0

Given the Riemann integral: $\int_a^b f(x)dx$

By definition: $dx = lim \frac{b-a}n$ as $n\rightarrow+\infty$. So, is it $dx$ a real number and a positive one as it seems apparent by defintion?
There seems to be a lot of confusion in literature about what $dx$ really is and none agrees that $dx$ is just a positive real number, but I really don't know why!

InanimateBeing
  • 3,074
Giack_89
  • 71
  • 8
    It is not. It's just a symbol. – Ted Shifrin Aug 02 '22 at 17:03
  • 6
    There isn't much confusion in the literature, the source of the confusion is that most ways of formalizing what $dx$ means are too technical for a basic calculus course. Anyhow, regardless of how you formalize it, $dx$ is definitely not a number. – Elchanan Solomon Aug 02 '22 at 17:03
  • No, it is an infinitesimal representing an infinitesimal change in the value of $x$. – bobeyt6 Aug 02 '22 at 17:16
  • I disagree with the downvote (upvoted to correct this). This was a sincere question about a confusing concept. – Annika Aug 02 '22 at 17:22
  • 1
    Note that although $\frac {b-a}n$ will be a positive real number for any specific $n$ the limit $\lim_{n\to \infty} \frac {b-a}n=0$ is not a positive real number. You're definition of $dx$ is not quite correct. $dx \ne \lim \frac {b-a}n$ but $\int_a^b f(x)dx =\lim_{n\to \infty}\sum_{k=0;x\in (k\Delta x,(k+1)\Delta x)} f(x)\cdot \Delta x$ where $\Delta x=\frac{b-a}n$. $dx$ represents the concept of the limit of $\Delta x$ but only in conjunction with it being within a limit of an infinite sum. The actual limit of $\Delta x$ is $0$ and is useless by itself...tbc.... – fleablood Aug 02 '22 at 17:42
  • 1
    ..cont... it's kind of a little bit like $n \times \frac 1n = 1$ so $\lim_{n\to \infty} n\times \frac 1n = 0$. If we want to express the concept of $\frac 1n \to 0$ we don't use $\lim \frac 1n$ because that's just $0$ and isn't useful. We can't break up a product of limits like that: $1= \lim n\times \frac 1n \ne \lim n \times \lim \frac 1n = \infty \times 0 =??????$. But we do want the concept of the two parts of the limit. $d\frac 1n$ is just what we use for the $\frac 1n$ part. – fleablood Aug 02 '22 at 17:49
  • @fleablood Why to write that in a comment? You should write an answer. – jjagmath Aug 02 '22 at 17:51
  • Because this is probably a duplicate. It's also fairly casual and doesn't actually define what $dx$ is. It just describes what it sort of is.... I see in my comment (which I can no longer edit) that I made a typo... I meant $\lim n\times \frac 1n =1$ (not zero). – fleablood Aug 02 '22 at 20:11
  • Thank you all guys for answers but would you please answer this question here? https://math.stackexchange.com/questions/4505249/work-in-elastic-force-what-sign-should-be-interpreted-for-dx – Giack_89 Aug 03 '22 at 09:45

1 Answers1

1

You can think of the $dx$ symbol as just part of the integral notation. The notation makes it looks like a number, because it takes the place of the actual small positive number $\Delta x$ that appears in each of the finite Riemann sums. But these $\Delta x$ values approach 0 in the limit, so there's no actual positive number that the $dx$ represents.

Karl
  • 11,446
  • Thank you for your answer, but would you please answer me this question here: https://math.stackexchange.com/questions/4505249/work-in-elastic-force-what-sign-should-be-interpreted-for-dx – Giack_89 Aug 03 '22 at 09:45