I am trying to learn about differential forms. I know that A $0$-form is just a scalar function $f$. My question is: How is the integral of a $0$-form defined?
In particular, if $f$ is a function of one variable (so the usual functions from basic calculus). From regular calculus I understand how to find $$ \int_a^b f(x)\; dx $$ but how does this look like when we use differential forms?
If $f$ is a $0$-form, then $df$ is a $1$-form, and then I think that $$ \int_a^b df = f(b) - f(a). $$ Is that correct?
In terms of differential forms, the standard integration you learned in calculus is actually about integrating $1$-forms: namely, the $1$-forms $f(x) \, dx$.
In general, you integrate $n$-forms over $n$-dimensional manifolds. So a $0$-form would be integrated over a $0$-dimensional manifold: that is, a discrete set of points $P$, taken with orientation (i.e., each point is assigned a sign). We then have
$$\int_P f = \sum_{p \in P} f(p) \sigma(p)$$
where $\sigma(p)$ is the sign assigned to $p$.
Remember that Stokes' theorem for differential forms says that
$$ \int_{\partial M} f = \int_M df $$ where $\partial M$ is the boundary of the manifold $M$.
If we consider the case where $M$ is the interval $[a,b]$, its boundary consists of the two points $a$ and $b$, where $b$ is given positive sign and $a$ is given negative sign. So Stokes' theorem along with the above definition of zero-dimensional integral says that $$ \int_{[a, b]} df = \int_{[a, b]} f'(x) \, dx = f(b)-f(a) $$ which is just the ordinary fundamental theorem of calculus.
Note however that this is not actually helpful for performing integrals! Given a function $g$ on the interval, you still need to find the $f$ such that $g(x) \, dx = df$: that is, such that $f'(x)=g(x)$. This is exactly what you needed to do in order to integrate $g$ before you knew anything about differential forms. The correct way to understand this fact is as the reason why people call the general version of Stokes' theorem an $n$-dimensional generalization of the fundamental theorem of calculus, not as any way to make concrete integration problems easier.