I'm trying to teach myself about differential forms, and my book says that functions can't be integrated on manifolds because the integral isn't coordinate independent, but if the manifold has dimension $n$, you can integrated $n$ forms in a coordinate independent way.
The example given is simple, but I'm struggling with it. when the manifold is just $(\mathbb{R},1_{i.d.})$. Let $x$ and $y=Cx$ be the coordinate, and coordinate change, and $f(x)=1$. Then $\int_a^bf(x)dx=\int_a^bdx=\int_{Ca}^{Cb}\frac{dx}{dy}dy=(b-a)\neq C(b-a)=\int_{Ca}^{Cb} f(y)dy$. So I guess you would say that the Riemann integral is not coordinate invariant. But the claim is that you can integrate one-forms on a one dimensional manifold in a coordinate independent way. But $f(x)dx$ is a one-form, so I'm a little confused.
When considering coordinate changes, one-forms and functions are different things, as you transform them differently when changing coordinates. The Riemann integral of a one-form does not change after a coordinate change in this context, which is what allows classical "u-substitution", but it does change if you try to transform the one-form $f(x)\,dx$ as the function $f(x)$.
Explicitly, the one-form $f(x)\,dx$ should be pulled back along the transformation $x=a(u)$ as $f(x)\,dx \mapsto f(a(u)) a'(u)\, du$, in correspondence with the chain rule, rather than simply $f(a(u)) du$.