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By Gregory, the total work done by all the external forces (that's the F

_{i}s ) is:

$$\sum_{i=1}^{N} \int_{t_A}^{t_B} \vec{F_i} \cdot \vec{v_i} dt = \sum_{i=1}^{N} (\phi_i(\vec{r_A}) - \phi_i(\vec{r_B})) $$

What I don't understand is how to go from the integral:

$$\int_{t_A}^{t_B} \vec{F_i} \cdot \vec{v_i} dt$$ to the potentials.

My idea is:

$$\int_{t_A}^{t_B} \vec{F_i} \cdot \vec{v_i} dt = \int_{t_A}^{t_B} -\nabla \phi_i \cdot \vec{v_i} dt = \int_{\vec{r_A}}^{\vec{r_B}} -\nabla \phi_i \cdot \vec{dr} = \phi_i(\vec{r_A}) - \phi_i(\vec{r_B})$$

My questions are :

Can we go from a time integral to a position integral without messing with the potentials? I know they are path independent, but are they time independent?

Does integrating from t

_{A}to t

_{B}do the same sum as integrating from r

_{A}to r

_{B}?