Line integal of vector field depends only on the shape and orientation of the curve? My question is about line integral of vector field in multi-variable calculus.
As you know well, the line integral of a vector field over some parametrized curve $X(t)$ is independent of reparametrization preserving same orientation.
It sounds that it does not ensure that the value of line integral is not completely determined by the image of curve, but equivalent class of parametrization.
I am wondering that if there is a theorem that for two $C^1$-parametrizations $X_1:[a,b]\rightarrow \mathbb{R}$, $X_2:[c,d]\rightarrow\mathbb{R}$ of some fixed curve which having same orienatation, is there some $C^1$-function $g:[c,d]\rightarrow[a,b]$ such that $X_1 \circ g=X_2$? 
If not, can you have some easy example which gives different values according to two different orientation preserving parametrization of some curve?
More precisely,
Let $F:\mathbb{R^n}\rightarrow \mathbb{R^n}$ be a continuous-function and C be a curve (the set of points) parametrized by $X_1(t):[a,b]\rightarrow\mathbb{R^n}$ and it is $C^1$-curve (differentiable and its partial derivatives are continuous).
Is there another $C^1$ parametrization $X_2(t):[c,d]\rightarrow\mathbb{R^n}$ of $C$ having the same orientation with $X_1(t)$ but $\int_{X_1}F\cdot d\mathbf{s}\ne\int_{X_2}F\cdot d\mathbf{s}$?
My question comes from the confusing notation that many author write the line integral over the curve(the set of points) not explicating any parametrization.  
Any comment will be greatly appreciated!
Similar question can be asked in line integral in complex analysis.
 A: Edit: The result you are looking for is stated here http://mathinsight.org/line_integral_independent_parametrization
although without proof.
No, that is not the case - I am blanking on the name, but from analysis there is the theorem that I believe goes something as follows: If $\gamma_1$ and $\gamma_2$ are two parametrizations taking $D \subset \mathbb{R}^k$ to $\mathbb{R}^n$, then if $\gamma_1 (D) = \gamma_2 (D)$ and $\Omega (\gamma_1) = \Omega(\gamma_2)$ (they possess the same orientation), then 
\begin{align}
\int_{\gamma_1 (D)} {\varphi} &= \int_{D} {\varphi \left( \gamma_1 ^1 (\mathbf{u}), \gamma_1 ^2 (\mathbf{u}), \dots , \gamma_1 ^k (\mathbf{u}) \right) \left| \mathrm{d}^k \mathbf{u} \right|} \\
&= \int_{D} {\varphi \left( \gamma_2 ^1 (\mathbf{u}), \gamma_2 ^2 (\mathbf{u}), \dots , \gamma_2 ^k (\mathbf{u}) \right) \left| \mathrm{d}^k \mathbf{u} \right|} = \int_{\gamma_2 (D)} {\varphi} 
\end{align}
where $\varphi$ is operating on the collection of $\gamma_1 ^i (\mathbf{u})$ and $\gamma_2 ^i (\mathbf{u})$, and the superscript is (unconventionally) denoting the derivative with respect to the $i$th component.
Unfortunately, I do not have my text with me right now, so I can't be sure, but I believe that was the gist of it.
