Answer
The integral $\mathop \smallint \limits_C^{} 1{\rm{d}}s$ represents the length of the line segment from the point $\left( {8, - 6,24} \right)$ to the point $\left( {20, - 15,60} \right)$.
Work Step by Step
We have the parametrization of the curve $C$ given by ${\bf{r}}\left( t \right) = \left( {4t, - 3t,12t} \right)$ for $2 \le t \le 5$.
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {4, - 3,12} \right)\cdot\left( {4, - 3,12} \right)} = \sqrt {16 + 9 + 144} = \sqrt {169} = 13$
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
$\mathop \smallint \limits_C^{} 1{\rm{d}}s = 13\mathop \smallint \limits_2^5 {\rm{d}}t = 13\cdot\left( {5 - 2} \right) = 39$
The curve $C$ parametrized by ${\bf{r}}\left( t \right) = \left( {4t, - 3t,12t} \right)$ for $2 \le t \le 5$ is a line segment from the point $\left( {8, - 6,24} \right)$ to the point $\left( {20, - 15,60} \right)$. So, the integral $\mathop \smallint \limits_C^{} 1{\rm{d}}s$ represents the distance of these two points, or in other words the length of the line segment.
