 Hello there and welcome to the screencast where we're going to work through an example of an application of the net change theorem. So here's the problem. We have cars that are passing through a very busy interchange and the rate at which those cars pass through the interchange is given by this function a of t, which is 4500 plus 1500t minus 300t squared. That a of t is measured in cars per hour and t is measured in hours in such a way that t equals zero corresponds to four o'clock in the afternoon. And the question is how many cars pass through this interchange between 4pm and 6pm. So first of all realize this is a classic setup for an integral. When an integral problem, remember it's like starting with a velocity and ending up with a distance travel. So here we're starting with a velocity, a rate of change here. We're given a rate of change in cars per hour and that's what this function tells us. And we're being asked for a total change, how many cars pass through between 4 and 6. If the situation were reversed and we were given a quantity and we were asked to find its rate of change, that would be a derivative question. But here since we're given the rate of change and asked to find a total change in the quantity or a net change in the quantity, that is going to be an integration problem. So that's where we're going to select integrals for our tool here. And the correct tool for the job here is the net change theorem, which in sort of a paraphrased form would say that the total change in the amount of cars that are passing through this interchange is the integral of the rate of change in that traffic flow. That's true for any function whatsoever, the net change theorem says that the total change in a function from a point A to a point B is given by the definite integral from A to B of the derivative of that function. So again, we're set up here for the net change theorem because we're given the derivative, a rate of change in the number of cars in the interchange, and we're asked to find a total change in that. So let's work with this and set up what the net change theorem would say. The total change in the number of cars that go through the interchange is going to be the integral of the rate of change. I'm drawing the integral sign here and I'm going to put the expression that I know represents the rate of change for 4500 plus 1500t minus 300t squared. I know this is a rate of change because over on the problem data, it says the cars are passing through at a rate of, and that rate of course is huge, and this is telling me a rate of change is measured in cars per hour. I'm going to need to put a dt here, and now what are the limits of integration? Well, let's look back at the problem data again, and it says how many cars pass through the interchange between the hours of 4 o'clock and 6 o'clock. So I'm given a time interval here. However, I don't want to just go over and put up 4 and 6 on my integral. I have to be a little nuanced with this and read the problem carefully. It says t is measured in hours and t equals 0 is 4 p.m. This especially is important. So the time on the clock is from 4 p.m. to 6 p.m. But the limits of integration I'm going to want to use are from 0 to 2. And now that's my integral that I want to compute. This is not going to be a very hard integral because we can use the fundamental theorem of calculus to help us out. So let's now compute this definite integral and see just how many cars do pass through this intersection. So we're going to use the fundamental theorem of calculus to compute this exactly. And remember, the fundamental theorem basically gives us a multi-step process. First of all, I need an antiderivative of the integrand. I need an antiderivative. I don't need to think about that. Let's call that antiderivative s. Then I'm going to have to, once I find that antiderivative s, I'm going to calculate s of 2. I'm going to calculate s of 0. And then I'm going to find the difference between s of 2 and s of 0. So step number 2 here is obviously very easy. It's step number 1 that we have to do a little bit of thinking. Let's go over here and start that thinking process. So what's going to be an antiderivative for this function here? Well, I can simply think about this using the reverse power function rule that we've seen in the section. So s of t here, an antiderivative for 4500 would just be 4500t. An antiderivative for 1500t, I'm going to raise the exponent up on t and then divide by 2. So that will give me 750. And then finally, there's a minus sign there I'm going to preserve. An antiderivative for 300t squared, I'm going to raise the exponent from 2 to 3 and then divide by 3 and that would give me 100t cubed. And it's probably a good idea on your own to pause the video for a moment and to stop and verify that s prime actually does give you this function that's in the integrand and it does. So now we're all set up to calculate the centric rule. Let's do that right here. So this is going to be equal to, I'm going to take my s of t and put in s of 2 or calculate s of 2 and then calculate s of 0 and subtract. s of 2 is going to be simply 4500 times 2 plus 750 times 2 squared, which is 4, minus 102 cubed, sorry, which is 8. Now I'm going to subtract off s of 0. But notice that in s, every term in s of t has a t on it. So this is going to be 0 plus 0 minus 0. Of course, the whole thing is going to go to 0. And now it's just arithmetic from here on out. This is 95, 9000, sorry, 2 times 4500 plus 3000, which is 750 times 4, minus 800. And you do the math on this and you get 11,200. And the units on this are in cars. The reason that the units are in cars is because you can actually look up in the integral itself and see this integrand's units are in cars per hour, if you recall that from the problem data. And dt has the same units as t does and that's in hours. So if I multiply cars per hour times hours, I'm going to get cars. So this really does have the look and feel of a total change. And that's what it is. So this integral tells us that through this intersection between 4 and 6 pm, 11,200 cars have passed. Thanks for watching.