Calculus Work Problem involving an integral?PLEASE HELP!?

Well, work is defined to be the integral of F dotted with dx.

W = ∫F ∙ ds

They beauty of this problem is that we're only moving in one dimension... Everything is always on the x axis!

So the integral becomes

W = ∫F ∙ dx = ∫F dx

So, what's the force between the two charges???

F = k * q1 * q2 / r².

Since k * q1 * q2 is a constant... we'll call it P. k is an actual physical constant. r is the distance between the 2 particles

So F = P/(5 - x)²

and

W = ∫P/(5 -x )² dx = P∫dx/(5 - x)² = P/(5 - x)

x ranges from 0 to 3, so

P/(5 - 3) - P/(5 - 0) = 3p/10

My p is your k.

W = ∫ F dx

"The force (Newtons) with which 2 electrons repel each other is inversely proportional to the square of the distance"

tells you that

F=k/d²

"if one electron is held fixed at the point (5,0), find the work done in moving a second electron along the x-axis from the origin to the point (3,0)"

tells you how to setup d and your limits of integration

one electron starts at 0 and moves towards five so,

d=(5-x)

F=k/(5-x)² so

Since, again, W = ∫ F dx and using "from the origin to the point (3,0)" to get the limits of integration

W= ∫ k/(5-x)² dx from 0 to 3

W = k ∫1/(5-x)² dx from 0 to 3

To verify that it is 3k/10

u = 5-x

dx = -1 du

W = -k ∫1/u² du from 5 to 2

integrating -k [-1/u from 5 to 2]

k [1/2 - 1/5]

3k/10

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This is an extremely completed equation. Are you like in AP Calculas?? Omg...I'm in Honors Calculas and I've never done anything like this..Umm..Sorry