Answer
$\mathop \smallint \limits_C^{} 1{\rm{d}}s = {{\rm{e}}^2} - {{\rm{e}}^{ - 2}}$
Work Step by Step
We have the parametrization of the curve $C$ given by ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t},\sqrt 2 t,{{\rm{e}}^{ - t}}} \right)$ for $0 \le t \le 2$.
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {{{\rm{e}}^t},\sqrt 2 , - {{\rm{e}}^{ - t}}} \right)\cdot\left( {{{\rm{e}}^t},\sqrt 2 , - {{\rm{e}}^{ - t}}} \right)} = \sqrt {{{\rm{e}}^{2t}} + 2 + {{\rm{e}}^{ - 2t}}} $
$ = \sqrt {{{\left( {{{\rm{e}}^t} + {{\rm{e}}^{ - t}}} \right)}^2}} = {{\rm{e}}^t} + {{\rm{e}}^{ - t}}$
Using Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$
where in this case $f\left( {x,y,z} \right) = 1$, we get
$\mathop \smallint \limits_C^{} 1{\rm{d}}s = \mathop \smallint \limits_0^2 \left( {{{\rm{e}}^t} + {{\rm{e}}^{ - t}}} \right){\rm{d}}t = \left( {{{\rm{e}}^t} - {{\rm{e}}^{ - t}}} \right)|_0^2 = {{\rm{e}}^2} - {{\rm{e}}^{ - 2}}$
