The "ds" which appears in an integral with respect to arclength is not a 1-form. What is it? The only reasonable way to interpret "$ds$" as a functional on tangent vectors has to be that it takes a tangent vector and spits out its length, but this is not linear.  So $ds$ is not a 1-form.  It still seems like a nice sort of object to think about integrating.  Does $ds$ fit into a larger class of gadgets generalizing differential forms?  Or it there some compelling reason that I shouldn't care about $ds$?
 A: It is an example of an absolute differential form, as defined by Toby Bartels here: http://ncatlab.org/nlab/show/absolute+differential+form. 
A: It is not a 1-form, it is a 1-density: a function that is continuous and homogeneous of degree 1 on the tangent space of the manifold. It also happens to be convex and positive in the complement of the zero section (actually, its restriction to each tangent space is a Euclidean norm). If the norm is not Euclidean, you have the arc-length element of a Finsler metric. The convexity is basically necessary and sufficient for the lower semi-continuity of the length functional (Busemann-Mayer theorem).
See my answer to this question for more on densities.
A: If you have a curve, also known as a 1-manifold, inside a Riemannian manifold, the Riemannian metric on the manifold restricts to a 1-dimensional Riemannian metric on the 1-manifold. The square root of this metric is a density (see alvarezpaiva's answer) that can indeed be integrated along the 1-manifold.
A: It is the volume element on the one manifold. It is a 1-form. 
