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Why can we say the following:

$u=\int\frac{\mathrm du}{\mathrm dx}\mathrm dx$,

where we treat $u$ as a function (e.g., see this proof).

Because as far as I know, $\int\frac{\mathrm du}{\mathrm dx}\mathrm dx$, as an indefinite integral, stands for a set of functions, whose derivative equals the integrand.

$\int f(x)\,\mathrm dx=F(x)+C$

So how can we say a function $u$ is equal to a family of functions?

Sha Vuklia
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    To make a joke out of it: to avoid forgetting to put the constant every time, you just never write it. – b00n heT Jan 27 '17 at 17:01
  • Correction: If $\frac{\mathrm du}{\mathrm dx}$ has an interval domain, then $\int\frac{\mathrm du}{\mathrm dx}\mathrm dx=u(x)+C.$ The claim is not that $u$ is a family of functions, but that the entire left (or right) side is a family of functions, or, to frame it another way, the general specification of the antiderivatives of $u'.$ – ryang Feb 02 '23 at 13:49

2 Answers2

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It is correctly written as $$ u(t) = \int_a^t \frac{du}{dx} \, dx $$ where $a$ is some chosen constant. You are correct, without the bounds, it is a family of functions. What you've written is an abuse of notation.

J126
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The antiderivate is never unique. If $f(x)$ has antiderivate $F(x)$, then $F(x)+c$, $c$ arbitary real, is an antiderivate as well.

The indefinite integral contains this constant and we can verify $F'(x)=f(x)$ no matter which value $c$ has. Therefore the indefinite integral can be considered to be the family of antiderivates.

Peter
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