Mean value theorem for line integral I am wondering if there is a mean value theorem for line integral. For example, let $f(x):\mathbb{R}^n\rightarrow \mathbb{R}$ be a continuous (not necessarily monotonic) function defined on smooth curve $C$, do we have the following theorem:
$F(x)=\int_Cf(x)dx=f(x_0)\cdot L_C$
where $x_0$ is some point on the curve $C$ (i.e., $x_0\in C$ ) and $L_C$ is the length of the curve $C$?
In addition, if $f(x)=\nabla F(x)$, do we have $F(a)-F(b)=||f(x_0)||\cdot L_C$, where $||\cdot||$ is some norm (e.g., $\ell_2$ norm)? 
Thanks.
 A: The answer is yes for continuous functions $f$ on continuous curves $C$. The reason for this is that the line integral (with respect to arc length) satisfies the inequality
$$ \min(f) \cdot L_C \leq \int_C f(x_1, \ldots x_n) \,ds \leq \max(f) \cdot L_C $$
which can be rearranged as 
$$ \min(f) \leq \cfrac{1}{L_C} \int_C f(x_1, \ldots x_n) \,ds \leq \max(f)  $$
We know that the minimum and maximum of the continuous function $f$ on $C$ exist because $C$ is compact, being the continuous image of $[0, 1]$. But $C$ is also connected for the same reason, so the range of $f$ must be an interval. The only option is that the range of $f$ must be the interval $[\min(f), \max(f)]$.
This is important because this number $\cfrac{1}{L_C} \int_C f(x_1, \ldots x_n)$ was just shown to be in that interval, and therefore is in the range of $f$. So, pick a value $\vec{c} \in C$ with $f(\vec{c}) = \cfrac{1}{L_C} \int_C f(x_1, \ldots x_n)$ and the theorem is proved.
EDIT: The added part of your question about the integral of a gradient vector field $f(x) = \nabla F(x)$ will have a negative answer. For example, a line integral of a gradient across a loop (initial point = endpoint) will give you zero, but the gradient vector field you are integrating might never be zero.
Example: Integrate the gradient of $F(x, y) = x + y$ along the counterclockwise oriented unit circle. 
