 Hi, I'm Zor. Welcome to Unizor education. I would like to continue talking about the Newton-Leibniz formula for integrating, for getting the value of definite integrals. Today I will just talk about a couple of examples. Now, this lecture is part of the course of Advanced Mathematics for teenagers and high school students. It's presented on Unizor.com. I suggest you to watch this lecture from that website because every lecture has very detailed notes and some lectures or some topics rather have exams. So you can just take exam and the site is completely free. So, first of all, let me just remind you Newton-Leibniz formula. So, if you want to find the value of definite integral of function f at x on the segment from a to b, which you can always consider as an area under curve, which represents the graph of function. Now, this is equal to where capital F is any indefinite integral or antiderivative from the function lowercase f. Okay, so I have derived this formula in the previous lecture and today I'm going to use it for a couple of concrete examples. I start with examples which I was using before when I introduced the concept of definite integrals as a limit of Riemann sums, if you remember. We divided the segment into little pieces, into n pieces and had this sum, let me just remind you limit, sum of f of xi delta xi from 1 to n where n goes to infinity and delta x goes to 0. So, we have divided our segment a, b into n parts and then we can calculate this expression and go to the limit and that's where I was actually using two particular examples to derive basically the value of definite integral. Now, I will do exactly the same two examples in a couple of more, but these two examples I will use to demonstrate how easy it is to use this Newton Leibniz formula to calculate versus whatever I was doing before when I did really, according to the definition of definite integral. So, I really divided the segment into n pieces, calculated the sum and then took the limit as the number of intervals goes to infinity. Okay, so let me just do these couple of examples. First, I have a function f of x is 10x and the segment is 0, 4. So, basically it's this. So, this is 4, this is 40 and this is 10x and we need the area under this graph. And obviously, I know that we all know geometry, this is a triangle, we can always calculate the area of triangle, it's 4 times 40 which is 160 divided by 2 is 80, yes, we know that. But in any case, I would like to use integral to calculate this area. Now, again, before when I was doing it this way, I divided this into n pieces, calculated the sum, went to a limit and got the same h, obviously. But now I will do it easier. I will just do this Newton Leibniz formula. Now, first of all, what is indefinite integral of 10x? Well, obviously this is 5x square. Well, I will not use the plus c because I will have this difference and c and minus c will obviously cancel out. So, this is indefinite integral from 10x. Why? Because the derivative of this is what? 5 times derivative of x square. The derivative of x square is 2 and x to the power of 1, so it's 2x, so it's 5 times 2 times x, which is 10x. So, my integral from 0 to 4 10x dx equals to 5x square, which is 4 square minus 5 0 square. So, this is function 5x square. I just substitute the upper limit of integration and lower limit, which is equal to 16 times 580. You see how easy it is and how fast? All you have to know is this. So, if the function is relatively known, simple, whatever else, and we can really come up with indefinite integral with anti-derivative of the function, so the function derivative of which will be equal to our function f, lower case f, then the whole thing is really easy. Next problem. Now, this is another example from the lecture where I was using this approach. And again, it was really kind of difficult, but anyway, this is what? From minus 1 to 1. So, it's this. It's a piece of parabola, and we had to calculate the area under parabola. Now, our answer, I do remember it because I think it was 4 third, this area. Okay. Now, this is a simple function, just a power function, and we know how easy it is to have an indefinite integral from this. Definite integral from minus x squared will be minus x cubed divided by 3, right? Minus will give you this. Derivative of x cubed would be 3x squared. 3 will cancel with this, so it will be only x squared. Now, I have to put plus x because derivative of x is 1, and that would give me 1, right? So, this is my indefinite integral, and all I have to do is substitute 1 and minus 1 into this formula to get the answer, right? So, the integral is equal to function of 1. It's minus 1 third plus 1, right? Minus function f of minus 1. Well, let me open parentheses here. Okay, so I have to put minus, minus 1 cubed divided by 3, plus minus 1, right? If I substitute minus 1, that's what I will have. Oh, God! So, what is this? This is minus 1, and this minus, it will be plus 1. So, it's 1 third minus 1, and this is minus in front of it, and I have minus, minus 1 third plus 1. So, if I will open the parentheses, it will be minus 1 third plus 1, minus 1 third plus 1, which is equal to 2 minus 2 third, right? Which is 6 thirds minus 2, which is 4 thirds. Right, exactly the same thing as I wanted to get. Again, what was the most difficult part of this problem to find the indefinite integral, to find antiderivative? But it's not really very difficult if the function is so simple. And then I have a couple of other problems which are not related to the previously solved problems. So, I don't really have any kind of answer which I have to match. But they are very interesting, I would say, because they have some physical sense. So, the problem number 3 is, there is a car which is moving with this speed. At some point, it starts slowing down to zero. So, the driver hits the brakes. Now, after 10 seconds, the car stops. Now, it's given that every second, the car is losing the same part of the speed. So, the speed is going from 20 to zero linearly. It's absolutely linearly slowing down. So, I didn't really specify even the function, right? And what's necessary to determine is the distance which the car actually went from this speed to a complete stop, if it took 10 seconds. So, my first part of the problem is basically put it into language of mathematics, right? So, if the car is losing its speed, gradually, equally every second from 20 to zero, and it took 10 seconds, it means that my function which describes speed, well, let's use V, usually in physics we are using V or T, is equal to 20 minus 10T, right? When T is equal to zero, I am at 20. When T is equal to 10, okay, I made a mistake, that's two. So, when T is equal to 10 seconds, it will be 20 minus 20 which is zero, and the function is linear, which basically gives us what we wanted to have as a gradual diminishing of the speed equal part of the speed is lost on each unit of time. Okay, that's given. So, we have the speed as the function of time. Now, what is the distance? Now, you remember that if the speed is constant and the time is whatever it is, then the distance would be multiplication of speed times time, right? But that's only if V is constant, what if V is changing? It means that the first second, the car will go longer than during the second second, because the speed goes down, down, down. And at the very end, the car during the last second will cover a very small distance, because the speed will be very small. Now, how to transfer it into integrals? Okay, here the definition of the integral actually helps, because I will do exactly the same as I was doing during the area under curve calculations to come up with the concept of the integral. So, basically I would like you to understand the following. We come up with some kind of an integral as a formula which represents the answer to our problem using the definition of the integral as a limit of certain sums, etc. But then we calculate this integral using the Newton's Leibniz formula. So, again, we are using the definition to formulate our problem mathematically. But then we are using the Newton's Leibniz formula to calculate the real value of this integral. So, how can I approach this and get the integral actually which represents this distance? Well, very simply, let's just divide this into n pieces. Now, during a certain interval of time we assume that the speed is not really changing significantly. So, if this is delta Ti, if I will consider the speed to be Vi in this particular interval, then this is the distance covered during this small interval. Then I have to summarize them and then I have to limit as number of intervals goes to infinity and each interval shrinks to zero, right? So, that's basically a definition of the integral. Now, what is Vi? Well, Vi is the speed on the i's interval, right? So, I know if my i is actually the number of seconds. So, I can basically say that Vi is equal to the formula basically 20 minus Ti, right? Times 2. So, I have a formula now that this is my V as a function of the time and this gives me as a limit integral from 0 to 10 seconds my speed which is 20 minus 2T dt. So, delta Ti is converted into dt and Vi is basically the function which represents my speed. So, we have mathematically expressed the distance which the car covers during this period of time. Now, all we need to do is let's forget about this preamble and let's just calculate it. So, what is this? Now, f of T which is indefinite integral of this is equal to how to get 20. I have to put 20T. How to get 2T? It's derivative of T square. So, it's minus T square. So, that's my formula for indefinite integral for antiderivative. And so, all I have to do is to put limits of integration. So, for 10, it would be 20 times 10 minus 10 square. And for 0, it would be minus 20 times 0 plus 0 square. This is 0, this is 0, this is 100, this is 200. So, the answer is 100 meters. And that's how long this car from 20 meters per second will cover until 0 speed. If the time it took is 10 seconds. Now, obviously, it's assumed that in this particular case the driver slows down gradually and equally on every period of time. So, the amount of speed which is lost during the time T is proportional to time T. So, the first second it will lose 2, the second second it will lose 2 meters per second. So, it will be, on the first second it will be, after first second it will be, from 20 it will go down to 18, then to 16, then to 14, etc., after each second. Okay. And my last problem is related to spring. This is the spring in its initial position. Now, if I would like to extend it, I have to basically have some force which is pulling the spring, right? And there is a Hooke's law which says something like this. Where F is the force, K is some kind of a coefficient which depends on the spring. Different springs have different coefficients, but it's just a constant which defines the spring. And X means extension of the spring. So, if you would like to extend it a little bit, then it's proportional to amount you would like to extend. So, if you would like to extend it by, I don't know, let's say by 1 centimeter, it's 1 force. If it's by 10 centimeters, it's 10 times greater force. Now, obviously it's not absolute, but in certain approximation it's the right law. So, we assume that this is the law. Alright. Now, my question is, let's just assume that K is equal to 0.5. Again, it's a characteristic of the spring, doesn't really matter, some kind of constant. And I would like to extend this spring by 1.1 meter. So, from X equal to 0 to X equals to 0.1 meter. I would like to extend it by this particular period. Now, what's interesting about this? In the beginning, when I'm extending only by 1 millimeter, for instance, my force should be very, very weak, right? Because I'm just a little bit extending. And the more I'm extending, the next 1 millimeter would be more and more difficult. Because the spring requires this type of force to hold it, let's say, at point 1 millimeter. It's 1 force, but at point 10 millimeters, it's 10 times greater force, right? Well, considering that the length would be 10 times more. So, my force is changing. Now, my task is to determine amount of work which I have to extend to extend the spring. I have to make this type of work, all right? So, this amount of work I have to calculate. Now, what is work? Well, if you remember your physics course, work is, if you have some kind of a force which is applied on a distance s, it's just the multiplication. So, if you have force of 1 Newton on 1 meter, the work will be 1 Newton times 1 meter, which in physics is called a Joel. That's unit of work. But again, that's in case f is exactly the same during entire distance s. Here, it's different, because in the beginning, force is very small. At the end, force is bigger. So, that's definitely an invitation for using integrals. So, how can we use integrals? Well, very simple, the same thing. We divide this into n pieces. And what can I say about work which is supposed to be produced on the ice interval? Well, it's equal to ice force times delta xi. Now, what is ice force? This is the force which is extending spring by xi, right? So, this is equal to k xi. Because, you know, the force is proportional to extension of the spring. So, this would be my xi. This little piece would be delta xi. And I assume that the force is not changing significantly from this to this, which means that I can use this formula for one individual amount of work which is supposed to be performed during this ice interval. Now, I have to summarize it together, and I have to limit as n goes to infinity. Same thing, and we have integral here, obviously. So, we have integral from zero to, what's my maximum extension? 0.1. Now, this is kx dx, right? Well, k actually, I know, it's 0.5. So, here is the integral which represents amount of work which we are supposed to perform to extend the spring. Now, what is this? How can I get x? x is one half of x square derivative, right? So, we have 0.5. Then, again, x square divided by 2, and that would be from 0.1. So, I have to put 0.1 here, square, minus 0.5, 0 square divided by 2, because the lower limit is 0. So, this is 0, this is 100, and this is 0.25, 0.0025, right? So, that's my answer. I think this is the unit. That's how it's spelled, I believe. Not sure. Joe, that's in the standard physics, we are using certain standard system of units. The force will be in newtons, for instance, the distance will be in meters. Amount of work will be in newton meters, which is, by definition, a Joe. In any case, that's the physics part of it, which might actually come handy in the future. In any case, I wanted to present you a few problems with one very important purpose. To state the problem mathematically, you need the definition of the integral as limit of the sum of individual pieces, which are multiplication of the value of the function times increment of the argument. To formulate. Once it's formulated, once you have obtained the integral, then the formula, Newton-Libnitz formula, is extremely handy, and it gives you the answer almost immediately, provided you know how to get the indefinite integral and the derivative of the function lowercase f. Like in this case, or actually all cases which I was considering were related to power functions, and for power functions, indefinite integral is very simple. So basically, let me think. Well, basically that's all I wanted to present for you today. Important are two things. Number one, you have to be able, using your knowledge about the problem, you have to be able to express it in terms of integral. And the second thing, you have to remember this very simple formula, Newton-Libnitz formula, which obviously can be used in case you know what's the indefinite integral from the function which you are integrating. Okay, that's it. I recommend you to read the notes for this lecture. The notes are on Unisor.com, right next to the lecture. And maybe you should actually do yourself these problems. Notes actually present the problem and solution, but don't pay attention to solution for a while. Try to do it yourself. That would be very helpful. Okay, that's it. Thanks very much and good luck.