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In the case of $\int_a^b f(x) dx=F(b) - F(a)$ this can be written as $\int_a^b \frac d {dx} F(x) dx=F(b)- F(a).$

Comparing to Stokes' theorem, $\int_{\partial\Omega} \omega=\int_\Omega d\omega$ where $d\omega$ is a differential form, I guess it would be something like

$$\int_{\Omega=[a,b]} dF=\int_{\partial\Omega=\{a,b\}} F$$?

So the manifold $\Omega$ is the compact interval $[a,b]$ over which the exterior derivative $dF$ is integrated, whereas the boundary consists of the set of the two limit points $\partial\Omega=\{a,b\}$ over which the differential form $F$ is integrated.


PS: Great explanation:

enter image description here

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  • $\begingroup$ $dF = F'dx$ and $\displaystyle \int_{\{a,b\}} = \int_{\{a\}^- \cup \{b\}^+} = \int_{\{b\}} - \int_{\{a\}}$. $\endgroup$
    – azif00
    Jan 24, 2021 at 4:07
  • $\begingroup$ What are you trying to learn about this case? Why it works or how the integrals on the boundary work? I would like to elaborate the answer a bit more without getting too long winded. $\endgroup$ Jan 24, 2021 at 4:45
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    $\begingroup$ Yes, your observation is correct. $\endgroup$
    – Deane
    Jan 24, 2021 at 4:48
  • $\begingroup$ @openproblem I see the idea, but I would like to understand the equivalence (to the extent that it is possible) between the manifold in Stokes', the differential form, and the most standard enunciation of FTC for a scalar univariable function. I don't understand your answer. $\endgroup$ Jan 24, 2021 at 4:50
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    $\begingroup$ I do not know how it is called, but the operator $\cup$ is called the union; in other words, it unites two subsets $A_1$ and $A_2$ to get their union $A = A_1 \cup A_2$. The $\{a\}^{-} $ denotes the set of one-single element $a$ oriented negatively. Hence, there arises the minus sign. The same story with $\{b\} ^{+} $ $\endgroup$ Jan 24, 2021 at 5:26

1 Answer 1

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By the choice of F, dF/dx = f(x). In the vocabulary of differential forms, this is saying that f(x) dx is the exterior derivative of the 0-form, namely the function, F. In other words, that dF = f dx. The general Stokes theorem applies to higher differential forms ω instead of just 0-forms such as F.

A closed interval [a, b] is a simple example of a one-dimensional manifold with boundary. Its boundary is the set consisting of the two points a and b. Integrating f over the interval may be generalized to integrating forms on a higher-dimensional manifold. In the more general case, the manifold has to be orientable, and the form has to be compactly supported in order to give a well-defined integral.

The Orientation of the interval can be thought of as the ordering of the elements, the closed bounded interval [a,b] is compact as you learn in analysis.

The two points a and b form the boundary of the closed interval. More generally, Stokes' theorem applies to oriented manifolds M with boundary. The boundary ∂M of M is itself a manifold and inherits a natural orientation from M. Intuitively, a inherits the opposite orientation as b, as they are at opposite ends of the interval. So, "integrating" F over two boundary points a, b is taking the difference F(b) − F(a).

This integrating over two points is really more of a degenerate case and maybe it is confusing as to why you are just plugging in the values.

You can understand it as integrating over a singleton using an atomic measure with the two point boundary. An atomic measure gives a non-zero measure to the singleton set so that taking an integral over a singleton makes sense in a way that is more than just hand waving.

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  • $\begingroup$ A singleton is a set with only one element? $\endgroup$ Jan 24, 2021 at 5:21
  • $\begingroup$ Yes. A singleton is a set with only one element. $\endgroup$ Jan 24, 2021 at 5:23
  • $\begingroup$ Ahh... This may be what Sergei just explained as $\{a\}^-$ and $\{b\}^+$? So two singletons with an atomic measure on each? $\endgroup$ Jan 24, 2021 at 5:31
  • $\begingroup$ @Numericallyilliterate Yes. $\endgroup$ Jan 24, 2021 at 5:48

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