Graduate Dx in an integral vs. differential forms

[Trying2Learn]

TL;DR

what is the dx in an integral

Good Morning

To cut the chase, what is the dx in an integral?

I understand that d/dx is an "operator" on a function; and that one should never split, say, df, from dx in df/dx

That said, I have seen it in an integral, specifically for calculating work.

I do understand the idea of "differential forms." However, I am trying to understand if there is a difference between how dx is used in an integral (effectively splitting out the denominator in the "operator" of d/dx).

I do understand that in calculating "work" in physics, we integrate the force over dx. In that case, we realize that force
is a vector and dx is a co-vector.

So, in a nutshell, i can see "through a dark glass" (so to speak) that something is going on here.

However, with that background, I hope I can rephrase my initial question as

"What is the difference between the "dx" in, say the integral leading to work done, vs. the dx in the denominator of the operator (which, should NEVER, in theory, stand alone).

I am not sure I am asking this properly. Please forgive me for the poor wording -- that is half the problem I face.

 

[sysprog]

My calc teacher said that the last infinitesimal in the infinite series was "treated as zero" and consequently discarded $-$ he didn't say that it was existentially nonexistent $-$ the Physics Forums aren't where to go on the net for discussions of Philosophy of Mathematics $\dots$

 

[Delta2]

Summary:: what is the dx in an integral

and that one should never split, say, df, from dx in df/dx

That is not correct, we can split them because they are actually stand alone quantities. You should read about how the differential of a function is defined. In short it is defined as $df=f'(x)dx$
In the Leibniz formalism the derivative is taken to be the quotient of two differentials: The differential of the function f, which is symbolized as $df$ and the differential of the identity function $i(x)=x$ which is $di=dx$
So to answer directly your question, the dx in the integral and the dx in the derivative are the exact same thing:it is the differential of the identity function.

 

[Trying2Learn]

My calc teacher said that the last infinitesimal in the infinite series was "treated as zero" and consequently discarded $-$ he didn't say that it was existentially nonexistent $-$ the Physics Forums aren't where to go on the net for discussions of Philosophy of Mathematics $\dots$

OK, I deleted the word "philosophical."

 

[andrewkirk]

There is no big difference. In neither case does the 'dx' stand alone. In the differential case it has to keep company with the dy or df(x) in the numerator of the fraction, in order to have a meaning. In the integral case it has to keep company with the integral sign in order to have a meaning. That's because in both case a limit is implied. Here are the full statements, which hold for reasonably well-behaved functions f(x).

$$\frac{df(x)}{dx} \triangleq \lim_{dx\to 0} \left(\frac{f(x+dx)-f(x)}{dx}\right)$$

$$\int_a^b f(x)\,dx \triangleq \lim_{dx\to 0} \sum_{k=0}^{\lfloor \frac{b-a}{dx} \rfloor} f(a+k\,dx)\,dx$$

where $\lfloor u \rfloor$ denotes the greatest integer not exceeding $u$.

 

[Trying2Learn]

There is no big difference. In neither case does the 'dx' stand alone. In the differential case it has to keep company with the dy or df(x) in the numerator of the fraction, in order to have a meaning. In the integral case it has to keep company with the integral sign in order to have a meaning. That's because in both case a limit is implied. Here are the full statements, which hold for reasonably well-behaved functions f(x).

$$\frac{df(x)}{dx} \triangleq \lim_{dx\to 0} \left(\frac{f(x+dx)-f(x)}{dx}\right)$$

$$\int_a^b f(x)\,dx \triangleq \lim_{dx\to 0} \sum_{k=0}^{\lfloor \frac{b-a}{dx} \rfloor} f(a+k\,dx)\,dx$$

where $\lfloor u \rfloor$ denotes the greatest integer not exceeding $u$.

Ah... thank you! That was what I was looking for!

 

[Arjan82]

Just as a fun fact, you can apparently actually extend calculus with 'infinitesimals', your dx would here then be an 'infinitesimal' if I understand correctly...(?) That is loosly defined a number smaller than any other number but not zero. This now can be used as a stand alone number (but not like any other number, special rules apply). Look e.g. here:
https://en.wikipedia.org/wiki/Nonstandard_calculus
https://www.math.wisc.edu/~keisler/calc.htmlI'm not a mathematician, so I would love to know if others have experience with this.