Nach Voraussetzung können wir

${\displaystyle {}F_{i}(x_{1},\ldots ,x_{i-1},x_{i},x_{i+1},\ldots ,x_{n})=f_{i}(x_{i})\,}$

mit stetigen Funktionen

${\displaystyle f_{i}\colon \mathbb {R} \longrightarrow \mathbb {R} }$

schreiben. Es sei ${\displaystyle {}g_{i}}$ eine Stammfunktion zu ${\displaystyle {}f_{i}}$. Das Wegintegral ist somit

${\displaystyle {}{\begin{aligned}\int _{\gamma }F&=\int _{a}^{b}F_{1}(\gamma (t))\gamma _{1}'(t)+\cdots +F_{n}(\gamma (t))\gamma _{n}'(t)dt\\&=\int _{a}^{b}f_{1}(\gamma _{1}(t))\gamma _{1}'(t)+\cdots +f_{n}(\gamma _{n}(t))\gamma _{n}'(t)dt\\&=\int _{a}^{b}f_{1}(\gamma _{1}(t))\gamma _{1}'(t)dt+\cdots +\int _{a}^{b}f_{1}f_{n}(\gamma _{n}(t))\gamma _{n}'(t)dt\\&=g_{1}(\gamma (t))|_{a}^{b}+\cdots +g_{n}(\gamma (t))|_{a}^{b}\\&=g_{1}(\gamma _{1}(b))-g_{1}(\gamma _{1}(a))+\cdots +g_{n}(\gamma _{n}(b))-g_{n}(\gamma _{n}(a)).\end{aligned}}}$

${\displaystyle {}\gamma }$