Can anyone please help me understand why the negative sign is being dropped on the second integral part?

Question

anonymous · Answer

attachment

anonymous · Answer

This is in relation to voltage

turingtest · Answer

apparently they just decided to make$$dx=-d\vec\ell$$and talk in terms of magnitude
you can do that in certain situations, and since there is no more context here that is all I can say

anonymous · Answer

"apparently they just decided to make
dx=−dℓ"

...it probably has to do with the actual problem.

anonymous · Answer

Well I have an answer _ this is done when CHRGE IS NEGATIVE

anonymous · Answer

When Charge is NEGATIVE $$  F = -E* q$$

turingtest · Answer

but this integral is for V

turingtest · Answer

You could also just be working on a straight line, in which case the vector $d\vec\ell$ is unnecessary, and you can just call it a scalar line segment $dx$ to which you can assign positive and negative direction as you wish.

turingtest · Answer

and it's along a line, like I said

turingtest · Answer

and look, you are even moving in the negative x direction, so $$d\vec\ell=-dx$$from the drawing

anonymous · Answer

This is the problem