Answer
The average value of $f$ along the segment is $Av\left( f \right) = \frac{1}{2}$.
Work Step by Step
The line segment $C$ can be parametrized by
${\bf{r}}\left( t \right) = \left( {2,1} \right) + t\left( {\left( {5,5} \right) - \left( {2,1} \right)} \right)$
$ = \left( {3t + 2,4t + 1} \right)$
for $0 \le t \le 1$.
So,
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {3,4} \right)\cdot\left( {3,4} \right)} = \sqrt {25} = 5$
The length of $C$ is $L = \sqrt {{{\left( {5 - 2} \right)}^2} + {{\left( {5 - 1} \right)}^2}} = \sqrt {9 + 16} = 5$.
Using Eq. (11) we calculate the average value of $f\left( {x,y} \right) = x - y$ along the segment $C$ as
$Av\left( f \right) = \frac{1}{5}\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s$
By Eq. (4), the integral becomes
$Av\left( f \right) = \frac{1}{5}\mathop \smallint \limits_0^1 f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$Av\left( f \right) = \mathop \smallint \limits_0^1 \left( { - t + 1} \right){\rm{d}}t = \left( { - \frac{1}{2}{t^2} + t} \right)|_0^1 = - \frac{1}{2} + 1 = \frac{1}{2}$
So, the average value of $f$ along the segment is $Av\left( f \right) = \frac{1}{2}$.
