Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{2}{{\rm{e}}^4} - \frac{3}{2}{{\rm{e}}^2} + 2$
Work Step by Step
We have ${\bf{F}}\left( {x,y} \right) = \left( {{{\rm{e}}^{y - x}},{{\rm{e}}^{2x}}} \right)$, and the piecewise linear path from $\left( {1,1} \right)$ to $\left( {2,2} \right)$ to $\left( {0,2} \right)$.
The path consists of two line segments ${C_1}$ and ${C_2}$:
1. The line from $\left( {1,1} \right)$ to $\left( {2,2} \right)$ is represented by ${{\bf{r}}_1}$. It can parametrized by ${{\bf{r}}_1}\left( t \right) = \left( {1,1} \right) + t\left( {1,1} \right) = \left( {1 + t,1 + t} \right)$ for $0 \le t \le 1$.
So, ${{\bf{r}}_1}'\left( t \right) = \left( {1,1} \right)$, and ${\bf{F}}\left( {{{\bf{r}}_1}\left( t \right)} \right) = \left( {1,{{\rm{e}}^{2\left( {1 + t} \right)}}} \right)$.
2. The line from $\left( {2,2} \right)$ to $\left( {0,2} \right)$ is represented by ${{\bf{r}}_2}$.
It can parametrized by
${{\bf{r}}_2}\left( t \right) = \left( {2,2} \right) + t\left( {\left( {0,2} \right) - \left( {2,2} \right)} \right)$
$ = \left( {2,2} \right) + t\left( { - 2,0} \right)$
$ = \left( {2 - 2t,2} \right)$,
for $0 \le t \le 1$.
So, ${{\bf{r}}_2}'\left( t \right) = \left( { - 2,0} \right)$, and ${\bf{F}}\left( {{{\bf{r}}_2}\left( t \right)} \right) = \left( {{{\rm{e}}^{2t}},{{\rm{e}}^{2\left( {2 - 2t} \right)}}} \right)$.
Evaluate $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using Eq. (8):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{{C_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$
$ = \mathop \smallint \limits_a^b {\bf{F}}\left( {{{\bf{r}}_1}\left( t \right)} \right)\cdot{{\bf{r}}_1}'\left( t \right){\rm{d}}t + \mathop \smallint \limits_a^b {\bf{F}}\left( {{{\bf{r}}_2}\left( t \right)} \right)\cdot{{\bf{r}}_2}'\left( t \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 \left( {1,{{\rm{e}}^{2\left( {1 + t} \right)}}} \right)\cdot\left( {1,1} \right){\rm{d}}t + \mathop \smallint \limits_0^1 \left( {{{\rm{e}}^{2t}},{{\rm{e}}^{2\left( {2 - 2t} \right)}}} \right)\cdot\left( { - 2,0} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 \left( {1 + {{\rm{e}}^{2\left( {1 + t} \right)}}} \right){\rm{d}}t - 2\mathop \smallint \limits_0^1 {{\rm{e}}^{2t}}{\rm{d}}t$
$ = \left( {t + \frac{1}{2}{{\rm{e}}^{2\left( {1 + t} \right)}}} \right)|_0^1 - \left( {{{\rm{e}}^{2t}}|_0^1} \right)$
$ = 1 + \frac{1}{2}{{\rm{e}}^4} - \frac{1}{2}{{\rm{e}}^2} - {{\rm{e}}^2} + 1$
$ = \frac{1}{2}{{\rm{e}}^4} - \frac{3}{2}{{\rm{e}}^2} + 2$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{2}{{\rm{e}}^4} - \frac{3}{2}{{\rm{e}}^2} + 2$.