 Now we saw in the previous video how one can calculate the work it takes to move an object If that force remains constant throughout the entire journey, it's fairly simple just force times distance now suppose that the force is Variable throughout the journey as we push something along Like let's for example say we have some object We'll just draw it as a box right here and we want to push it forward and And then of course we're trying the frictional force is the thing we have to be worried about right as we push it along the ground There's rubbing and such, but what if like the friction Changes with time right maybe the material or it gets just kind of we have to go up and down or whatever It could very well be that as we push this thing along our force function changes based upon our location So if we think of it as like we started X equals zero and we're trying to move over to some other location Down the road. We look at a specific location X The force it takes to push it at that moment could be a function of its location f of X So if we wanted to calculate the force or the work excuse me to move the object from Location zero or we could call location a over here to location B How might we do that? Well, if the function of force is variable what we could do is we could slice up our Domain into teeny tiny little pieces, right? We call the first one the first slice x1 x2 x3 Continue on we'll have some place in the middle x i minus 1 x i And so then we think of just just look at one subinterval right here x i minus 1 to x i let's on this piece assume That force is constant for this small section, right? We have to pick a representative. We'll say the force will be equal to x f of x i star or x i star some number that sits inside this interval, right? So we choose it somehow x i star and so the force is Equal to f of x i star just for that interval for the next interval we pick a different force for the next one Pick a different force, so we'll assume that it's constant on that little chunk and which case then the work The work of pushing it from we'll call this the work w i the work from pushing it from x i minus 1 to x i Would then be the force which as we're assuming is constant to be f of x i star And then we times it by the distance how far do you have to go? Well the distance you have to go is from x i minus 1 to x i and as all of these things are equally spaced The distance you would have to go is delta x right here So this w i gives you the work to do one of these little segments All right, but there's a lot of these segments There's the first one the second one the third one Etc. All right if we add up the work to do the first one with the one to do the second one with the One to do the third one and go until the end if we take the sum of those things that gives us Approximation of the work that's necessary to be done But there's going to be error to it because we're assuming right there's an assumption there that the work The force is constant from one point to another well to have less error and to be more accurate We could shrink these intervals to be smaller and smaller and smaller smaller kind of sounds like the area problem We've done before this is our technique of accumulation here We just make the subintervals get smaller and smaller smaller and thus we get more accurate to calculations The best we could do is to make these intervals be Infantessibly small that is we could take an infinite number of these subdivisions if we take the limit as the number of subdivisions goes to infinity This would calculate the work to move from point a to point b But as this is now the limit of a Riemann sum this becomes an integral and so this is the takeaway We want to get from this is that the work Work is equal to the integral of force with respect to distance We're here. We're denoting your position function as x so if we integrate force with respect with respect to position We get our work quantity here, and that's a very nice way of taking the simple Constant work problem we saw before and we can make it a variable work by using this integral And this is what we've seen before with volume problems with area problems This is what we call this strategy of accumulation or strategy of integration If we can solve a problem when as we can assume things are what we call linear if we can solve the linear problem Then we can approximate variable situations using linear approximations, and we can integrate those linear approximations Let's see an example of such a thing suppose When a particle is located at distance x feet from the origin It's given by the following force function f of x equals x squared plus 2x pounds That's the force acting on it. Well, how much work is done in moving it from distance 1 to distance 3? Well from the previous slide the work is going to be the integral from 1 to 3 of our force function f of x dx for which we saw In the in the description here that our force function is x squared plus 2 x and And so integrating that we get x cubed over 3 plus x squared as we go from 1 to 3 And that's all there really is to it. Now. We just kind of plug and chug We're going to get 3 cubed over 3 plus 3 squared And then we subtract from that 1 cubed over 3 plus 1 squared Simplifying these things if we can of course There's a 3 here. The divides is one of the threes in the top leaving a 3 squared Three squared is 9. There's two of those and so we end up with 18 and then we're subtracting from that one third and into one 18 take away one of course is 17 so we get 17 and a third. I guess that's a minus, isn't it? minus a third right there, so that gives us 16 and two-thirds or 16.666, you know forever the devil's fraction right there if you prefer we could write this as fifty-thirds As this is sort of like a I mean this is a physics problem a decimal approximation might be more appropriate But we can write the exact answer as a fraction Whichever you kind of prefer in this situation So if you have the force function given to you work is fairly simple You just integrate force function with respect to position that gives you the work What can be a challenge and we'll see this in the next couple of videos here? What can be a challenge is that? What if the force function isn't given to you? I mean where does this come from right? How does one model such a thing like this now? This is not going to be a physics course So we're not going to go in every possible application of work problems using calculus But to illustrate this technique of accumulation We are going to look at a few examples from physics and show where these force functions come from and then Show you how you can integrate that to calculate the work