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[-] marcos@lemmy.world 9 points 2 weeks ago

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 5 points 2 weeks ago
[-] marcos@lemmy.world 4 points 2 weeks ago* (last edited 2 weeks 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 1 points 2 weeks ago

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 3 points 2 weeks 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.

[-] OrganicMustard@lemmy.world 1 points 2 weeks ago

In the case of dx as an infinitesimal it makes sense. You are making a sum of all the values of the function in the integral range and multiplying with a constant dx.

[-] kogasa@programming.dev 2 points 2 weeks ago

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

this post was submitted on 14 Jul 2025
813 points (95.0% liked)

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