 So, let us recall what we had done last time. We started looking at integration. So, given a function f on a interval a, b to r, f bounded, we defined the integral of f via upper sums and lower sums. So, we defined for every partition p, we defined what is the lower sum, what is the upper sum and we said f integrable if and only or integrable definition if the supremum of least upper bounds of lower sums is equal to greatest lower bound of upper sums and that was denoted by integral f dx. So, we gave some examples f monotone implies f is integrable. Then we gave some more examples namely the function f of x is equal to 1 if x belongs to rational say in 0, 1 and 0 otherwise is not because upper sum is always 1, lower sum is always 0 is not integrable. We also gave example of Thome's function for the popcorn function which has this. So, this function has the special property namely it has infinite number of discontinuities. In fact, this discontinuities are all rational points in every sub interval of 0, 1 but is integrable. So, there is various examples of integrable functions keeping in mind the number of discontinuities the function can have. So, as we were pointing out that this function is discontinuous everywhere, monotone is discontinuous at all rational countable at the most countable in many and Thome's function has infinite number of discontinuities in every sub interval and is integrable. So, let us prove an important class of functions which is integrable namely theorem let f be a b to r, f continuous. So, let f be a continuous function on an interval close bounded interval a, b then f is integrable. So, let us prove that. So, basically what we want to show it is integrable we are not interested in computing the integral. So, we will use a criteria that given any epsilon. So, we will use a criteria that for every epsilon bigger than 0, there exists a partition p of a, b such that upper sum minus the lower sum is less than epsilon. So, this is what we will show and we had observed that this is a necessary and sufficient condition for a function to be Riemann integrable. So, let us start with an epsilon. So, fix. So, fix. So, what does continuity imply? f continuous implies two things. One, f is bounded. In fact, it attains maximum and we know that. And second, we proved that f is uniformly continuous. Every continuous function on a close bounded interval is uniformly continuous. So, what does uniform continuity say? So, hence we are already given an epsilon. So, given epsilon, so for epsilon bigger than 0, there is a delta such that whenever x minus y is less than delta, that implies f x minus f y is less than. Whenever two points are closed, their values are closed. That is what uniform continuity says. So, mathematically that says that for any epsilon bigger than 0, there is a delta so that whenever x and y are closed by a distance delta, the values are closed by epsilon. So, that is quite useful in the sense that now we can select a partition so that the points are always inside a distance delta. So, choose a partition. So, let p be a partition of a, b such that, so it divides into sub intervals. So, such that I did not introduce this notion is less than delta. So, what is the norm of the partition? Norm of p, a partition is defined as the maximum of the length of sub intervals. So, x i minus x i minus 1, i between 1 and n. If p is the partition, say a equal to x 0 less than x 1 less than x n equal to b. This is the definition of a length, definition of the norm of a partition. So, every partition divides the interval a, b into sub intervals. So, look at the maximum length of the sub intervals. And that maximum length is called the norm. So, this is what is called, so this norm, this is called of the partition. So, uniform continuity says given epsilon, there is a delta, say that whenever two points are closed, fx minus fy is closed. So, if we choose a partition whose maximum, whose norm is less than delta, then that will mean that for any two points in a sub interval, the length will be less than, the distance will be less than delta and hence the values will be closed by epsilon. So, now also observe, also note, f attains maximum and minimum on each sub interval, because it is continuous function. So, let capital M i, which is the maximum of f of x, which is x i minus 1, we attained at some point. So, let us call it as x belonging to x i minus 1 to x i. Let us say this value is attained at some point, say x i dash and where x i dash is a point in between and small m i is the infimum of fx. So, x belonging to x i minus 1 to x i and let that be attained at some point, say x i double dash, where x i dash and x i double dash belong to. So, we are just saying that f is a continuous function. So, look at this restriction on the closed bounded interval x i minus 1 to x i. It must have a maximum value somewhere in that interval. It should have a minimum value in that interval and must be attained. So, those points we are calling as this. So, why we are doing all that is because now let us look at the upper sum, upper sum with respect to y minus the lower sum with respect to f. So, what is that? So, that is equal to capital M i minus small m i 1 to n, maximum minus the minimum into the length of the interval x i minus x i minus 1. Now, this value is taken at that point. So, this is i equal to 1 to n. So, f of x i dash minus f of x i double dash into the length of the interval. So, that was the. Now, these two points x i dash and x i double dash are inside this interval. So, that means the distance between them is less than delta because norm of the partition p is less than delta. Whenever that happens, we know that by uniform continuity these values are small. So, that is all the reason we did that. So, let us imply. So, this is less than or equal to delta times sigma i equal to 1 to less than delta. So, this is less than epsilon times x i minus x i minus 1 because x i dash x i double dash both belong to the interval x i minus 1 to x i, implying that the distance between them is less than or equal to 1 is less than delta. That should be okay to say that implying that f of x i dash minus f of x i double dash is less than epsilon by uniform continuity. So, what we have done is the interval a, b, we have found a partition, say that the distance between any two points in the sub interval is less than delta and that delta is corresponding to the uniform continuity of the function. So, the values for any two points will be less than epsilon. So, all that is used to bring it here. So, what is this quantity? So, consecutive terms will cancel out. So, this is epsilon times b minus a. So, given epsilon bigger than 0, we have found a partition p, say that the upper sum minus the lower sum is less than epsilon times b minus a. So, constant times something does not matter. We could have started with epsilon divided by b minus a. So, hence, f is integrable. So, that proves to the theorem that every continuous function is integrable. At least, historically, this was first proved by Cauchy. You will find Cauchy coming in your various courses. He was a French mathematician who contributed a lot of things and a lot of branches in mathematics. The real analysis, complex analysis, algebra, statistics, you have Cauchy's distribution coming in statistics and probability. So, he was the first one who proved this theorem, gave a proof that every continuous function. In fact, he was the one who actually defined rigorously what is the notion of function being integrable. And he all, at that, that was a time when mathematicians did not bother much about continuity or discontinuity. They thought every function is continuous kind of a thing. So, he assumed, not only continuity, he even assumed the fact that we are using that it is uniformly continuous. So, that was only, I would not say mistake, but that is oversight that what we call now as uniform continuity. He thought that this continuity at that time and he gave a rigorous proof of this. So, the class of integrable functions is quite large. It includes all continuous functions on the interval a, b. The problem is, we had started looking at the two questions that what is the class of integrable functions. So, we have given a lot of examples. Now, second problem is, how does one compute integral? So, let us look, answer that question, how does compute, can we have a way of computing for a given function? What is that integral? So, there comes a theorem which is very important and that is why this integral becomes important that is called the fundamental theorem of calculus. So, let us state and prove what is called fundamental theorem of calculus. So, theorem, let me call this as part one. There are two parts of this. So, let us, so it says let f and capital F be two functions on defined on the interval a, b to r with the following properties. One, f is integrable, capital F is continuous on at least the open interval a, b and f is differentiable, oh sorry, continuous in the closed interval, I am sorry, we should say f is continuous in the closed interval and is differentiable at least in the open interval, differentiable, capital F is continuous and differentiable in interval a, b such that its derivative is the function f of x. So, small f and capital F are related with each other in the following way that the derivative of capital F is equal to small f in the interval open interval a, b and of course, we are putting conditions like small f is integrable, capital F is continuous on the interval a, b, so it will be integrable. Then, if all these conditions are satisfied, then f, b minus f of a is equal to integral a to b f x dx. So, the function capital F and small f are related with each other by this equation that if small f is the derivative, so look at this. So, this integral if small f is the derivative of capital F, then integral of small f is just f, b minus f of a. That means what? There is a big advantage that for a given function if you are able to find that it is a derivative of something, then integral of that function is just f, b minus f of a. You do not have to go to partitions, you do not have to go to limits or anything. So, this is and this was again proved by Cauchy rigorously and historically it was a very significant theorem. At least it relates differentiable functions derivative with the integral. So, in that sense people started looking at derivative as an integral and integral as the process is of reverse of each other kind of thing. There is a reason because of this theorem. But the importance of this theorem lies in the fact that it helps you to compute integrals. If you are able to recognize that small f is the derivative of a function capital F, then the integral is. So, let us prove this theorem, give a proof. So, we want to compute f of b minus f of a and we know what is f of b minus f of a and we want to show it is an integral. So, essentially that means, that means essentially that this is a number we should be trying to show lies between the upper and the lower sum. Whatever be the partition, we can show that for every partition p of the interval a b, f of b minus f of a is in between, then that must be the integral. So, that is what we are trying to show. So, we will show for every partition p of a b lower sum is less than or equal to f of b minus f of a is less than or equal to the upper sum or that is enough to show that because if this is the only number between upper and lower, they are shrinking. So, this number must be equal to the integral of the function. So, let us start with the partition. So, let p be a partition and so let us say a equal to x 0, x n equal to b, b any partition. So, now let us, what we are trying to do is I want to bring in inside the upper and the lower sums. So, I have to go to the upper and the lower sums with respect to the partition p. Somehow, I have to steer my arguments towards that idea. I should bring in this. So, now look at f b minus f of a. This is what we want to compute. Now, this is how the partition comes in. This is same as f of x i minus f of x i minus 1 i equal to 1 to n. I add and subtract the consecutive values of f at the consecutive points. So, now, partition points are inside now. So, let us call this as 1. Now, the next step should be that this partition, this f of x i minus f of x i minus 1 should be related with small f. Then, we will get some summation kind of, I think, in terms of small f. Now, small f is the derivative of capital F. So, what is the which is the theorem which relates a function with the values of the derivative? So, that is all one thinks about. So, f b minus f of a in terms of the derivative and that is Lagrange's mean value theorem. So, by Lagrange's mean value theorem on x i minus 1 to x i, there is a point c i belonging to x i minus 1 to x i such that f dash at c i is equal to f at x i minus 1, sorry, f at x i minus f at x i minus 1 divided by x i minus 1 to x i minus 1 for every i. So, once that is the case, that means what? So, that implies that f of x i minus f of x i minus 1 is equal to f dash of c i at and the length of the interval. So, that is Lagrange's mean value theorem essentially. You see how it Lagrange's mean value theorem is playing a part at all these places and this derivative of small f or capital F is small f. So, it is f of c i x i minus x i minus 1. Now, keep in mind, I said apply Lagrange's mean value theorem on this interval. One should check the conditions of Lagrange's mean value theorem are applicable. Capital F is continuous everywhere. So, it is continuous on the interval x i minus 1 to x i. It is differentiable in the open interval. So, it will be differentiable in the open intervals x i minus 1 to x i, each one of them. So, Lagrange's mean value theorem is applicable for each closed bounded interval x i minus 1 to x i. So, that is why those conditions were put both of these ones. And now, so let us, we want to go to one, we need summation. So, implies summation of x i minus f of x i minus 1 i equal to 1 to n is equal to sigma f of c i i equal to 1 to n x i minus x i minus 1. And that by 1 is equal to f b minus f of a. So, let us call this as 2. So, what we have done is, we have computed f b minus f of a in terms of the function small f. It says this summation is equal to f at some point c i. c i is a point in that interval into the length of the interval. So, now, what is f of c i? That is the value of the function at the point c i. So, it will be always bigger than or equal to the minimum value and less than or equal to the maximum value. So, using that implies that if I look at the minimum value, small m i x i minus x i minus 1 will be less than or equal to summation f of c i. So, this is not imply note. This is the observation minus x i minus 1 i equal to 1 to n and that is less than or equal to capital M. I have put summation. So, let me put summation everywhere. I put summation in between. So, less than or equal to summation capital M i of x i minus x i minus 1. Is that okay? Because this is the value at some point in between in the interval. So, f of small f of c i is bigger than M i is less than capital M i. So, now, just combine. What is that in between quantity? 1 and 2 say that is f b minus f of a. So, 1 plus 2 plus you can call it as 3 if you like, imply that. So, this is a lower sum is less than or equal to f b minus f of a is less than or equal to the upper sum. Because this thing is a lower sum left hand side in 3. This is lower sum. This is the upper sum. So, lower sum is less than this summation and that summation is by 2 equal to sigma of capital F i and that by 1 is f b minus f of a. So, putting these three equations together, we get this and that completes the proof. So, this is one of the important theorem in integral or in whole of calculus or analysis as well as differentiation and integrations are concerned. It relates differentiation and integration in the way that if you know that the derivative of a function, then you can compute the integral of that function. So, as a consequence of this, all you get every derivative formula gives you a integral of formula. Derivative of sin is cos. So, if we integrate cos, we must get back sin. So, integral of derivative of any function, any derivative formula will give you a corresponding and that is one of the reasons that as soon as differentiation is done in undergraduate courses and integration is started, immediately 500 formulas appear and you are sort of supposed to remember them and start computing integrals. The culprit is fundamental theorem of calculus. And historically, this was very important because not only it said that for every continuous function, you can have the integral and if you know that this function is the derivative of something, then you can compute the integral and that led to lot of research in what is called theory of Fourier series. So, it gets related historically with the theory of Fourier series. So, let me not go into that. So, basically the important thing is as far as computation is concerned, so this is what is important. You can compute integral of a function, integral of f. If you know that, keep in mind f dash is equal to small f, say, automatically say small f is continuous because capital f is continuous, small f is differentiable.