If $M$ is connected and $\int_M \omega = 0,$ then $\omega = 0$? Let $M$ be a connected, orientable, and compact $(r+1)$-dimensional manifold and $\omega$ a $(r+1)$-differential form on $M$. If $\int_M \omega = 0$, then $\omega =0$?
I want to say that the answer is yes. The intuitive reason behind it is that if $\omega\neq 0$, since $M$ is connected then "it must be $\omega>0$ or $\omega<0"$, so we have $\int_M \omega \neq 0$. I've been trying to formalize this intuition by stating what means $\omega>0$ first. Since we can write $\omega(x) = a(u) du_1\wedge \cdots \wedge du_{r+1},x\in M,$ for $x = \varphi(u)$ and some chart $\varphi$, then $\omega>0$ would mean that $a(u)>0$ for every $x\in M$. But i'm struggling to use the fact that $M$ is connected to ensure that $a(u)>0$ should happen.
How can I proper formalize this intuition?

Let $f:[0,2\pi]\to\mathbb{R}$, note that $M=[0,2\pi]$ connected, orientable and compact, and define $f(x)=\sin(x)$ a $0$-form, then
$$\int_M f(x) \,dx=0$$
