Differential forms: How do I establish uniqueness of this function? Conclude that if $\eta$ is an exact $1$-form, and $p \in U$, that there exists a unique function $g$ such that $dg = \eta$ and $g(p) = 0$. $U \subset \mathbb{R}^n$ is open and connected. 
By definition of exactness, I know such a $g$ exists but I am unable to prove the uniqueness bit. Was considering expressing $g$ as in integral in generality. 

Since $\eta$ is exact, $\eta = df$ for some function $f$ on $U$.  Let $g(x) = f(x) - f(p)$.  Then $dg = df = \eta$ and $g(p) = f(p)-f(p) = 0$.
Now suppose $\tilde g$ is any function on $U$ satisfying $d\tilde g = \eta$ and $g(p) = 0$.  Let $h = g - \tilde g$.  Then $dh = 0$, which means that $h$ is constant on connected components of $U$.  Since $U$ is connected, $h$ is indeed constant.  Since $h(p) = 0$, that constant is zero.  Hence $g$ and $\tilde g$ are identical.
