Answer
We prove that $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = cd - ab$
Work Step by Step
Let $C$ be any path from $\left( {a,b} \right)$ to $\left( {c,d} \right)$ parametrized by ${\bf{r}}\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right)$.
Calculate $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using Eq. (8):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{\left( {a,b} \right)}^{\left( {c,d} \right)} {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{\left( {a,b} \right)}^{\left( {c,d} \right)} \left( {y\left( t \right),x\left( t \right)} \right)\cdot\left( {x'\left( t \right),y'\left( t \right)} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{\left( {a,b} \right)}^{\left( {c,d} \right)} \left[ {x'\left( t \right)y\left( t \right) + x\left( t \right)y'\left( t \right)} \right]{\rm{d}}t$
Since ${\rm{d}}\left( {xy} \right) = \left[ {x'\left( t \right)y\left( t \right) + x\left( t \right)y'\left( t \right)} \right]{\rm{d}}t$, the integral becomes
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{\left( {a,b} \right)}^{\left( {c,d} \right)} {\rm{d}}\left( {xy} \right) = \left( {xy} \right)|_{\left( {a,b} \right)}^{\left( {c,d} \right)} = cd - ab$
Hence,
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = cd - ab$
