Answer
$\mathop \smallint \limits_C^{} \left( {3x - 2y + z} \right){\rm{d}}s = - \frac{9}{2}\sqrt 6 $
Work Step by Step
We have $f\left( {x,y,z} \right) = 3x - 2y + z$ and the parametrization ${\bf{r}}\left( t \right) = \left( {2 + t,2 - t,2t} \right)$ for $ - 2 \le t \le 1$.
So,
$f\left( {{\bf{r}}\left( t \right)} \right) = 3\left( {2 + t} \right) - 2\left( {2 - t} \right) + 2t = 7t + 2$
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {1, - 1,2} \right)\cdot\left( {1, - 1,2} \right)} = \sqrt {1 + 1 + 4} = \sqrt 6 $
By Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$\mathop \smallint \limits_C^{} \left( {3x - 2y + z} \right){\rm{d}}s = \sqrt 6 \mathop \smallint \limits_{ - 2}^1 \left( {7t + 2} \right){\rm{d}}t = \sqrt 6 \left( {\left( {\frac{7}{2}{t^2} + 2t} \right)|_{ - 2}^1} \right)$
$ = \sqrt 6 \left( {\frac{7}{2} + 2 - 14 + 4} \right) = - \frac{9}{2}\sqrt 6 $
So, $\mathop \smallint \limits_C^{} \left( {3x - 2y + z} \right){\rm{d}}s = - \frac{9}{2}\sqrt 6 $.
