this post was submitted on 14 Jul 2025
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[–] marcos@lemmy.world 0 points 16 hours ago (2 children)

Hum... I don't think the integral "operator" applies by multiplication.

You can put the dx at the beginning of the integral, but not before it.

[–] LeFrog@discuss.tchncs.de 0 points 15 hours ago (1 children)
[–] marcos@lemmy.world 0 points 14 hours ago* (last edited 14 hours ago)

Nobody on your link is treating the integral "operator" as multiplicative.

dx \int f(x) is blatantly different from \int f(x) dx

[–] OrganicMustard@lemmy.world 0 points 13 hours ago (2 children)

If you were using nonstandard analysis with dx an infinitesimal you could put it outside I guess. Maybe with differential forms too?

[–] marcos@lemmy.world 0 points 13 hours ago

Switch it with a summation operator and see if it makes sense. The problem isn't the operation by itself, but the fact that the operator implies an argument application, like a function.

[–] kogasa@programming.dev 0 points 12 hours ago

In the context of differential forms, an integral expression isn't complete without an integral symbol and a differential form to be integrated.