Answer
Prove equation (11):
(11) ${\ \ \ \ \ \ \ }$ $Av\left( f \right) = \frac{1}{L}\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s$
Work Step by Step
Define the average value $Av\left( f \right)$ of a continuous function $f$ along a curve $C$ of length $L$:
$Av\left( f \right) = \mathop {\lim }\limits_{N \to \infty } \frac{1}{N}\mathop \sum \limits_{i = 1}^{} f\left( {{P_i}} \right)$
Write
$Av\left( f \right) = \frac{1}{L}\mathop {\lim }\limits_{N \to \infty } \frac{L}{N}\mathop \sum \limits_{i = 1}^{} f\left( {{P_i}} \right)$
Let $\Delta {s_i}$ denote the length of ${C_i}$. So,
$Av\left( f \right) = \frac{1}{L}\mathop {\lim }\limits_{N \to \infty } \mathop \sum \limits_{i = 1}^{} f\left( {{P_i}} \right)\Delta {s_i}$
As $N \to \infty $, we get $\Delta {s_i}$ approaches zero. So, we can write
$Av\left( f \right) = \frac{1}{L}\mathop {\lim }\limits_{\Delta {s_i} \to 0} \mathop \sum \limits_{i = 1}^{} f\left( {{P_i}} \right)\Delta {s_i}$
By definition, the term $\mathop \sum \limits_{i = 1}^{} f\left( {{P_i}} \right)\Delta {s_i}$ at the right-hand side is the Riemann sum approximation to the line integral of $f$ along $C$. Taking the limit as $\Delta {s_i}$ approaches zero, the Riemann sum converges to the line integral of $f$ along $C$. Hence,
(11) ${\ \ \ \ \ \ \ }$ $Av\left( f \right) = \frac{1}{L}\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s$
