 Let's see another example of using integration to calculate the work. It takes to move an object when the force is variable Have a variable force here. So imagine we have a 200 pound cable that's dangling from the side of a building And the cable itself is 200 feet or 100 feet long now the building is sufficiently tall that the entire 100 feet of The cable is dangling and there's not like some coil that's bunched up at the bottom right here How much work does it take? How much work is required to lift the cable to the top of the building now? The the temptation might be be like oh take 200 pounds the weight times 100 feet the distance But that would suggest you have a 200 pound weight right here, and you have to move that up to the top So that that would be good in that situation, but the issue here is you start winding up your cable When you wind up the cable the the part that you've already lifted up You don't have to continue to lift it right once it's to the top of the building You just drop it on the ground you keep on pulling the stuff that's dangling so the stuff that's really close to the top It only has to go a short distance as opposed to the stuff at the bottom which has to go a really far distance and So as we lift the cable up the building the cable in some respect We can think of it's getting lighter along the way And so since its weight is variable as we lift it up How does one compensate for something like this and so what we're going to do is we're going to subdivide the length of the cable along The an axis which is parallel to the building itself So this is something we often do when these type of physics problems. We need to introduce a geometry Geometric coordinates of some kind now these are single variable problems So I only need one variable and let's say the top of the building is the location x equals 0 And we're going to orient the x-axis as going downward That's the positive direction and so we can do that if we want to that's just for a simpler integral You could place the origin at the bottom like at the bottom of the rope if you wanted to but I again I think this will be a little bit easier if we set up the top to be where x equals 0 represents and so what we're going to do is we're going to subdivide our x-axis into small pieces and So consider a typical piece x i minus 1 and x i So what we're going to do is when you subdivide the problem the basic idea We do in this these integration type applications is you're going to subdivide You subdivide the the problem along some axis and then you're going to assume You're going to assume that we have a quote-unquote easy problem Right and so For work functions are for work problems We assume things are easy when like we have a constant a Constant force or some kind so if you subdivide right here and let's make sure that each distance is a thickness of delta x So that's how big that piece is and So if we look at the subdivision that is location x from the origin Which is at the top there so we have the chunk of rope that is at x and so what we're going to do is work and idealize this chunk Of rope as a single point Right, so just imagine all of the mass of the all the weight of this slice of the rope is Concentrated to a single single point. How far does that point have to go to get to the top of the building? Well, that's going to be a distance of x that's your distance What is going to be the mass of the thing or this case was the weight the force of this thing? Is going to be well, how do we figure that one out? We know how long it is its length? Is going to be delta x, but how heavy is it this is where we introduce the idea of the linear density of our rope here If the rope has a density of row which row will be pounds per foot if we times pounds by foot by feet That would give us the weight of it And so how do we figure out what this linear density row is? We'll look at the total rate the total length and the total weight row will be this ratio You're going to get 200 pounds per 100 feet and Assuming that it's a uniform cable, which that's what we're talking about right here You divide that we're going to end up with two pounds per foot That's our linear density. We put it in right here. And so the force function is going to be two times delta x Like so and so as we integrate this thing The work we integrate remember we integrate force times distance That's what work turns out to be now in this situation Our force function is going to be two times dx that infinitesimal amount, right? Because we want these slices of rope gets smaller and smaller smaller because we were assuming the rope was a single point Well, if we make the subintervals gets smaller and smaller smaller that assumption becomes less erroneous The smaller and smaller we slice the rope. So the force is a two Two dx and then the the distance it has to travel is x feet to get to the top of that And so now we have to consider how far do we have to go like what's the range here? Well, if you're at the very top of the building, that's x equals zero You don't have to go any distance at all that would be a lower bound if you're at the very bottom of the rope That's going to be x equals 200 ups are those 200 pounds 100 feet And so we're going to integrate this from 100 and so the integral we're tasked with is two times integral from zero to 100 of x dx That's it. The anti-derivative of 2x of course is x squared as you go from zero to 100 Plug it in zero makes everything vanish plugging in 100 We didn't get 100 squared or there's one with four zeros behind it 10,000 Foot pounds was the units we were using here And so that's how much work you would take to lift this rope and So notice right here that when it comes to these work problems, it's not about memorizing a specific form of her specific problem The idea is applying a strategy What was the strategy here? The strategy was to to Align your problem along an axis Subdivide the axis into these small pieces and then on the axis make an assumption that makes the problem trivial That makes the problem linear and then solve the linear problem and then add those linear solutions together And take the limit which becomes an integral and then that would give us the work to lift this cable here