The meaning of $dx$ in an indefinite integral

Question

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

Answer 1

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*}

Answer 2

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: $$ \int dx \;f(x) $$