Question: I think I am confusing myself too much on this. 

Let $ f(x)=0$ for $x<1$, and $f(x)=1$ for $x\geq1$. What is $\int\_0^1f(x)\,dx$?

I am worried because $f$ is discontinuous at $1$. Does that make the answer different from zero, or make the function not integrable in that interval?

For example, $\int\_0^2f(x)\,dx=\int\_0^1f(x)\,dx$+$\int\_1^2f(x)\,dx$. For the second interval $f$ is constant at 1, so it can be taken outside the integral sign, but, that does not really apply for the first interval $0\leq x\leq1$, does it?

Answer: 
Let $E/F$ be a finite extension (which is separable as the characteristic is zero), we will prove that $E=F$, hence $F$ is algebraically closed:
Let $a$ be a primitive element of $E/F$. Then $a$ is algebraic over $K$, so there exists a finite Galois extension $N/K$ containing $a$, in particular 
$$
E\subseteq NF.
$$
(One can take $N$ to be the Galois closure of $K(a)$, for example.)
Let $f\in K[x]$ be the minimal polynomial of a primitive element of $N$ over $K$. Since $N/K$ is Galois, any of the roots of $f$ generates $N$ over $K$. 

Now, by assumption $f$ has a root in $F$, so $N\subseteq F$, hence $E\subseteq NF=F$. This proves that $F$ has no non-trivial extensions, hence $F$ is algebraically closed.
The question asked is more specific than the title: in the question you say $K$ has characteristic $0$, but the title is a broader question. The answer to the broader question is also "yes". That is, even when $K$ has characteristic $p$ the field $F$ is algebraically closed. This is a theorem of Robert Gilmer. See [here](https://www.jstor.org/stable/2315743?seq=1#metadata_info_tab_contents) or Theorem 2 [here](https://kconrad.math.uconn.edu/blurbs/galoistheory/algclosure.pdf).