Understanding of electric potential's integration form I already known that the potential difference when a charge moves from A to B is

But I still have confusions about what does the infinitesimal of vector $s$ refers. I mean when you change the movement of the charge from B to A, the $\Delta V$ should be opposite number of it. But if the E and S's direction is opposite, the dot product of E and S should be negative. Since the range of integration is reversed, the out come of the delta Vab is same to delta Vba.
Help is really appreciated.
 A: If I understand your question, I think you are assuming that when you integrate from B to A two things change: the direction of ds (and therefore the dot product of E and ds) and the integral (because you changed the limits) and therefore you end up with the same as when you integrated from A to B. But integrating from B to A just changes ds to the negative of what it was when integrating from A to B. That's what changing the limits does. So the dot product is the negative of what it was when integrating from A to B and the end result is the negative of the original potential difference.
A: you appear to have accounted for the change in direction twice. You need to first recall why reversing the limits change the signs in good old integration of functions in the cartesian plane. Although we simply put the minus sign, the reasoning behind it is that the new  or infinitesimal change in x is negative of what it was before (since you are varying x in the opposite direction) and you directly account for the new  with a minus sign outside the integral and the old dx inside
Here, since you are already accounting for the new displacement vector in the proper way, you no longer need to use the "reversing limits reverses signs rule" as that would be repetitive
