# Can a function be an antiderivative? $u=\int\frac{\mathrm du}{\mathrm dx}\mathrm dx$

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?

• To make a joke out of it: to avoid forgetting to put the constant every time, you just never write it. Jan 27, 2017 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 $u'.$ Feb 2, 2023 at 13:49

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.
• Wouldn’t the integral actually be the following? $$\int_a^t\frac{du}{dx}\,dx=u(x)\Bigr|_{x=a}^{t}=u(t)-u(a)$$ Oct 25, 2017 at 21:41
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.