I'm studying differential forms and I know how to manipulate all the equations. On trying to find a pictorial understanding, I am a bit stuck on the following. A one-form is suppose to assign a measurement to a line and it must also be linearly dependent on the tangent vectors at that point.
Consider the line integral $\int f(s)ds$. At a point $p$, consider a path to be integrated over, say $p +\phi(t)$. We have $ds=\|\phi'(t)\|dt$. Obviously, $ds$ does not depend linearly on the tangent vectors at point $p$, i.e., the paths that go through $p$. Does this mean $ds$ is not a one-form?
Thanks in advance for any explanation.
No, $ds$ is not a $1$-form. It is technically a density (or the "absolute value" of a $1$-form). Note that if you reverse the orientation of the curve, you get the same integral, whereas for the integral of a $1$-form you will get the negative of the integral.