Answer
We prove that $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{2}\left( {{c^2} - {a^2}} \right)$
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):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{\left( {a,b} \right)}^{\left( {c,d} \right)} {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{\left( {a,b} \right)}^{\left( {c,d} \right)} \left( {x\left( t \right),0} \right)\cdot\left( {x'\left( t \right),y'\left( t \right)} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{\left( {a,b} \right)}^{\left( {c,d} \right)} x\left( t \right)x'\left( t \right){\rm{d}}t$
Since ${\rm{d}}x = x'\left( t \right){\rm{d}}t$, the integral becomes
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^c x{\rm{d}}x = \frac{1}{2}{x^2}|_a^c = \frac{1}{2}\left( {{c^2} - {a^2}} \right)$
Hence,
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{2}\left( {{c^2} - {a^2}} \right)$
