00:01
Okay, so since we're talking about a variable force in this example, we will be using the integral, as opposed to this equation right over here that relates to a constant force.
00:18
Basically, all the integral takes into account is a varying force, and how we do that is we break up the force into infinitesimally small pieces, and that accounts for any deviations.
00:34
Okay, so in this case, the force is equal to 2 over x squared neutons, and that's with respect to x, and we're going to do this along a distance from 1 to 3 meters, and that's going to give us our work.
01:05
So now this becomes just an integration problem.
01:08
We know how to set it up, and now let's integrate.
01:13
So 2 over x squared, well this could be rewritten as this.
01:19
Sometimes it makes it easier to have it in a non -fraction form...