By Poincare lemma, for contractible manifold , say $\mathbb{R}^n $, every closed form is exact.\
In addition, all $n^{th}$ forms on $\mathbb{R}^n$ are closed. How come that not all Riemennian standard integrals nonzero?
We can always take positive bump function on the sphere, which holds non zero integral.
There's no contradiction because Stokes's theorem only holds for forms with compact support: given your $n$-form, Poincaré's lemma doesn't tell us anything about the support of the $n-1$ form.
For example, let $n=1$; $\omega=f(x)dx$ be our $1-$form, where $f$ is the positive bump function. As Poincaré's lemma predicts, defining $h(x)=\int_0^xf(t)dt$ gives us $dh=\omega$, but $h$ does not have compact support. If we restrict ourselves to a compact submanifold of $\mathbb{R}$, e.g. $[a,b]$ Stokes' theorem holds, and it is simply the FTC.
What you found is exactly the reason for which, in the hypotesis for Stokes's thm, we ask that either our manifold is compact or the support of the form is. If you look closely to a proof of ST, you will see that this hypotesis is crucial in the proof.