# How inexact differential shows work is a path function.

How inexact differential shows work is a path function.
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A 1-form $F={F}_{i}\text{d}{x}^{i}$ can be integrated along a path $\gamma :t↦x\left(t\right)$ giving :
${\int }_{\gamma }F=\int {F}_{i}\frac{\text{d}{x}^{i}}{\text{d}t}\text{d}t$
If $\gamma$ is a loop encircling a surface $\mathrm{\Sigma }$, by Stokes theorem, we have :
${\int }_{\gamma }F={\int }_{\mathrm{\Sigma }}\text{d}F$
Now, if $\text{d}F=0$, the integral of F around any (null homotopic) closed path vanishes. If ${\gamma }_{1}$ and ${\gamma }_{2}$ are two path with the same end points, then we can build a loop $\gamma$ by first following ${\gamma }_{1}$ and then ${\gamma }_{2}$ in the reverse order and we get (if ${\gamma }_{1}$ and ${\gamma }_{2}$ are homotopic) :
$0={\int }_{\gamma }F={\int }_{{\gamma }_{1}}F-{\int }_{{\gamma }_{2}}F$
Therefore the integral of F is depending only on the path. On the other hand, if $\text{d}F\ne 0$, then this integral is path dependent.
Another way, maybe more familiar in physics is the following : using the Poincaré lemma, if $\text{d}F=0$ then, there is (at least locally) a 0-form V (ie a function) such that dV. In that case :
${\int }_{\gamma }F={\int }_{\gamma }\text{d}V=V\left(b\right)-V\left(a\right)$
In other words, if $\text{d}F\ne 0$ then F does not derive from a potential.