Prove: $\omega$ Is Exact

Question

Let $\omega=\omega_1dx_1+...+\omega_ndx_n$ be a continuous form on an open and connected set $\Omega \subset \mathbb{R}^n$.

Let assume that for all $x,y\in \Omega$ there is $c\in \mathbb{R}$ such that $\int_{\gamma}\omega=c$ for every $\gamma$ that connects $x$ to $y$.

Prove: $\omega$ is exact in $\Omega$

$\int_{x}^{y}\omega=c$ for any $\gamma$ so $-\int_{y}^{x}\omega=-c$ for any $\gamma$ summing them both we get $\oint \omega=0$ and therefore $\omega$ is exact

Am I missing out something here? I have used the theorem:

$\omega$ is exact $\iff\oint \omega=0 \iff$ any $\gamma_1,\gamma_2$ with same start and end $\int_{\gamma_1}\omega=\int_{\gamma_2}\omega$

Answer 1

You don't need the middle part of your theorem, or to think about closed curves at all. The right part

any $\gamma_1,\gamma_2$ with same start and end $\int_{\gamma_1}\omega=\int_{\gamma_2}\omega$

is exactly the property you've been given, only worded differently (this is the only thing I can see that is of any merit to "show"). The left part

$\omega$ is exact

is what you are after. Connect the two using your theorem, and you're done.