Given an a.l.p parametric curve $r(t)$ we know that $r''(t)$ is the curvature scaled normal to the curve at $r(t)$.
Thus $\|r''(t)\|$ is the actual curvature. Consider then a discretization of the domain of $r$ such that the domain is $U = \bigcup\{u_i \mid u_i \cap u_j = \emptyset \text{ if } i \neq j\}$ and let $S = \{t_i \in u+i\}$. Or simply put we grab a lot of points through the domain.
Then $\frac{\sum \|r''(t_i)\|}{N}$ where $N$ is the total of all intervals is a discrete approximation of the global average of the curvature of the curve.
What would the analytic formulation be? i.e. what is the global average of the curvature of a curve?
What does "a.l.p" mean in the question? I would note that in general, $||r''(t)||$ is not the curvature; this is only true when $r$ is an arclength parametrization, or equivalently $||r'(t)||\equiv1$. In general, we have
$$\kappa=\left|\left|\frac{dT}{ds}\right|\right|=\frac{||T'(t)||}{||r'(t)||}$$
where $T$ is the unit tangent vector. Of course, $\kappa$ is a scalar, and so the best way to capture a meaningful "average" (keeping in mind that different parametrizations can give different results) is to use an arclength line integral, and then divide by the total arclength $s$:
$$\kappa_{avg}=\frac1s\int_C\frac{||T'||}{||r'||}ds$$
But remember that $ds=||r'(t)||dt$, so:
$$\kappa_{avg}=\frac1s\int_0^T||T'(t)||dt=\frac{\int_0^T||T'(t)||dt}{\int_0^T||r'(t)||dt}$$
(EDIT: I've been informed in the comments that "a.l.p" means arclength parametrized. In this case, many of the points I make about parametrizations are moot. Nonetheless, I'll leave the original here for posterity. Note that the result I give is true regardless of parametrization, which may be useful in general.)