-a-boo I'm not familiar with the term 'quantifier'—I looked up it and it says that it has something to do with mathematical logic. Is it possible if you give a layman explanation of what you have said, given that I have never studied mathematical logic before? Don't worry if you can't—you've already been very helpful, thank you. – Joe Jan 10 '21 at 21:48
  • anyway that was only a paranthetical remark, not central to your main confusion (which I hope I've addressed with the other comments and answers). If you really want to know, a quantifier means something like "for all" (universal) or "there exist" (existential). A statement like "$x^2\geq 0$" by itself is completely meaningless, because it doesn't specify what $x$ is. The correct statement is that "for every $x\in\Bbb{R}$, $x^2\geq 0$". If we don't say what $x$ is then I can take $x=i$ to get $x^2=i^2=-1<0$. This just goes to show the importance for specifying "quantifiying" the variables. – peek-a-boo Jan 10 '21 at 21:56
  • @peek-a-boo Great. To summarise, the formula $$\int f(u) , du= \int f(g(x))g'(x) , dx, \quad u=g(x)$$is a little inaccurate because the LHS denotes a primitive of $f$. Even if we 'substitute $g(x)$ back in for $u$', this is still formally incorrect because in that case $f(g(x))$ does not refer to the composite function $f \circ g$ but rather the function $f$, evaluated at a number which just so happens to be equal to $g(x)$. If $u$ is 'truly' simply a shorthand for $g(x)$, then $\int f(u) , du$ would be the same as $\int f(g(x)) , dg(x)$, which doesn't make any sense. – Joe Jan 10 '21 at 22:17

1 Answers1

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To be precise, and under some conditions, you can write

$$\int_{g(a)}^{g(x)}f(u)du=$$

$$\int_a^xf(g(u))g'(u)du=$$

$$\int_a^xf(g(t))g'(t)dt$$

the substitution made is $$u=g(t)$$

hamam_Abdallah
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