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So, an integral is notated like this:

$$\int_a^bf(x)dx$$

And from my understanding, it's an operator that is defined for three operands: $a$ and $b$, which can be anything, and an integrand of the form $f(x)dx$.

$dx$ is just an infinitesimal number, so $f(x)dx$ is simply $f(x)$ multiplied by $dx$. This gives an infinitesimal based on $x$, and the integral is the infinite sum of all of those. I found a good page that explains this idea nicely.

But what about integrands that are not of the form $f(x)dx$? For example, things like:

  • $\int_a^b2$
  • $\int_a^bf(x)dy$
  • $\int_a^b(f(x)dxdy)$
  • $\int_a^bf(x)$

Certainly, if the intuition explained above is true (i.e: $f(x)dx$ is simply an algebraic expression; therefore the integral operator must accept any algebraic expression as its integrand), then these things must be syntactically legal. Is that wrong? How should these things be interpreted?

I suspect that my interpretation of $f(x)dx$ is wrong. $dx$ is part of the integral's notation, and plays a special role in what the integral operator does (defining the variable of integration). But $dx$ is also supposed to represent a numerical value which is multiplied by $f(x)$. How can it be both of those things at once?

I'm probably overthinking this. I just want to understand the notation and the intuition behind it. I hope someone can recognize what my confusion is and rectify it for me.

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    $\begingroup$ Where are you seeing those sorts of expressions? You should never omit the differential $\mathrm d \cdot$ with Riemann integrals, and those are the integrals you learn about first. $\endgroup$ – GFauxPas Dec 1 '16 at 1:42
  • $\begingroup$ Are you in an introductory calculus course or something at a higher level? $\endgroup$ – andars Dec 1 '16 at 1:46
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    $\begingroup$ The integrand is not $f(x) \ dx$, rather $f(x)$. You should think of an integral as taking in a function, not a function times a differential. The differential is there to tell you what you are integrating with resepct to. The only expression of the four you listed which is "legal" is $\int_a^b f(x) \ dy$, and this is simply $f(x)\int_a^b 1 \ dy=f(x)(b-a)$ since $f(x)$ is constant with respect to $y$. $\endgroup$ – kccu Dec 1 '16 at 1:47
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    $\begingroup$ @GFauxPas "never" is a strong word. For instance, Abbott always omits the differential in Understanding Analysis, and I don't see any problem with that. In certain contexts it is just notation and can be omitted without any ambiguity. $\endgroup$ – andars Dec 1 '16 at 1:55
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    $\begingroup$ @GFauxPas I've never seen integrals that look like that. I'm just wondering about them, for the sake of completeness and intuition about how integrals work $\endgroup$ – Vivian Cox Dec 1 '16 at 3:02
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The integral symbol is simply a "nick name" for writing a more complex limit. $ \int_a^b f(x) dx = \lim_{n \to \infty} \sum^n_{i=1}f(c_i)\Delta x_i $. The "dx" bit tells us which variable we are integrating against it isn't really a number or a variable in its own right.

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    $\begingroup$ I know it's a shorthand for that limit expression, which is basically the limit of a Reimann sum. That's the part that gets me. $dx$ is supposed to represent the limit of $\delta x$ as the number of rectangles in the sum increases—so $dx$ is the base of an infinitesimally thin rectangle, and $f(x)$ is the height. So multiplying them gives the area. The integral is the infinite sum of these areas ...isn't it? That's how it's always been explained to me. $\endgroup$ – Vivian Cox Dec 1 '16 at 2:57
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    $\begingroup$ Except $dx$ is not a number, not even an "infinitesimal" number, if it were the limit of the $\Delta x$, we'd have $dx=0. I think it's okay to let notation just be notation here. $\endgroup$ – YoTengoUnLCD Feb 16 '17 at 6:41
  • $\begingroup$ @VivianCox: It is simply logically invalid to think of an integral as an infinite sum of infinitesimal areas. For those who want to claim non-standard analysis, I have two comments for you: (1) I don't buy the existence of an ultrafilter on the naturals; (2) Asymptotic analysis does not require any unjustifiable set-theoretic assumptions and yet is more powerful than NSA, so there is no point to NSA. $\endgroup$ – user21820 Aug 24 '17 at 15:36
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The pure mathematicians in the group seem to be leaning heavily toward the idea that this is just an artifact of notation and that there's no real factor being multiplied there. There is something to this view, so I don't want to come off as going against it completely. I'm not echoing points from other answers, but see, for example, the answer by Q the Platypus that talks about the Reimann sum.

From a physics point of view, however, I think we have to assign at least a modicum of meaning to it from the point of view of dimensional analysis. Take a simple equation like

$x(t) = x_0 + \int_0^t v(t') dt'$

for velocity $v(t)$ and position $x(t)$ with initial position $x_0$. The units on that will only work out if you assign units of time to the $dt$ "factor" that are consistent with the units used to measure $v$. At minimum it's also representing a requirement for consistency in measurement between the integrand, the limits of integration, and the units of measure in the result. In some sense it is still just notation, but - insofar as you asked about intuition - this definitely showing you a bit more than just which variable is integrated.

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  • $\begingroup$ Even in the pure mathematical sense, I think the $dt$ in the integral shows something more than just which variable to integrate over. It also invokes a measure on that variable. Measure in pure mathematics tends to be dimensionless, but if it had a dimension, it seems to me that the dimension would also be invoked by the $dt,$ as desired. $\endgroup$ – David K Feb 18 '17 at 14:50
  • $\begingroup$ @DavidK I didn't dare say "measure" here as a I thought it might attract negative attention if used less than rigorously, but that's definitely part of what I had in mind too. $\endgroup$ – Brick Feb 18 '17 at 21:04

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