How common is it for a densely-defined linear functional to be closed?

How common is it for a densely-defined linear functional to be closed?

I've always held the vague belief that any densely-defined operator
encountered "in nature", if it isn't bounded, is probably at least
closable. But, today I noticed the following thing:
Consider the Banach space $C_0(\mathbb{R})$ of continuous, complex-valued
function vanishing at $\pm \infty$ in the uniform norm. We have a
densely-defined linear functional $\int : C_c(\mathbb{R}) \to \mathbb{C}$
given by Riemann integrating compactly supported functions. This
functional is not closable. Indeed, choose $f \in C_c(\mathbb{R})$ with
$\int f = 1$ and put $f_n(t) = (1/n) \cdot f(t/n)$. Then $f_n \to 0$
uniformly, but $\int f_n = 1$ for all $n$. Thus $(0,1)$ belongs to the
closure of $\operatorname{Graph}(\int) \subset C_0(\mathbb{R}) \times
\mathbb{C}$, and $\overline{\int}$ is not single-valued.
This surprised me because integration against an infinite measure is one
of the most fundamental examples of an unbounded linear functional. So, if
integration is not closed, how ubiquitous could closed linear functionals
possibly be?
Question: Can you come up with any good "natural" examples of
densely-defined linear functionals on Banach spaces which are closable? Do
you have a sense for how important, or common, such examples are? Or are
non-closable functionals the rule rather than the exception?
Thanks.