# B On understanding the dx

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1. Sep 15, 2018 at 6:34 AM

### archaic

Hello!
Let $I$ be an interval of size $L$, suppose we divide it into bits of $dx$ then $L=dx+dx+... =\alpha.dx$
Since $dx$ is by definition infinitesimally small is it correct to imply that for each $x$ there's a corresponding $dx$ hence, here, $\alpha$ would be, theoretically, the total number of $x$s?
Because, as I see it, that's how the average value of a function is calculated :
$$\frac{\sum{f(x)}}{\alpha} = \frac{1}{L}.\sum{f(x).dx} = \frac{1}{L}. \int f(x).dx$$

2. Sep 15, 2018 at 7:05 AM

### verty

$${1 \over L} \int f(x)\ dx$$
The integral is the area, you always want to remember that. The area divided by the length of the interval is the average height.

I don't think you will find a book that says there are $\alpha$ many $f(x)\ dx$'s. There are uncountably many, too many to ever count.

3. Sep 15, 2018 at 7:40 AM

### archaic

Hence the "theoretically", I think the logic I used is quite sound, and it's just like the concept of the integral, you add the area of a bunch of rectangles of width $dx$ and that would give you the area under the curve $\Leftrightarrow$ for each rectangle of height $f(x)$, i.e for each $x$, we assign a width $dx$ which seems like saying that for each $x$ there's a corresponding $dx$.

4. Sep 15, 2018 at 7:53 AM

### Staff: Mentor

Your imagination might be o.k. for a physicist and a while, i.e. as long as $dx$ doesn't become a basis vector or linear form; for a mathematician it is misleading, since it indicates an entire procedure of a limit, i.e. it is always a context sensitive abbreviation of something more general and as such should not be treated as an independent entity.

Last edited: Sep 15, 2018 at 8:08 AM
5. Sep 15, 2018 at 7:56 AM

### FactChecker

A comment on notation:
I don't think that there is any official notation convention. I normally interprite $dx$ as an infinitely small change in $x$. When I want to indicate an arbitrarily small, but not infinitely small, change in $x$, I would use $\Delta x$. In any case, it is sometimes important to distinguish between the two concepts.

6. Sep 15, 2018 at 8:29 AM

### archaic

I'll restate the question differently :
With the integral, the concept is to sum up the area of rectangles of height $f(x)$ and width $dx$. Considering that the area under the curve covers up all $x$s then is it safe to assume that, since each all $f(x)$s possible (theoretically) are being used in the calculation of the area under the curve, the sum of all $dx$s is equal to the interval on which we're applying the integral, i.e if $L$ is the length of that interval, is $L = \alpha.dx$ a safe assumption?
Because explaining the average value formula with what I'm saying seems logical; average = sum of everything / how many is everything.

7. Sep 15, 2018 at 8:39 AM

### Staff: Mentor

8. Sep 15, 2018 at 9:14 AM

### PeroK

My approach to this is that $\Delta x$ indicates a (small) change in $x$ and $dx$ is notationally used in the limit as $\Delta x \rightarrow 0$. In a sum, you are always summing with respect to $\Delta x$; and, the $dx$ makes its appearance in a integral, which is defined as the limit of these sums.

In particular, $L = \alpha dx$ has no meaning.

$dx$ can also have a meaning as a "differential", which is discussed here:

http://tutorial.math.lamar.edu/Classes/CalcI/Differentials.aspx

In any case, it is essential that you distinguish between the finite quantity $\Delta x$ and the differential $dx$.

9. Sep 15, 2018 at 10:37 AM

### archaic

I, in fact, do distinguish between the two.
It's just that the $L = \alpha.dx$ seems logical to me.
If $\int f(x).dx =$ area under the curve then it follows that we calculated the area formed by each point $x$ of the interval, hence, somehow, I see it as though for each $x$ there's a corresponding $dx$, i.e as if there's this imaginary space between two neighboring $x$s (even if numbers are infinite, it's just a supposition) of length $dx = x_2 - x_1$ with, in some sens, $x2$ and $x1$ atom like real numbers and that difference between them is some kind of microscopic indivisible distance.

10. Sep 15, 2018 at 10:42 AM

### PeroK

Unfortunately, this is not the way the mathematics of the Riemann integral works.

Within the Real numbers, there is no no such thing as "neighboring" points. Any two points have a finite distance between them, and an (uncountable) infinity of other points between them.

If you want to understand Real Analysis, you will need to revise your ideas about the real number line.

11. Sep 15, 2018 at 11:27 AM

### verty

Let $F'(x) = f(x)$ and $dx = g(t)\ dt$. Then: $$\int dF = \int f(x)\ dx = \int f(x(t))\ g(t)\ dt$$ The first step is justified because $dF = f(x)\ dx$. How would you apply your theory to this?

12. Sep 15, 2018 at 11:28 AM

### archaic

Of course I know that.
He explains it better than me (in the context of $\bar{f}$) :

13. Sep 15, 2018 at 11:35 AM

### PeroK

That last sentence is interesting. I would suggest that the following is better:

"Now take the limit as $\Delta t \rightarrow 0$ and you get the integral."

14. Sep 15, 2018 at 11:43 AM

### archaic

excuse me I don't have an answer

15. Sep 15, 2018 at 11:48 AM

### verty

Keep learning and don't worry about what the stuff means. Later you'll figure it out.

16. Sep 16, 2018 at 12:31 PM

### nuuskur

It is the differential of $x$. If you are asking for a definition - a differential is 'vanishingly small'. There you go. You can also have something like $\int g(t) d(f(t)) = \int g(t)f'(t)dt$. If you start thinking what the **** $dx$ means ..you won't get very far. Riemann integrals are evil. Learn measure theory and Lebesgue(-Stieltjes) integrals at your earliest convenience and everything becomes much clearer, none of this infinitesimally small mumbo jumbo.

You can also view it as a differential form (vector analysis) - it represents some tiny tiny tiny piece of a curve or something.

The differential business happens to be very useful at least as a heuristic. So, don't take it for more than a heuristic and don't try to get too rigorous.

Reminds me of a quote by, I think it's von Neumann's: you don't understand it, you just get used to it.