good afternoon so today we'll cover section 5.2, definitive integral so we'll start from a very important definition we'll define the definitive integral let's assume that we have a function F defined on an interval AB, okay, then we'll take this interval and form a partition that is with sin points x_1, x_2 x_3 and so on x now coincides with A and x n coincides with b, okay then so make a partition x_0, x_1, x_n, then we'll pick sample points. Pick sampling points which we'll call x_1*, x_2*, x_n*. Basically inside each sub-interval the cycle x_1*, x_2* and so on. remember these points they could be either the leftmost point of this subinterval, the rightmost point of this subinterval or somewhere in between. Okay, then we say that define an integral from a to b f(x) dx to be equal to the limit as n goes to infinity of the following sum. It goes from 1 to n f(x_1)* delta x. So here delta x normally if they're equal length it's b-a over n. That's the length of individual intervals. okay, so this definition is valid if the limit exists and is the same for all choices of sampling points. So for instance it could happen if I choose the sampling points to be the left side of each interval, the left hand rule, and I take the limit, I get a number then I choose them to be the right side of each interval and if the limit is different, then this definition doesn't make sense so it only makes sense if the limit exists, and if it's the same for any choice of all of the sampling points so I'm going to start by making an introduction so let's look at this new symbol and talk about its components. First of all, this guy here is the integral sign and it looks like an elongated s why s? Because it's equal to a sum so it's the limit of a sum. So the symbol s comes from that word. and and b are the limits of integration limits of integration this heref is called the integrand and finally this differential here tells me what is my independent variable okay my horizontal axis is called X in this case and this variable has to match, you know, the variable upon which the function f depends. We'll call this variable the silent variable of integration so let me show you. So the limit from a to b f(x)dx actually can be written as the integral
of f(y)dy. Or I could insert any lette r and the answer won't change. Why is that? well if you look look at the right hand
side of this definition okay and if you take f to be x^2 remember, we actually did this exercise last time we said let's consider the parabola f=x^2 and we actually calculated this sum. What was it equal to? 1/3, right? So we got a number. Does this number depend on x? no, it's a number. It's not a function. So whatever you get as an answer here does not depend on x so x cannot appear in the final answer. So you can replace it by anything else the answer won't change. So we'll call it a silent variable of integration so x or y or anything else questions? question? student: Yeah, I'm a little bit confused most of the time we plug in x as the independent variable and not y, so in the case of the right hand side are we plugging in a y value of a and a y value of b? Instructor: y is just a variable. I can call it alpha. I can call it... t. Absolutely, yeah. So you're right, usually the convention is that usually when you draw a graph, this is y, this is x so not to get confused, we don't often use y there but we might. We can. Okay. So finally this guy here also has a name this is called the Reinmann sum and we'll practice to evaluate such sums today. Okay? so what is the meaning of this new object? let me do it here so if the function of x is non-negative on the interval [ab] then the integral of f(x)dx what is the meaning of that? do you remember? Exactly because we already saw this right hand side. We only did it for non-negative functions and we said, "that's the area." Remember, we split it into a number of vertical rectangles. We calculated the area . So this is the area under the curve f f(x) between a and b. So it has a very clear meaning for non negative f okay now what happens if f changes sign so for instance we can
consider this kind of function see this takes both positive and
negative values so if you look at the definition you can
see we can do the Reinmann sum we can do these, so first we have these rectangles and we have these rectangles, okay, and if you look at the formula they all
have the same base they all have different height, and the height of these is actually negative so we can do these separately. We can group these together, and we get the area area under this part of the curve, and then this here gives me also an area, but it has a negative sign because each of the the components here has a
negative height. So if we call this area A1 and this area A2 then the integral of f(x)dx is equal to A1 minus A2. This is called the net area or the signed area net area or signed area so normally we will assume that the function f that appears under the integral is either continuous or it has a finite number of
discontinuities okay and next I'm going to show you an example of using this definition to calculate the integral. And this example is such that we will know the answer before we did any calculations so in this example we said that f(x) is x-3 we'll say that x belongs on the interval from 1 to 4 evaluate f(x) dx by using the definition. Before I use the definition I'm going to give you an answer so this function is a straight line and this is what allows me to give you an
answer or you are going to give me an answer, okay? so let's plot f(x) which is x-3 as a function of x. At 0, we'll get -3. It crosses the horizontal axis at 3, okay, its value at 4 is equal to 1, it's value at 1 is given at 2 so we're interested in the integral of this function between 1 and 4. Okay? so we talked about that, it actually does change sign, so to calculate it simply, we can calculate this area and you can
calculate this area and remember that this area comes with a
negative sign, right? so let's call this A1 the positive area, let's call this A2 and let's calculate them from geometric considerations. This is easier what's the area of this object? It's a triangle 1/2. It has a base of 1, a height of 1 so it's 1x1 over 2 is 1/2. This one- what's the area of this? so 2. so this is 2 times 2 over 2 is 2. So what's the answer? we have A1 minus A2 is 1/2 minus 2 is -3/2. So whatever magic we're doing here, we're expecting to
get this answer so let's keep that in mind and perform the calculations. We're going to use this definition in order to use the definitions I have to know delta x, x_i*, and f of that amount. So let's do it step by step the easiest one is delta x we have to go b minus A over n, so it's 4 minus 1 over n is 3 over n. Questions? now we have to create a partition so A is 1, B is 4. What is x1? so I'm going to have x1, x2 x3 and so on so each of these intervals have length 3/n 3/n 3/n. So what is the value of x1? it's to the right of 1, the distance is 3/n, so it's one plus 3/n, right? now what's the value of x2? it's one plus twice that thing, so 2*(3/n) this guy is 1 + 3 times 3/n So we can see the pattern. The pattern is that x_i equals 1+ so for each i I have to step i times. So I have i times delta x, 3 over n. Questions? so these are the values of x1 x2 x3 these are not quite what I need, I need sampling points and this is really my choice. I am going to use the right hand rule use the right hand rule. What does it mean? that if I have an interval xi-1 xi, I'm going to take my sampling point to coincide with the rightmost border of that interval so xi* is going to be equal to xi which is 1 plus plus i3/n the last step before I can write down the definition is to evaluate the function at this point f(xi*) so I go here f(xi*) f(xi*) what's my function it's 3-- it's x-3. So I have to take the argument minus 3 and that's equal to 1+3i over n minus 3, and that's equal to 3i/n-2 okay so I evaluated my function at all the sampling points and now I'm ready to use a definition please ask me questions at any time so we write down from 1 to 4 (x-3)dx this is the integral we're trying to evaluate. This is the limit as n goes to infinity of the Reimann sum and under the sum I have f(xi*), that's the expression I got. 3i/n minus 2 times delta x. That's 3 over n. So again, this is f f(xi*) and this is delta x so now I have concrete numbers in the formulas and I have to evaluate first the sum and then the limit, okay so I continue to write limit so to evaluate this sum this is a constant. It doesn't depend on i so this guy can actually come out here so I have 3/n and here I have a sum of 3i/n minus 2 okay. The next step I'm going to split this in to two sums. I'm going to have a sum of 3i over n minus... like this next I notice that in this sum I have i times a constant, 3/n. So this constant. This constant can come out so I have 3/n the sum of i, and this I can rewrite as 2 the sum of one questions? So I simplify that as much as I could. And now I have two sums, and both of them we're going to evaluate. The first sum so let's do it here what does this stand for? This stands for one plus two plus three plus four, and so on all the way to n, that is called an arithmatic progression and we have a formula for it, right? you don't have to remember this formula, but you must have seen it before in some class or another. Okay and this one, I want you to tell me what this one is equal to So either one or n. So what do we do here? this is a machine. We have to take it and add it for each i i goes from 1 to n, so we are going to add this thing together n times. So we'll go 1 +1 +1 +1 +1 n times. So this is n questions? So I evaluated both of these objects. So this n(n+1) over 2, this is n so I continue here limit as n goes to infinity I have to collect everything I have here
so 3/n 3 over n, n(n+1) over 2 minus 2 n. Okay. so something is going to simplify. I am going to divide through by n. One second so 3 this n cancels, and I'm dividing by n so I have 1+1/n over 2 minus 2 okay, dividing through by n all the terms. And now I can take the limit I do this in order to take the limit. The limit is easy because 1/n goes to zero as n becomes large so this is 0 so I have 3 times 3 halves -2 so this is the answer I get and we'll go here. See? It's the same thing. So you can see two methods, right? This is done very quickly this is long and
painful but we could only do this because these were triangles right? This new definition eventually will give us a tool to evaluate the area under curves. Not just straight lines. Question? student: how did you divide everything by adding it? So here, let's write this I have 3, this n cancels so I have 3, and I divide by n here. n+1 over n 2, right? And n from here, I get this. more questions? okay very good so it is actually very rare when we can complete this calculation and get the exact numerical answer again, this was possible because it's such a simple function so sometimes it make sense to
approximate let's not calculate this...the limit but we can get an approximate answer Okay. So in order to do this, we introduce the midpoint rule so simply put this rule allows me to approximate this
integral by a finite sum both of f of xi bar delta x, sorry, delta x. What's xi bar? So let's draw one of the intervals from xi-1 to xi. Remember the right point rule we take our sampling point here. Left point rule, we take it here. midpoint rule, we use the middle. So so we say that xi* simply equals xi bar where it's the middle point. xi bar xi okay is it me? no okay fine. So this is the midpoint, and thus the name was a question? okay. was that music? so this is the midpoint and let you let me give you an example of how we can apply this rule to approximate a particular integral. So approximate the integral from 0 to 1 x^2 dx with n = 4 by using the midpoint rule. So let's draw every time they ask you to evaluate the Reinmann sum. The best thing to do is to draw your interval actually right down the partition find the sampling points. Do it systematically, and then it's very easy so we go from 0 to 1 they specify n = 4. So this means that we divide it in to 4 subintervals. This is 1/2 one-quarter, three quarters. Four intervals. Now we have to find the sampling point Which has to be the middle point, because we are using the midpoint rule so this one here, x1 bar what is it equal to? 1 over 8 okay, x2 bar is 3 over 8 x3 bar 5 over 8 and x4 bar is 7 over 8 questions? So we know our points, just put it here. we have to evaluate the function at these four points. And the function is x^2 so f(x) is x^2. So f(xi) is just xi^2 so we can say that the integral from 0
to 1 x^2dx is approximately equal to okay, each term here will contain the multiplier delta X so I'm going to take it out of the bracket and I forgot to calculate delta x. So what's delta x here? what is the distance between my points in the partition? One quarter, right? n=4, the total interval is one, so it's one quarter. so I have one quarter that comes out and here I have the sum f1-- oh, f x1, x2 x3, x4. So one quarter of one over eight squared, 3 over 8 squared, 5 over 8 squared, seven over 8 squared. So if you put it in your calculators, maybe one of you can do it here, if you have a calculator we're going to evaluate this number and see how good this approximation is to this integral what is the exact value of this integral? 1/3. We did it last time, right? So this is going to be something like 0.32 I believe something like this. So it's not going to be precisely the exact value, but it's going to approximate it and if we take a higher n, we will get a better answer okay. Questions? Question? Student: "the 1/8, 3/8, 5/8, where do they come from?" Where do they come from? So these are these values Okay. These are the sampling points for each interval I have to take the middle point and evaluate my function at that point so if I use different
color inside each interval I calculated the middle of that interval. So my first interval goes from 0 to one quarter I only have four intervals, this, this, this and this. for the first interval, the middle corresponds to 1/8 I have to evaluate my function at that point the function is x^2, so the function evaluated at that point is 1/8 squared. That's the first term. Then I have to take the second inerval its middle is given by 3/8, the function at that point is given by 3/8 squared and so on. I add them up. More questions? So next let's go over some properties of definite integrals so the first property I write down here the integral from a to b f(x) dx. The limits of integration, a to b can be...can be swapped and we actually can go from b to a so if a is less than b, we can go this way or we can go the other way but the corresponding integral acquires a negative sign. Why is that? Because when we go in the positive direction, the increment, delta x, is a positive quantity. When we go back, the value decreases .So the increment is negative. Okay? so if you look at the Reimann sum you end up multiplying by a negative value, delta x. so we have a minus sign here. Something that immediately follows from this is the following: the integral of f(x)dx from a to a-- what's that equal to? 0. Two reasons for this: the first reason comes from this formula if I take them
both to be a then this is equal to minus the same thing, so this can only be zero the other reason is that we're considering a the area
under a curve of an infinitely thin base. So it has to be zero, right? okay so then let's list a bunch of properties that will help us evaluate very many integrals and Riemann sums, too so property number 1 states the integral of a constant and it's equal to this so let's suppose that my function is the constant, c which can be negative or positive so then this object on the left its either the area under this curve if this is a positive constant or it's the area above that curve with a minus sign, so it's the same thing, and this is a rectangle so the area is equal to b-a times C. and that's what we have here on the right hand side. Okay? So it's very easy to evaluate the definite integral of a
constant because it comes from this geometric consideration. Questions? Okay. The next one talks about the integral of a sum of two functions. So here, both f and g are functions of x. Question? Student: just to make sure, I can't see. Is that an equal sign? here yes. Student: So that only works it works if-- oh, I see. nevermind. It works for both positive and negative values of C and also for C=0 So if I have a sum of two functions, then I can evaluate the integral separately. It's fdx plus gdx gdx. How do I know this? this is very easy to prove all I need to do is recall the definition okay so this is the definition of the integral f(x) so I can replace f with f+g then I'll have f+g here so I can split this sum in to two sums one was f, the other one was g. Then I can evaluate the limits separately. The limit for the f sum plus the limit for the g sum and that's equal to these two things Okay? So I'm sorry, I shouldn't forget to put the limits of integration. Because what I just wrote actually as a very different meaning. That's the antiderivative, that's not what I'm talking about here Okay. Property number 3: the integral of a constant times a function is given by the same constant times the integral over the function Proof? You go here again. If we have c times f here then we have c times f here. c comes out of the sum c comes out of the limit, in order to have a c times this which is given by the integral times c okay, and finally property number 4 I'm not sure why it's listed as a
separate property in the book but i's the integral of the difference between two functions it actually follows from properties 2&3 but I'll write it down nonetheless so the integral of a difference is the difference between the two integrals and next we apply all these rules to calculate an example so let's look at the integral from 0 to 1 5+2 x^2dx so one way to approach it will be to do it by definition. Write down the limit of a sum of this thing, create a partition calculate sampling points and then do the sums and then the limit it's very, very tedious. We're going to do the same thing by using the rules so we notice that the integrand is a sum of two functions so by rule number two I'm going to
present it as the sum of two integrals. So this is rule number 2. I can evaluated separately. Okay, now the first one, we have to look at rule number 1 which is a constant, right? we have five here so we have five times 1 minus 0. And that's by rule number one so I dealt with this one nicely. Now this one we have to look at rule number 3. we have a constant inside, 2 comes out. So we have 2 so this is rule number 3. So what's this equal to? I already asked you like five times today. 1/3, right? so we're going to use this fact to write down the answer Okay? Very easy. So if we can decompose it somehow and follow different rules, we can evaluate integrals easily. Questions? now rule property number five let's suppose that I have are an interval from a to b and I look at an integral of fdx-- of x dx, and I define another point, C which may or may not be between a and b actually, it could be anywhere else. Then I can rewrite this integral as the sum of the integral from a to c plus an integral from c to b. I will illustrate this with a picture let's suppose for simplicity that f is positive and the quantity on the left is has the meaning of the integral under the curve. Question? Student: "just want to make sure we're on the same page. C is between a and b?" It doesn't have to be. So in this example it is but in general it doesn't have to be. It's just another constant. So... and you can prove it, of course. Here we illustrated it with an example that is more obvious on the left we have the area under this curve. the whole thing. Now on the right, we have two integrals which is this area and this area you add up the two, and you get the full thing. Okay? So this is the the meaning of this property as an
example for instance we can say that we can write this down as the integral from 0 to 1 1 to five. So this is an example Okay. Question? Student: can you draw an example where c is not between a and b? Yes. So let's consider the following geometry a, b, c so we have to assume in this case that f is actually defined for the whole interval. Okay, so if f only exists here, we cannot do this. but if c is in the domain of f, then we can potentially define c to not belong to this interval. so, so this is my function f okay so I have an integral from a to b fdx is equal to from a to c fdx from c to b dx. Now let's simplify this. So first of all, this first interval I'm going to use this rule and flip the limits so I get abfdx is minus from c to a fdx now the second integral spans the interval from c to b. okay? so I'm going to split this one in to this plus this so instead of this integral I'm going to write from c to a fdx plus from a to b fdx. Right? now you can see that this the first and the second terms cancel
out and I get that identity the left hand side is equal to the right hand side so it's a little bit more... less obvious
but it also works more questions? Okay. Finally, the last group of properties is called comparative properties they are numbered from 6 to 8, So let's suppose that f is non-negative then the integral from a to b f(x) dx is non-negative. This is obvious because for non-negative fs the integral is nothing but the area under the curve, and that's a non- quantity. Okay? So we already know this. fine number seven let's suppose that we have
two functions and one of them is bigger than the other on the same, you know interval. Then we can compare the integral of f was the integral of g. The function is bigger and then the integral is bigger too. Let me illustrate this with an example Let's suppose that both functions are non-negative, okay? For simplicity. So this is the larger of the functions, f and this is g okay? Between a and b so the integral of f is this whole area and the integral of g is part of it right so you can see that the area under
the function g is smaller than the area under the function f, and that's exactly what this says. Okay? this only works if f is greater than g. If they intersect somewhere, then we cannot make this conclusion
finally the last property property 8 Is like this. Let's suppose that we have two numbers m little and M capital Which work like this. so this is my function from a to b and there's some value, m and another one, M capital such that they enclose this function. One is always smaller than my function, and the other one is always larger for all values of x on the interval. Okay? Let's suppose I know that these two numbers exist. make an estimate of the integral of f okay? I know what it's bigger than and I know what it's smaller than what are these things? So the integral of f is this area now look at this box here this is completely contained inside this bigger area, and the area of this box is m times b-a. well I can see that the it into love F youth gonna the smaller done that this box on the other hand but those are the books this on the
aerial the box you buy em capital you minus a in a
completely contains then the growth so I can have there's bound the upper
bound on the bill so a I will say this was a an example
and then I let you go game example like this fifth find the %uh so the integral from 0 to 1/4 0 he to the Linus affects squared yes so at this stage we have no idea what this alias but got but I can read
it down a double in the quality that defines what it is so let's blow this function it some taking
functional X day because you're a and one a when ethical 0 itself is equal to 1 in 1x chemical 21 it's now the is E to the -1 Pratt so for sure I can write down the bowling double
inequality he to the -1 is almost less or equal to you to the minus
excluded and thus always less or equal to what so this function that lot is bounded between if -1 which is
something like one quarter and one someone to use proper 28 to a evaluate I the bounds for this and I'll on the left I'm gonna have you to the -1 times 110 and on the right they have one so to conclude well I know that this integral is between one
already and one and this is much more
information than I get by just looking at this rate thank you much
