Let $u:\Omega\times {]0,T[} \to \mathbb{R}, f:\mathbb{R}\to \mathbb{R}$, now consider: $$\int_0^t f(u(\mathbf{x},s))\operatorname ds.$$
Instead of having to write this down each time I would like to call this thing $F$. But what are the arguments of $F$? Is it something like? $$F(u, t)$$ If $u\in H^1(\Omega)$ would then $F: H^1(\Omega)\times {]0,T[} \to H^1(\Omega)$?
Is the following example right?
To get some feeling with what this thing actually is, I tried it out with an example.
If I take some nice function $u: (x,t)\mapsto x^2+t$, and $f = \text{id}$ then $F = \int_0^t (x^2 + s) \operatorname ds = x^2t+\frac{t^2}{2}$ which would mean $F(u,t) = u\cdot t- \frac{t^2}{2}$. However my example functions were very nice behaving, and I wonder if any wild function $f$ would still result in an $F$ to be written as a function of $u$ and $t$.