I am trying to find the work done on a particle when a varying force acts on it. I am familiar with W = ∫F dot ds but am running into a sign question. Suppose a particle is moving in the -x direction because a force is acting on in that direction. Suppose it moves from x = 4 to x = 0, and suppose the force is given by F = -5 i. Then from W = Fs cos θ, the work would be (5)(4) cos 0 = 20 J. It makes sense that the work is positive since the force and the displacement are both in the same direction. Now I know an integral would not be needed for this problem (but another problem I am working on does, hence my question). When I used ∫F dot ds (with unit vectors) to do this, I got a negative answer for the work done. I used F = -5i (since the force is in the -x direction) ds = -dx i (since the path is from 4 to 0, the length element points towards -x) and when I integrated, I did so from x = 4 to x = 0 since that was the path. Then ∫F dot ds = ∫(-5i) dot (-dx i) = 5∫dx = 5x, but when I plug in the limits of the interval and take 5(0) - 5(4) = -20 J. Where am I putting in a negative sign that does not belong?