 Hi, I'm Zor. Welcome to user education. We continue developing the theory of definite integrals, and the first application I started with, to basically to introduce definite integrals, was the area under the curve of the function. Now, so the previous lecture was purely theoretical, and let me just remind you that, we have defined the concept of area under the curve. This is the curve, this is the function f of x, this is a, this is b. We have decided to divide the segment a, b into small intervals, and build rectangle which has the base, the corresponding interval and the height, some value within this particular set of values within this interval. It can be maximum, it can be minimum, it can be everything in the middle, it can be left border or right border of this, of this interval, doesn't really matter, because as number of these intervals goes to infinity, while the width of the largest goes to zero, the limit will exist and will be always the same, and that limit will be by definition the area under the curve. So today, I would like to basically present two concrete cases of calculation, what exactly this area under curve is, if we will follow the algorithm, if we will follow the process of dividing our segment into pieces and then increasing the number of intervals. Okay, so by the way, as far as the formula is concerned, I will put this formula here. So that's f of c i times delta x i, i from one to n, where c i belongs to x i minus one x i, and delta x i is x i minus x i minus one. So delta x i is basically the width of the i's interval. x i is here, x i minus one, and this is x i. So this particular interval is x i, where a is x zero and b is x n, right? So c is any point within this particular interval can be left, right, maximum, minimum, middle point, whatever it is. Doesn't really matter, we have proven in the previous lecture that the limit of this expression, as long as delta x i are going to zero, well, the largest among them goes to zero. The limit will exist and that would be the area under the curve and the examples which we will consider are two actual examples. So let me start with the first one. So the first example is, function is 10x on interval from zero to four. Okay, so it looks like this. This is four, this is 10x, which means this is 40, right? This is zero. Now this is area under the curve. So this is our function. This is interval from zero to four, the segment. Now, obviously from geometry we know that if this is four and this is 40, then the area of this triangle, which is actually an area under the curve, would be four times 40 divided by two, which is 80. So let's just keep in mind this, and this is obviously the right way of doing this, except that that's not the one which I would like to introduce you to. I would like to use this formula and come up with the same number, if possible. I hope I will be able to. Okay, so let's just remember this result and what we will do is we will do exactly what our process is telling us. So we will divide into n pieces. Well, let's divide first into two pieces. So this is x is zero, this is x2, and the middle one is x1. So this is one interval, this is another interval. So we will build two intervals on our segment. Now, let's take, let's say, a left point on each interval. So the left point here is zero, so the rectangle will have zero height. And so it will be zero times the length of the interval is two, right? And this is the rectangle, so that's basically equal to zero. The second rectangle would be here, and as I said, I'm adjusting by the left boundary. So if this is two, this is 20. So the height of this rectangle is 20. So it will be two times 20 is equal to 40. And the total will be 40. So some of this rectangle which has zero height and this rectangle would be 40. Well, it's obviously less than the area which we are talking about. Okay, now let's do it more finely divided segment. Let's divide it into four pieces. So this is one, this is two, this is three. So this would be x1, x2, and this is x3. And this is x4. So we have four different intervals. And let's do exactly the same. Build a rectangle based on each interval with the left value as the height. So again, the first will be zero. The second would be, okay, this is 10 obviously and this is 30. So the first will be zero. The second would be 10 times 2. Next it would be 20 times 2. And next would be 30 times 2. Equals 2, 10, 20, 30, 20. No, sorry, not by 2, by 1. Let's see something is wrong. 2 was the previous. Now I have only one. The width of each interval is 1. So that's what, 60. So it was 41. My next is 60. Should we continue? Well, let's just go through one more iteration. So we have divisions by one half. This is one and one half. This is two and one half. This is three and one half. Now, so this height, this height would be five. Now this would be 15. This would be 25. And this would be 35. So what we have is zero times half, which is zero. Then five times half, 2.5, 2.5. 10 times half is five. 15 is 7.5. 20 times half is 10. 25, 12.5. 30, 15. And 35 is 17.5, right? That's my last rectangle. 1, 2, 3, 4, 5, 6, 7, 8. 8 rectangles. This is two rectangles. This is four rectangles. And this is eight rectangles. So what is the sum? This and this is 20. And another 20 is 40. And another is 60. And 10 is 70. So you see, our first iteration was 40, then 60, then 70. And you remember, the whole area is equal to 80 from the geometry, right? So it looks like we are actually moving to the right direction. Okay, so now it's time to do it precisely. So let's consider that we have divided this into n pieces, n equal pieces, right? So my xi is equal to what? So we divide four into n pieces, which means my xi would be equal to i times four divided by n, right? So the length from zero to xi, this is coordinates, right? So xi minus xi minus one is four divided by n. So that's my delta xi. Now, f of xi is equal to this times 10, right? Function is 10 times x. So this would be 40i divided by n. So now I have to summarize function value, 40 times i divided by n, times four divided by n from i from one to n, right? So this is my height of the rectangle. This is my widths of the rectangle. So it's equal to 160 divided by n square i from one to n. Well, we all know what is the sum of i's. It's one plus two plus three, et cetera, plus n. Very easy to count. I do remember that the formula is this. n times n plus one divided by two equals 280 n plus one divided by n, right? And what's the limit as n goes to infinity? Well, this is obviously goes to one. So the limit is 80. And that's exactly what we had to really obtain, if you remember, right? From geometrical considerations. So it looks like it works fine. So this particular process. And now I have divided into n equal parts. Basically, it's not really necessary because as I was saying in the previous lecture, the theorem was proven that it doesn't really matter how I divide into n pieces as long as with n increasing to infinity, my largest segment goes to zero. So it doesn't really matter. I chose the easiest way, right? When all intervals are equal to each other. Now then the height of the rectangles, I have chosen by left boundary. I could have chosen by the right boundary. It will probably be exactly the same. Should I demonstrate it? Well, it's very easy, actually, I think. Because then, I think in this case it would be i plus one. And it would not really change the limit. Okay, I'm not going to do it, but basically it's kind of obvious. Okay, next example. So the first one we have proven that we are going to the right number using this particular process. Now let's consider a slightly more complicated example because now I will have, as a function, I will have y equals to one minus x square on interval from minus one to one. Now, x square is, as you know, parabola. Minus x square would be inverted parabola. And plus one, I have to raise it by one above the x-axis, right? So it would be something like this, from minus one to one. That x is equal to minus one or one, y is equal to zero, right? So this is basically my segment. And this is the area which I would like to basically calculate the area of. All right, just to make things a little bit simpler, let me, instead of considering from minus one to one, let's consider only from zero to one, and then I will double the result because the parabola is symmetrical. There is absolutely nothing wrong with going all the way from minus one to one and in the description of this lecture and the notes on the Unisor.com, I do consider from minus one to one. It's just a little bit easier for me to do it here from zero to one. So let me just do it from zero to one. So I don't really have to worry about negative numbers and stuff like this. Let me just consider this. How is this part? Okay, so again, let's do exactly the same thing. We will divide this particular area into n equal parts. So my xi is equal to one over n, right? So this is one and divided by n parts. So that's my xi's. Now x, well, I'm sorry, it's i divided by n. It's delta xi, which is equal to xi minus xi minus i, which is one over n. Okay, now my f of xi. So let's talk about the left boundary of each or right boundary or whatever. Delta xi, this definitely belongs to interval, i's interval, right? It's equal to one minus xi square, right? Which is one minus i square divided by n square. Now if I sum f of xi times delta xi, what I will get? I will get one minus i square divided by n square times one n equals to sigma one n minus sigma i square divided by n cube, right? If I open these parenthesis, it will be one n's and minus i square divided by n square times n, which is n cube. Now this is from one to n, so this is equal to n times one n's, which is one minus sigma i square divided by n cube. Okay, so it's one minus one over n cube. It's a constant so far of i square of one to n equals to... Now I happen to remember, well not remember, I wrote it down, the formula for sum of the i square. You can prove it by induction or there are different ways to do it, let's not waste our time, I'll just use it as it is, n cube. So sum of i square is equal to n times n plus one times two n plus one divided by six, which is equal to... Well, obviously if you will factor out two, you will have n, n plus something and again n plus something, well in this case it's n plus one half. And this is n cube, so one minus two over six and here I will have n cube n times n plus one times n plus one half, correct? And obviously this thing goes to one as n goes to infinity because this is polynomial, this is n cube, this is n cube. So if you divide n cube by n cube is one and everything else would be n square and n etc. Which dividing by n cube will give you n in the denominator and it will go to zero. So the whole thing is equal one minus two six or one minus one third, which is two-third and the whole area if you will add this half it will be four-third. So that's basically the answer to this problem. Again, if you do not remember this formula, just try to prove it by induction if you want to. It's very, very easy to prove that the sum of squares from one to n is equal to this. By induction it's just really in no time at all. So these are two examples of how to approach the area under curve using the classical definition of what actual area is. Now, do people do exactly what I have just done? Absolutely not. They are calculating the area under the curve in a completely different way. And to demonstrate you how exactly it can be done, which I will do in the next lecture, that's exactly how I will introduce the definitive integral. So the definitive integral is actually an answer how to easier calculate something like area under the curve using our apparatus which we have already known, which is derivative and indefinite integral, which is sometimes called antiderivative because it's a reverse operation. But that would be the subject of the next lecture. Meanwhile, what I might actually suggest you as an exercise if you really are up to this, I have just used the value on the right side of the interval of xi. Try to use it on the left side, function of xi minus 1. And see if you will get exactly the same result. It's a very nice little exercise and you will also have some kind of summation problem. It would be really interesting to see if you will get the same. I mean, you must get the same because I have proven the theorem in the previous lecture that you have the same thing. But it's just always nice to basically make sure that everything is going. Alright, so that's it for today. Thank you very much and good luck.