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This semester I'm taking integral calculus for the first time. We started with the differential (i.e. $dy=f'(x)\,dx$) and right after that with the indefinite integral. Since then, I've been trying to make sense of the $dx$ when it's part of an indefinite integral (i.e. $\int f(x) \, \boldsymbol{dx}$). I know there are already a bazillion answers regarding this question, but all of them refer to the definite integral and the ones that do touch on indefinite integrals only say things like "it's just a syntactical device to tell you the variable to differentiate with respect to or the integration variable" (Ihf, 2012, web). I don't like this answer, especially because I was taught that given two functions $f(x)$, $g(x)$ and the antiderivative of their product $\int f(x)g(x)\,dx$, if I were to assign $f(x)$ to $u$ and $g(x)\,dx$ to $dv$ in order to integrate by parts, then I would have to integrate $dv$ to find $v$. This only makes sense if $dx$ means something by itself and is not just a quirk of the notation, otherwise we would end up with something like the following: $$\int g(x)\,dx\space dx$$

After thinking about it a lot, I believe I finally found a way to make sense of the $dx$ as something that isn't purely and simply a notation device. My reasoning is as follows:

  1. Given a function $f(x)$, let $y=f(x)$. Then $$\frac{dy}{dx}=f'(x)$$
  2. Then from the fundamental theorem of calculus, we know that $$\int\frac{dy}{dx}\,dx=y$$
  3. Let $dy=f'(x)\,dx$. Then the equation $$dy=\frac{dy}{dx}\,dx$$ holds and $$\int\frac{dy}{dx}\,dx=\int dy=y$$
  4. Finally, integrate both sides of the equation $dy=f'(x)\,dx$ in order that $$\int dy=\int f'(x)\,dx\implies y=f(x)$$

From this I conclude that, by integrating a function, what we are really doing is integrating the differential of that function. This makes perfect sense to me, although I'm aware that seemingly logical things aren't necessarily logical. That's why I would appreciate it if someone could tell me whether the above is mathematically correct or pure gibberish.

PS, I had never written a mathematical proof before and I have taken no proofs courses yet, so any suggestions are welcome.

Reference: lhf. (2012, May 9). What does $dx$ mean?. Mathematics Stack Exchange. https://math.stackexchange.com/q/143262

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    $\begingroup$ I would have to say this is a very logical and methodical analysis for someone who has never done a proof before. This is how professionals sketch out their ideas before trying to poke holes or prove cases of them. $\endgroup$ Mar 22, 2021 at 0:26
  • $\begingroup$ That's pretty good. Whenever I have trouble understanding $d-$something-or-other, I put it over $dt$ to make it a derivative function and put $dt$ next to it. It seems to work for dealing with differentials in general. Reduce everything to the same simple case if possible. $\endgroup$ Mar 22, 2021 at 0:26
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    $\begingroup$ I am of the conviction that indefinite integrals as a whole is just a notational quirk that means "antidifferentiate". It is only meant to look kindof like a definite integral because the fundamental theorem of calculus tells us the two operations are closely related. Of course, that's not too say meaning can't be imposed upon $dx$ in an indefinite integrals. Just that that's not something I personally worry about. $\endgroup$
    – Arthur
    Mar 22, 2021 at 0:31
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    $\begingroup$ You can always look at an indefinite integral as a definite integral with an unspecified lower limit. $F(x)=\int^x f(y)dy$ is the indefinite integral of $f(x)$. $\endgroup$ Mar 22, 2021 at 1:33
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    $\begingroup$ Related question: Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio? This ties to this question because you are using this here and the stuff you do demonstrates that this notation for the integral is very useful as it allows you to do these kinds of "sketchy" manipulations we all do treating derivatives as fractions in the integral when doing substitutions and similar and get the correct result. $\endgroup$
    – Winther
    Mar 22, 2021 at 12:41

2 Answers 2

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An explanation is given in section 2.9 Fundamental Theorems of the Calculus in Introduction to Calculus and Analysis I by R. Courant and F. John.

It is quite customary to use a notation which is not perfectly clear without comment: we write \begin{align*} \color{blue}{F(x) = \int f(x)\,dx} \end{align*} when we mean that the function $F(x)$ is of the form \begin{align*} F(x) = c + \int_a^xf(u)\,du \end{align*} for suitable constants $c$ and $a$, that is, we omit the upper limit $x$, the lower limit $a$ and the additive constant $c$ and use the letter $x$ for the variable of integration.

Strictly speaking, of course, there is a slight inconsistency in using the same letter for the variable of integration and the upper limit $x$ which is the independent variable in $F(x)$. In using the notation $\int f(x)\,dx$ we must never lose sight of the indeterminacy connected with it, that is, the fact that the symbol always denotes one of the primitive functions of $f$ only. The formula $F(x) = \int f(x)\,dx$ is just a symbolic way of writing the relation \begin{align*} \color{blue}{\frac{d}{dx}F(x)=f(x)}. \end{align*}

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Remark
In mathematics it may be that "$\int$" and "$dx$" are considered to be brackets, inclosing the integrand: $$ \int f(x)\;dx $$ In physics it may be that "$\int dx$" is used as an operator, so it precedes the integrand: $$ \int dx \;f(x) $$

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    $\begingroup$ This is a comment, not an answer. $\endgroup$
    – Christoph
    Mar 22, 2021 at 12:33

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