Answer
$\mathop \smallint \limits_C^{} \left( {2{x^2} + 8z} \right){\rm{d}}s = \frac{2}{3}\left( {{{\left( {{{\rm{e}}^2} + 5} \right)}^{3/2}} - 2\sqrt 2 } \right)$
Work Step by Step
We have $f\left( {x,y,z} \right) = 2{x^2} + 8z$ and the parametrization ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t},{t^2},t} \right)$ for $0 \le t \le 1$.
So,
$f\left( {{\bf{r}}\left( t \right)} \right) = 2{{\rm{e}}^{2t}} + 8t$
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {{{\rm{e}}^t},2t,1} \right)\cdot\left( {{{\rm{e}}^t},2t,1} \right)} = \sqrt {{{\rm{e}}^{2t}} + 4{t^2} + 1} $
By Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$\mathop \smallint \limits_C^{} \left( {2{x^2} + 8z} \right){\rm{d}}s = \mathop \smallint \limits_0^1 \left( {2{{\rm{e}}^{2t}} + 8t} \right)\sqrt {{{\rm{e}}^{2t}} + 4{t^2} + 1} {\rm{d}}t$
Write $u = {{\rm{e}}^{2t}} + 4{t^2} + 1$. So, $du = \left( {2{{\rm{e}}^{2t}} + 8t} \right)dt$. The integral becomes
$\mathop \smallint \limits_C^{} \left( {2{x^2} + 8z} \right){\rm{d}}s = \mathop \smallint \limits_2^{{{\rm{e}}^2} + 5} {u^{1/2}}{\rm{d}}u = \frac{2}{3}\left( {{u^{3/2}}|_2^{{{\rm{e}}^2} + 5}} \right)$
$ = \frac{2}{3}\left( {{{\left( {{{\rm{e}}^2} + 5} \right)}^{3/2}} - 2\sqrt 2 } \right)$
So, $\mathop \smallint \limits_C^{} \left( {2{x^2} + 8z} \right){\rm{d}}s = \frac{2}{3}\left( {{{\left( {{{\rm{e}}^2} + 5} \right)}^{3/2}} - 2\sqrt 2 } \right)$.
