Definitions
Consider a generic smooth closed curve $c(t):[0,1]\mapsto \mathbb{R}^2$ and define its barycenter $g$ as \begin{equation*} g \triangleq \frac{\int_0^1 c(t)\,\lVert \dot{c}(t)\rVert \text{ d}t}{L} \end{equation*} where $L\triangleq\int_0^1 \lVert \dot{c}(t)\rVert \text{ d}t$ and \begin{equation*} \dot{c}(t) \triangleq \frac{\text{d}c}{\text{d}t}(t) \end{equation*} is the tangent vector at point $t\in[0,1]$. Now, for some $\bar{t}\in(0,1)$ split $c(t)$ in the two segments $c_1(t)\triangleq\{c(t)\}_{t\in[0,\overline{t})}$, $c_2(t)\triangleq\{c(t)\}_{t\in[\overline{t},0)}$ and define their barycenters as \begin{equation*} g_1 \triangleq \frac{\int_0^\overline{t} c(t)\,\lVert \dot{c}(t)\rVert \text{ d}t}{L_1} \qquad g_2 \triangleq \frac{\int_\overline{t}^1 c(t)\,\lVert \dot{c}(t)\rVert \text{ d}t}{L_2} \end{equation*} with $L_1\triangleq\int_0^{\overline{t}}\lVert \dot{c}(t)\rVert \text{ d}t$, $L_2\triangleq \int_{\overline{t}}^1 \lVert \dot{c}(t)\rVert \text{ d}t$.
Problem
I'm wondering if the following relation holds \begin{equation*} g=\frac{L_1}{L}g_1 +\frac{L_2}{L} g_2 \end{equation*} I believe that is true because, if I'm not wrong, \begin{equation}\begin{aligned} g &= \frac{\int_0^\overline{t} c(t)\,\lVert \dot{c}(t)\rVert \text{ d}t+\int_\overline{t}^1 c(t)\,\lVert \dot{c}(t)\rVert \text{ d}t}{L} \\ &=\frac{L_1}{L} \frac{\int_0^\overline{t} c(t)\,\lVert \dot{c}(t)\rVert \text{ d}t}{L_1}+ \frac{L_2}{L} \frac{\int_\overline{t}^1 c(t)\,\lVert \dot{c}(t)\rVert \text{ d}t}{L_2}\\ &=\frac{L_1}{L} g_1 + \frac{L_2}{L} g_2 \end{aligned} \tag{$\star$} \end{equation} The problem is that I've tried to see if this is true numerically. I've wrote a script in which I approximate the curve $c(t)$ with a finite, ordered and wrapped sequence of $m=10^4$ points $c_i \triangleq c(t_i)\in \mathbb{R}^2$ (stacked in a vector) \begin{equation*} \hat{c} \triangleq \left[\begin{array}{cccc}c_1' & \cdots & c_m' & c_{m+1}'\triangleq c_1'\end{array}\right]' \end{equation*} so that its barycenter is approximated as \begin{equation*} \hat{g} \triangleq \frac{1}{\hat{L}} \sum_{i=1}^{m+1} c_i \qquad \hat{L}\triangleq m+1 \end{equation*} In the same way, for some $\overline{m}\in\{1,\dots,m\}$ I've defined \begin{equation*}\begin{aligned} \hat{c}_1 &\triangleq \left[\begin{array}{ccc}c_1' & \cdots & c_{\overline{m}}'\end{array}\right]' \\ \hat{c}_2 &\triangleq \left[\begin{array}{ccc}c_{\overline{m}+1}' & \cdots & c_{m+1}'\end{array}\right]' \\ \end{aligned} \end{equation*} and \begin{equation*} \begin{aligned} \hat{g}_1 &\triangleq \frac{1}{\hat{L}_1} \sum_{i=1}^{\overline{m}} c_i \qquad \hat{L}_1 \triangleq \overline{m}\\ \hat{g}_2 &\triangleq \frac{1}{\hat{L}_2} \sum_{i=\overline{m}}^{m+1} c_i \qquad \hat{L}_2 \triangleq m-\overline{m}+1 \end{aligned} \end{equation*} but I was not able to verify that \begin{equation*} \hat{g} \approx \frac{\hat{L}_1}{\hat{L}}\hat{g}_1 + \frac{\hat{L}_2}{\hat{L}}\hat{g}_2 \end{equation*} thus, I have to understand if I've made some mistakes in my code or if my derivations written here are wrong.
Question
Is it correct $(\star)$? If the answer is yes, do make sense my finite approximations of the quantities of $\{c(t)\}_{t\in[0,1]}, g, \{c_1(t)\}_{t\in[0,\overline{t})}, \{c_2(t)\}_{t\in[\overline{t}, 1]}$?