I had a question about the integration of differential k-forms and if there are any easy ways to parameterize them. For example, if we have a one form $\alpha$ and we want to integrate over some curve $\gamma$ then we have that $\int_{\gamma}\alpha=\int_{\gamma}\alpha(\gamma'(s))ds$. However, if we want to integrate arbitrary k-forms over some domain, is there a way to parameterize in a similar way where we end up pulling in k-vectors into the integration?

yeah you need to use the canonical map over the canonical form in a canonical manner

@YousufSoliman Yes, integrating a form on a domain is the same as integrating pullback of that form by a parameterizaton of that domain.

That's the content of the change of variables theorem.

That makes a lot of sense. I guess now I really just am having an issue understanding integration in general. So let $M$ be an $n$-dimensional manifold, $\gamma:[0,1]\to M$ be a curve, and $\alpha$ some 1-form on $M$. So we have that $\int_{\gamma([0,1])}\alpha = \int_{[0,1]}\gamma^*\alpha$. Once I'm here, how do I make the jump to $\int_{[0,1]}\alpha(\gamma^*(s))ds$.

I understand the definition of the pullback, by pushing forward tangent vectors, but my confusion comes from how do we show that $\gamma^*\alpha=\alpha(\gamma'(s)) ds$. Why does plugging in a unit tangent vector give us the proper real valued function in front of the 1-form $ds$?

Nevermind, I figured it out now. Sorry about that and thanks @BalarkaSen and @Juan. I just realized that if I take any tangent vector $v\in\mathbb{R}$ that we have that $\gamma^*\alpha(v)=\alpha(\gamma'(v))$ and since $dv(v)=1$ we can find the real valued function in front of $dv$ or $ds$ simply by plugging in any "test" vector!