 Welcome back to our lecture series, Math 1210 Calculus I for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Misledine. It's good to have everyone. In lecture 48 in our series, we're going to talk about the net change theorem that shows up in section 5.4 of Jane Stewart's Calculus textbook. This section is entitled, Indefinite Integrals and the Net Change Theorem, but admittedly, we had started talking about indefinite integrals at the end of chapter 4, so we've been using this notation for a while. That really poached a lot of the topics that are in this section. I want to really focus on the stuff that we haven't talked about yet in section 5.4, which is the net change theorem. What you can see this illustrated on the screen, the integral of a rate of change is the net change, that is the integral from A to B of f prime of x dx equals f of B minus f of A. You might be wondering, really, is this something different? Isn't this just the fundamental theorem of Calculus Part 2? Of course, the answer is, yeppers, it absolutely is the fundamental theorem of Calculus. Really, the net change theorem is not saying anything different than what we've already seen in the last lectures 47 and 46 specifically about the fundamental theorem of Calculus. Basically, what we're gonna do in section 5.4 is we are going to apply the fundamental theorem of Calculus to story problems. That is, can we look at uses of rates of change in science? Physics, chemistry, biology, economics, what have you, right? Because after all, back in section 3.7 of Stuart's book, we had talked about how rates of change are used in the sciences, the physical sciences, financial sciences, life sciences, all the sciences, right? We talked about examples of how rates of change are used in that context. With the understanding that rates of change, of course, just means, it just means you're taking a derivative, right? Whenever we see rate of change, we think of a derivative. And so with that context, the integral of a derivative is, we'll use the fundamental theorem of Calculus, of course. But when we realize that sometimes scientists don't know that rates of changes are derivatives. So with that perspective, since rates of change show up all the time in science, we can integrate those things to find the net change. What we mean by net change, of course, is that if you take the first value and you subtract it from the second value, this here gives us our net change in the quantity that we're measuring. So really, I just wanna take this as an opportunity to look at some story problems, some applications of integration to help us with scientific questions. And so a lot of these examples are drawn out of Stuart's textbook. You can look there to see them and there's some other ones as well. And it's not hard to engineer these type of examples. They're all over the place. Because rates of change are ubiquitous in science, this means that derivatives are relevant to scientific study. Well, as we could integrate rates of change to find out relevant information about scientific problems, that is the net change problem. Well, then that makes integrals relevance to scientific problems as well. So we just wanna explore that for a few minutes. So just a few examples we can look at here. If we consider the volume of water in a reservoir with respect to time, maybe we call that V of T, right? What's the volume with respect to time here? Well, then it's derivative, V prime of T is the rate of which water flows into the reservoir at time T, right? And so this rate of change, of course, would be the net flow, right? Cause you might have some water entering, some water exiting, but V prime T is measuring the flow of water into the system. So that idea is the derivative measures the flow of the water. And so if we were to integrate the flow from time one to time two, if we integrate the flow, this will equal the volume at time two minus the volume of time one. And that difference would give us the change in the amount of water in the reservoir at a given time. So if we integrate the flow of water, we get the change of volume. That's the takeaway from this example here. Let's look at another example from chemistry here. If we take C of T to represent the concentration of, the concentration of a product of some chemical reaction over time, then the rate of the reaction, so how quickly is this reaction happening? How quickly is it changing? If we look at the derivative of concentration with respect to time, well, that's the derivative here, the rate of the reaction. Well, if we integrate said derivative, if we take the integral, take the integral here of the rate of the reaction with respect to time from time stamp one to time stamp two, then this gives us the net change in the reaction there. It's the change in the concentration over time. So if you integrate the change of concentration with respect to time, you'll get the net change in concentration. We did linear density earlier this semester when we were in section 3.7. So if we have the mass of a rod, we measure it from left to right and then M of X measures the mass of the rod. And so kind of draw a picture of that where we have our rod, it's like Gandalf's staff right here, some crooked looking rod thing or whatever, right? And we're trying to measure here, here's our x-axis, whoops, x-axis where x equals zero is over here. And so if we pick a specific x-value right here, we're trying to measure what is the mass of our rod at this moment M of X. Well, then the linear density, remember, rho of X, this linear density, this was just the derivative of mass with respect to position. Well, if you integrate your density function, if you integrate density, then you'll capture M of B minus M of A. This difference of mass is the mass of the segment that lives between A and B. So if we have A right here, we have B right here. This right here would measure the mass of this segment. That's what we're trying to figure out. Well, you can accomplish that by integrating from A to B, your density function, rho DX. So integrating linear density gets you back the mass function, but it gives you the net change of mass. Here's another one about population growth. If we know how a population is growing or if we know how a virus is spreading or so many other things, if we have the rate of growth, DNDT, so N measures the population, T is time. So this is the change in population over time of say, again, some organisms are growing in an ecosystem, people are growing in a city, some infectious disease is spreading, right? If we know the rate of growth, then the integral of that rate of change gives you the net change of growth, right? The integral of DNDT with respect to time gives you the population at time T minus the population at time one and that difference. So it'll tell you how much has the population increased from T one to T two. And this is the net increase. So it has bursts built into there, it has deaths built into there. It's the net population, the net population change, I should say there, the net change. And again, if you look at something like this, when you use this notation, you're almost tempted like, oh, I'm a college address student who doesn't know anything about calculus, DT, DT, it sort of cancels out and you should be getting a sum of the ends, which is exactly what's happening right here with this net change formula. Again, these are just some examples just pulling from different sciences here. If we take the cost function C of X, it's the cost associated with producing X items, X commodities. So we could be talking about the cost of producing X rolls of toilet paper or whatever we wanna do. C prime of X represents then the marginal cost, it's the derivative cost, the rate of which cost is changing as production increases. Well, if we integrate the marginal cost, we'll get the net cost. That is the cost to increase from production level X one to production level X two. All right, so I have, I have essentially one last example, which kind of is two examples, but it's really one example at the same time. It's a comparison. So if we come back to this original problem from physics, the motion problem, if S of T is our position function, which you can see it labeled right here, if F of T is our position function, then velocity is its derivative. If you integrate velocity, you will get the net change of position, which what we call displacement. Displacement represents the net change of position. So if you have a particle that starts right here, it goes to the right, then to the left, and to the right, then to the left, then to the right, and it plays sort of like a game like this. Displacement, it doesn't care about the journey. Displacement is just like, well, how, what is the distance between where you started and where you stopped? This is what we mean here by displacement. It's only looking at the net change of position. And if you integrate the velocity function, it gives you the change in position, the displacement. Turns out we actually did a problem like this much earlier before we learned the fundamental theorem of calculus, but we can use FTC to help us calculate this displacement. This is in contrast to what we might call the total distance, right? Because we talked about that squiggly line a moment ago, right? Our particles go in something like this, goes all over the place, right? What if we wanna find not the displacement, but the total distance? The total distance would be the length of this really obnoxious curve right here. Like how long is that thing? This would be the total distance traveled. So if you want the total amount, the total change and not just the net change, the gross change you might call it that way, you can actually calculate total distance or total change, gross change, but this is accomplished by, all one has to do is take absolute value of your rate of change. If you take the absolute value of the rate of change, then this will actually give you the total distance because there's no distance that kind of nixes out with another one. Going forward doesn't cancel out with going backwards. And so in the next segment of this lecture, look in the link there right now to see it if you want to. But in next section, we'll actually compare in contrast with a motion problem, how we can use integration to find the net change of position, aka displacement, or how we can use integration to find the total distance of a particle that moves. I will see you then.