 Hi, I'm Zor. Welcome to a new Zor education Today I would like to spend some time for something which is called the fundamental theorem of calculus That's a great name. Well, actually the first fundamental theorem of calculus even greater Well, this lecture is part of the course of advanced mathematics for High school students. It's presented on unizor.com together with every lecture there is a very detailed notes and In some cases there are exams for you to basically Control how you study the subject. All right, so now let me start with a couple of things basically reminder first of all reminder of what is the definite integral We define it as limit of some of these expression where number of intervals goes to infinity and All Intervals which we are dividing segment a b Into our shrinking to null intervals. So that's the definition now graphically I Would like to remind you that integral is actually the area under the curve now What I would like to do right now is I would like to talk about a particular function T is some intermediary Point between a and b So the function f at x let's assume it's a continuous function And if I will write this particular integral in the limits from a to t It actually becomes a function of t, right? So if f lowercase f function is given Interval a b is given then I'll just cut my Interval somewhere in between and in this case the area From point a to point t under the curve is actually a function of upper limit t okay now if we assume that the function is positive by the way then this function f at t is Increasing from zero when t coincides with a well the area is equal to zero obviously And then it's growing and growing up to the point when t is equal to b to a full integral from a to b But it's just an observation doesn't really matter right now So let's just consider this function f at t now the first fundamental Theorem of calculus is the following Derivative of the function capital F of t Where t is actually an upper limit of this integral is the same as the function f at t Well, maybe from the first side. It doesn't look obvious, but actually it is and here is why Let's just think about what is a derivative Of t it's basically a limit of f of t plus delta t minus f of t Divided by delta t as delta t goes to zero right Now what is f of t plus delta t? Well, this is the area from a to t plus delta t So this is area from this to this What is the area? What is capital f of t? That's area from a to this So what's the difference? Well, the difference is this area It has the widths From t to t plus delta t. So the widths is equal to delta t and it has a height Equal to well, this is function f of t, right? So it's in this case. It's lower case of t in this case is lower case of f of t plus delta t But if t is shrinking then you can obviously assume that the lengths The height excuse me the height of this rectangle Can actually be The good measurement to evaluate the area under the curve, right? So area under the curve would be approximately this Yes, this function is changing from t to that to t plus delta t But if the function is continuous then the change is really very very small and if delta t is shrinking as We are receiving here obviously The difference between the area under the curve and area of this particular rectangle, which is this one Is meaning it is really miniscule, right? So whenever you do this difference, which is approximately this and Divided by delta t You will get only Function f of t. So this is basically what is the Kind of a geometrical sense of this particular formula And let me just prove it more or less formally Without resorting to nice features Especially considering they're not Artistically nice anyway So how can we prove this more or less formally? It's just an interesting point actually So let's consider it this way Again, I will use this Limit f of t plus delta t minus f of t divided by delta t Okay, so what is f of t plus delta t? its integral From a to t plus delta t f of x the x What is f of t? That's integral from a to t f of x dx now I Hope you remember this particular property of definite integrals if you have a segment from a to c and b is Somewhere in between Then integral from a to c can be broken down as a sum of two intervals from a to b and then from b to c Well, basically it's like if you have certain area You you divide it in two different halves and some of these two halves is equal to the total area right from a to c This is a to c is this from a to b which is this part and then from b to c the second part, right? So this is kind of an obvious thing. We did Presented this particular property in the previous lecture So I will use this in this particular case where my point c would be t plus delta t and My point b would be t So what does it mean then? Well, it means this First we go from a to t then from t from t to t plus delta t and That's the same as going from a to t plus delta t all the way From which follows that the difference between these two This numerator difference between this and this It's a difference between this and this Which is equal to? It's integral from t to t plus delta t f of x this This part so if this minus this equals to this right and that's what exactly we are talking about here So numerator is this Now how can I evaluate this now? Don't forget that the function f of x is assumed to be continuous Which means on every segment it has its minimum and it has its maximum, right? so that means that Integral from t to t plus delta t f of x the x would be less than m times delta t and greater than Lowercase m times delta t again. That's one of the properties of the integral Which we were discussing last During the last lecture when integral of some function on some segment is always Not greater than maximum of this function on this segment Multiplied by the lengths of the segment and greater or equal than minimum multiplied by the length of the segment now Which means that my? expression this one Which is basically this So what what do we have right now? We have that this particular expression Delta t is canceling great where capital M is maximum on this particular interval from t to t plus delta t this one and Lowercase m is minimum, but now function is continuous. So what happens if my? lengths of this interval if delta t goes to zero if delta t goes to zero and The function is continuous obviously minimum and maximum. They're all Coming together into the value of the function at this particular point So as soon as this point is coming closer and closer to t Well delta t is going to zero right minimum and maximum on on this particular segment This particular segment are all going to the value of the function at point t Which means that this also will go right remember this from the theory of limits we had this theorem I Called the theorem about two policemen and a drunk man And a drunkard so this is a drunkard and these are two policemen and they're going to the same point He doesn't have any choice, but to go to the same point, right? So the limit of this is this and that's exactly what this is all about so what's interesting here as an important observation for a Continuous function f of t Well, we know that there is always integral right from this function since it's continuous We have proven that there is an integral there is a Existence and uniqueness of that limit of the sums right and we have constructed this So we know that this thing is Exists what does it mean? It means that for any continuous function there is a function Derivative of which is equal to our function which means there is an indefinite integral or anti derivative So continuous functions always have these indefinite integrals or derivative Okay, so this is something like the first part of the Fundamental theorem of calculus or the first fundamental theorem of calculus and What it what it also implies a very very simple Calculation well simple not exactly simple, but at least at the right approach to calculate integrals and here's how we will do Okay, so what do we have as a method of calculating integrals Well, this is what we had as a method and I even try to illustrate it in a few different cases I really went through this partitioning of my Segment into a certain number of parts then increased or doubled whatever number of number of parts and Went to the limit and basically Came up with the result. What's that? That's exactly what my integral is from like from one to four of some function Y is equal to X square or whatever else Now I would like to suggest to a different approach based on this fundamental theorem of calculus and here is how Let's just assume for a second that we know how to find an indefinite integral from f of x So this is indefinite integral. You see there are no limits here or Anti derivative as it's sometimes called so gfx is anti derivative of f of x or in the indefinite integral Well, but I know that this is also Anti derivative right the indefinite integral why because it's the derivative is equal to f of t so I know that G of t is equal to f of t as well. I Assume we know it we don't know this function because it's again. It's some integral Which we don't know how to calculate right, but for instance, we do know some kind of indefinite integral so these two unknown to us function capital F and Known to us function g. They both have their derivative equal to the function. We know which is lowercase f so Now let's go back to derivatives We know that if two functions Have the same derivative Then they are differ by a constant, right? Well, if you will subtract this from this you will see That derivative minus derivative is derivative of their difference, right? It was the zero right f minus f and if this is zero if derivative is zero then the function Which we are differentiating is constant. So the difference between f of t and g of t is constant Can we determine this constant? Actually, yes, and it's very easy to do Look at it this way. I Know that function f of a If our upper limit is equal to a it's equal to zero, right? Again, one of the properties of definite integrals So let's substitute a here. So what do we see f of a minus g of a is equal to c now this is zero from which we go g of C is equal to minus g of a Right, okay, fine. So let's just use it this way f of t is equal to G of t Plus c now c is this so it's this and what have we done well We have expressed unknown to us integral in terms of Indefinite integral and the antiderivative of function lowercase f In particular if t is equal to b for instance, we have a complete formula f of b which is Integral from a to b f of x dx is equal to g of b minus g of a Where g is some kind of Antiderivative of lowercase f indefinite integral. So what have we accomplished? We have expressed the definite integral of some function in terms of its indefinite integral and This is a justification for using the same word integral for both Definite and indefinite integrals So that was basically the reason because from the surface They seem to be completely unrelated to each other because what is an indefinite integral well indefinite integral is a Function derivative of which is equal to this What is definite integral? Well, it's some kind of a limit of sums of multiplication of this function times the increment of argument This big formula etc. So this is Related to function f just by differentiating and this thing is expressed as a rather complicated definition and yet they are related using this particular Formula, which is called Newton Leibniz formula. All right, so How people calculate integrals? Well, they are calculating using this formula. They don't do this I did it in one of the lectures just as a demonstration that we can do it this way Again in some cases in most of the cases people are trying to do it this way So if this is some kind of a function, let's say it's X square, right? Then they take the indefinite integral from X square, which is X cube divided by 3 and We substitute a and b and we calculate the result very simple Significantly simpler than if I will do something like this All right. Okay, but that probably would be some exercise for one of the future lectures for now I would like to point out that this is really a fundamental Serum of calculus probably one of the most important or the most important I don't know and this formula Newton Leibniz formula is basically the most important formula which you have to remember about definite integrals and that's the relationship between definite integrals and Indefinite integrals or entire derivatives That's it for today. Thank you very much and Go to this website. I do recommend you to read all the notes for this lecture. Okay. Good luck