Answer
$\mathop \smallint \limits_C^{} x{{\rm{e}}^{{z^2}}}{\rm{d}}s = \frac{1}{2}\sqrt 3 \left( {{\rm{e}} - 1} \right)$
Work Step by Step
We have $f\left( {x,y,z} \right) = x{{\rm{e}}^{{z^2}}}$, and the piecewise linear path from $\left( {0,0,1} \right)$ to $\left( {0,2,0} \right)$ to $\left( {1,1,1} \right)$.
The path consists of two line segments ${C_1}$ and ${C_2}$:
1. The line from $\left( {0,0,1} \right)$ to $\left( {0,2,0} \right)$ is represented by ${{\bf{r}}_1}$. It can parametrized by ${{\bf{r}}_1}\left( t \right) = \left( {0,t, - \frac{1}{2}t + 1} \right)$ for $0 \le t \le 1$.
2. The line from $\left( {0,2,0} \right)$ to $\left( {1,1,1} \right)$ is represented by ${{\bf{r}}_2}$.
It can parametrized by
${{\bf{r}}_2}\left( t \right) = \left( {0,2,0} \right) + t\left( {\left( {1,1,1} \right) - \left( {0,2,0} \right)} \right)$, ${\ \ \ }$ for $0 \le t \le 1$
${{\bf{r}}_2}\left( t \right) = \left( {0,2,0} \right) + t\left( {1, - 1,1} \right)$
${{\bf{r}}_2}\left( t \right) = \left( {t,2 - t,t} \right)$
So,
$f\left( {{{\bf{r}}_1}\left( t \right)} \right) = 0$, ${\ \ \ }$ $f\left( {{{\bf{r}}_2}\left( t \right)} \right) = t{{\rm{e}}^{{t^2}}}$
Since $f\left( {{{\bf{r}}_1}\left( t \right)} \right) = 0$, the scalar line integral along ${{\bf{r}}_1}\left( t \right)$ is zero. So, we do not need to evaluate $||{{\bf{r}}_1}'\left( t \right)||$.
Evaluate $||{{\bf{r}}_2}'\left( t \right)||$:
$||{{\bf{r}}_2}'\left( t \right)|| = \sqrt {\left( {1, - 1,1} \right)\cdot\left( {1, - 1,1} \right)} = \sqrt {1 + 1 + 1} = \sqrt 3 $
By Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_{{C_1}}^{} f\left( {x,y,z} \right){\rm{d}}s + \mathop \smallint \limits_{{C_2}}^{} f\left( {x,y,z} \right){\rm{d}}s$
$ = 0 + \mathop \smallint \limits_0^1 f\left( {{{\bf{r}}_2}\left( t \right)} \right)||{{\bf{r}}_2}'\left( t \right)||{\rm{d}}t$
$ = \sqrt 3 \mathop \smallint \limits_0^1 t{{\rm{e}}^{{t^2}}}{\rm{d}}t$
$ = \frac{1}{2}\sqrt 3 \mathop \smallint \limits_0^1 {{\rm{e}}^{{t^2}}}{\rm{d}}\left( {{t^2}} \right)$
$ = \left( {\frac{1}{2}\sqrt 3 {{\rm{e}}^{{t^2}}}|_0^1} \right)$
$ = \frac{1}{2}\sqrt 3 \left( {{\rm{e}} - 1} \right)$
So, $\mathop \smallint \limits_C^{} x{{\rm{e}}^{{z^2}}}{\rm{d}}s = \frac{1}{2}\sqrt 3 \left( {{\rm{e}} - 1} \right)$.
