Answer
$\mathop \smallint \limits_C^{} \left( {{{\rm{e}}^{x - y}},{{\rm{e}}^{x + y}}} \right)\cdot{\rm{d}}{\bf{r}} \simeq - 4.5088$
Work Step by Step
Write the vector field ${\bf{F}}\left( {x,y} \right) = \left( {{{\rm{e}}^{x - y}},{{\rm{e}}^{x + y}}} \right)$ and the path ${\bf{r}} = \left( {x,y} \right)$, such that
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} \left( {{{\rm{e}}^{x - y}},{{\rm{e}}^{x + y}}} \right)\cdot{\rm{d}}{\bf{r}}$
The curve can be parametrized by ${\bf{r}}\left( t \right) = \left( {t,\sin t} \right)$ for $0 \le t \le \pi $. So,
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{{\rm{e}}^{t - \sin t}},{{\rm{e}}^{t + \sin t}}} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( {1,\cos t} \right)dt$
Using Eq. (8), the vector line integral becomes
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} \left( {{{\rm{e}}^{x - y}},{{\rm{e}}^{x + y}}} \right)\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^2 {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} \left( {{{\rm{e}}^{x - y}},{{\rm{e}}^{x + y}}} \right)\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^\pi \left( {{{\rm{e}}^{t - \sin t}},{{\rm{e}}^{t + \sin t}}} \right)\cdot\left( {1,\cos t} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} \left( {{{\rm{e}}^{x - y}},{{\rm{e}}^{x + y}}} \right)\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^\pi \left( {{{\rm{e}}^{t - \sin t}} + {{\rm{e}}^{t + \sin t}}\cos t} \right){\rm{d}}t$
Using a computer algebra system, we compute the integral and the result is
$\mathop \smallint \limits_0^\pi \left( {{{\rm{e}}^{t - \sin t}} + {{\rm{e}}^{t + \sin t}}\cos t} \right){\rm{d}}t \simeq - 4.5088$
So, $\mathop \smallint \limits_C^{} \left( {{{\rm{e}}^{x - y}},{{\rm{e}}^{x + y}}} \right)\cdot{\rm{d}}{\bf{r}} \simeq - 4.5088$.
