Okay? Alright, so we are trying to find the area under the curve y equals one over x plus one over the interval three to four by subdividing into four equal parts, right? So I'm going to sub-divide 234. And so it's this rectangle, this rectangle, this rectangle, and this rectangle. Four rectangles. Okay? So I need to find the subdivision points. Are we okay with this? Alright. So you find the subdivision points and you use those height, okay? This is important. You use the height coming from the subdivision points. I'd be okay with this. Ok, just to show you, earlier here, I had these subdivision points and I use the heights at subdivision points. Okay? So the height of the, if you'd like to face the right endpoint of the intervals, sorry. It's always the right endpoints of those intervals. Okay? So, alright. So you'd have figured out these subdivision points. So you first, you figure out the length of the sub interval, which is the length of the interval, three to four, which is 14 minus 12 divided by four. So what are the subdivision point? Here is P. Here is for sort of the 1.1234th. And can anyone tell me what's the first subdivision point? So what are the what are the, this number here? The first 125, right? Okay. T, the length of the subinterval is this much 0.25. So this is three plus 0.25. Ok? So this is 3.25. Ok. That's the first one. The next one. So it will be 0.25 more. Right? That's the width. Do you all understand this? Of course, you'll have to do this later. So you know the starting point and you know your width, and you keep adding to it. And so the next one will be 3.75 and then the next one is four. These are the four points. Figure them out. Okay? So I'm going to label them here. Now. This is 3.253.53.75. And this function one over x, I'm just going to call it F x for the moment. Ok? So approximate area, some of areas of rectangles. Okay, so what's the area of the first rectangle? How would you compute the area of the first rectangle? That's one. You do 0.5 as the width. And then once I that by inputting pointy five into the Y equals equation, correct? Right. Sorry, who's that? I'm trying to figure out who it is. Hmm, I couldn't see who's Microsoft's sorry, who was that, isn't it? Oh, they're up there. Okay. All right. Okay. Right. So you have to dig the width then multiply my height. Okay? So the first one, the width is 0.25. And exactly what you said, right at the height of the height at 3.25, your function f. And then the next one. So the area of the second one, it's like the same day is 0.25 times f hat x, at which point they play five. Right, at the second height, when? The second height. Yeah. Okay. And then the third one will be this width, which is 0.5 times the height at 3.75 and 0.25 and f at the last rectangle. Okay. Width times height, height of that four, Sorry, I didn't whatever. Right. Yeah. I have with me on that. If someone had doubt because you will have to do this for your homework. Okay, then the rest is all calculated. Well, there's one more step here, which is I always recommend. You always pull out the width outside. It saves you number of multiplications. Right? X is one over x plus one. Okay? So you use a calculator and you come up with the answer, and you get this. How are you okay with this? Do you understand this? So let us go back and look over this again, because we will have to, you'll have to do this in the homework. Alright? So you have some curve, some function y equals f x over some interval, three to four. And I want to find the area under the curve. You don't need to know the shape of the curve. You can draw any shape you like, doesn't matter what shape it is. First thing is there to figuring out the length of the subinterval. That's when you're going to cut into four pieces or whatever number of pieces. So you figured out the length of the interval, which is four minus three and divide by four. Length of each. Then what are the subdivision points? So you have three to four and every 0.25 you go 0.25 more, 0.250, and you get these points. Okay? And then the area of the rectangles as width times height of each of the rectangles. Which really even notice it comes out always to this, it's always this width of the interval times the value of the function at the subdivision points added up. This subdivision points. Every time this is what we're doing is it's clear what I'm saying here, like this, this thing here. Approximate area as you've figured out the into subintervals. And then you add up the value at the subdivision points of the function. Right? The thing here. So you don't have to, it's good to go through this exercise, this whole thing, but I did here. But you will see it will always be width of the sub interval times the value of the function at the subdivision points added up. Is it clear what I'm saying here? Like this is what we're doing. Right? Because the advantage with this thing is it doesn't say how many subdivision points u hat. This is true of this formula is true for any number of subdivision points. Alright? Okay. So now I'm going to take this thing here and define. So suppose your interval was okay. Before I go ahead, I want to make sure this is this part clear. Like the approximate area is always this. This is what we are doing. That we are taking the width of the sub-intervals and multiplying by the value of the function at the subdivision points. Ok, this is that, this thing here. Look at this. The width of the sub interval times the value of the function at 3.253.53.754. Those are your subdivision points. 3.253.53.754 subdivision points. Okay? So I'm saying this formula here, width of sub interval, value of the function at subdivision points. Make sense there? Or it's not clear yet. Matthew, is this okay? But I'm saying okay, because I'm extracting a general principle here. That's why I'm saying this. So whenever you're doing an approximation, it's always you have an interval, you have a function, and you'll have subdivision point. So okay, here, here's what I'm. So suppose interval is a2 b and subdivide into n parts. Then could we, for 10 million? I cut it into n parts. What is the width of each sub? And I'm pointing to a general thing. If a to B and I divide it into n parts, how wide is each subinterval? Hawaii does this. Can you get a formula? Matthew? What are the b minus a divided by N? Perfect, right? It's always figuring out the length of the interval b minus a divided by n. Okay? They've cut it into n parts. Here's a first subdivision point. I call it x 1, second, 1, third, 1, fourth 1. And the last one is always that. Alright? So if you have some curve, y equals f x over the interval a b, approximate area. With n subdivisions of AB is equal to width of sub-interval multiplied by the value of the function. Had this subdivision points. Okay? That's the general formula. You want to find the area under the curve y equals f x over the interval a, b divide into n parts. That till width times the value of the function at b. Subdivision point added up. Okay? It doesn't matter whether you do four parts, but you do ten parts per million parts. It's always the same formula. Okay? So now you know that if I want better and better approximation of the area, I just take larger and larger n. I take a million, I take a billion, I'd take a trillion. So what is the exact area? I get better and better approximations. What is the exact area? Right? For example, I told you, I wrote to I told you this the other day. I said one over three. I did it some other way. I didn't do it by approximation. So there must be some exact area. What is the exact area? These are all approximations. How would I define the exact area? Could you do like the interval at one minus the interval at 0? I can get, I was showing you how to get approximations, right? But I told you the correct answer is one over three, which means I must have a meaning for what the Octavia is. See if I gave you a shape like this, you can find the area. There is no definition of area for shapes which are not rectangles or triangles. All I'm doing is approximating rectangles and triangles. So there has to be an exact definition. Okay, let me ask you this question, but you'll see where I'm headed. Right? I gave you this, say here, this curve y equal to x square, right? And let's say this point is two, x is two. So this is two comma four. What's the slope here? Slope of this curve. Let's make it 39. Actually. What the slope of this curve at this point tonight. You all know that. What's the slope? Greg, Kayla, you when it says something, right, you just find the derivative 2x and six. But what is the definition of slope there? Right? The derivative, right? You did not say to find the slope, you did not say, you can get an approximation. You can take 3.1 and whatever it is and take that line and find the slope of that line. Right? You see it? I'd probably not. Right. If you wanted the slope there, you could just take another point and slope of the line. We know that. And you can take out another pointer to get the slope. Those are all approximations. What was the exact slope? The derivative. The derivative is the limit as this point come closer and closer to three. That what the exact definition of the slope? It was the limit of the oldest approximate slopes. Same here. The area under this curve is the limit of these approximate areas. As the number of subdivision goes to infinity. To flag the slopes, they were all approximate slopes. What was the exact slope? It was when the subdivision, like here you had you had three and you have three plus h, you found the slope of that line. And as h goes to 0, that was the exact slope. Alright? Same area. So I'm going to write this definition for any function. Effects on a b. Integral a to b. F x dx is the limit as the number of subdivisions go to infinity. That is the exact video. It's the number of subdivisions go to infinity. What do you get? The approximation is a better and better and that's what is called the integral. And this is equal to the area under the curve y equals f x on the interval a to b. Ok? So this integral here of a to B is this is the definition. And it has a meaning it has one of the meetings as it's the area under the curve. Okay. Now you say, well, what are you, why are you emphasizing all this and doing this? I've shown you how to find the approximate areas. But what is the correct answer? What did the exact answer? If you don't have a meaning for the exact answer, you're not approximating anything. Okay? This is the definition of the exact area of the curve, which is denoted by this notation. Okay? Notice there's integral a to b and the function and so on. Okay? So point here is this. Ok? So whenever you want to compute the area under a curve or you know, for example, this shows up when you're trying to find total calories burned your mother applications. The way to get better and better approximation is to do this, right? Subdivided into 100 parts. This is how Archimedes was doing integrations, right? If you think about it, you all know this formula. Write area of a circle is pi times the radius squared. Why is that true? Right, we know the area of a rectangle we can agree on when I can define length times width, right? So this is x times y, yes, area is x, y. But it's so trivially obvious that the area of a circle is pi r-squared. Who don't like, where is this coming from? If you took a square piece of paper of radius R and you see how much material did you use? You really think it'll come out that who came up with it? Where is this coming from? Right? The only thing we can be certain about the area of a rectangle that say nothing else we can be certain about. So how is it that the area of a rectangle is this gives you that the area of the circle is pi r square, right? We believe, does you know it's, it's not a God-given law. I'm sorry. It's not a law of some physics laws. Something this is true. And this is discovered by Archimedes. How did he do it? He did it by cutting this up into rectangles, finding the areas of the rectangles, and doing finer and finer subdivisions. And showing that this sum adds up to this. That's how he did it. He did integration. What I'm doing here, what I'm trying to show year, that's how the Greeks figured out that the area of a circle is pi R square. By approximating the circle buys finer and finer rectangles, just like what I'm doing here for the area. And taking the limit. He took this limit actually. How do you do that will not do it in this course. He actually did this limit. And that's how he got pi r-squared. Right? So, but then my question is, how did I tell you that this area is one over three? And I did it my head, I didn't take any limit. Right? And the answer is that there is a connection between this way of doing thing, which is the limit and the antiderivative, which Archimedes did not know. That's why our committee's kept doing the limits. Archive. It is 300 BC. This is how we found the area of circle. You probably know the area of the volume of a sphere. You probably know the volume of a cylinder. All these were computed by Archimedes using this technique, this limit technique. Right? So the question is, how did I do this? And this was discovered by Newton and Leibniz. He said, this thing can be done with antiderivatives. There is no kind of, right now you can't see where the derivative shows up here anywhere. This approximating my rectangles. There's nowhere derivative shows up this area. This area for circle. You're approximating the rectangle, bury the deliberated. There's no derivative here. But it was Newton's genius and Leibnitz that they said, oh, I can do these areas sums using a derivative in a very clever way. And that is called the fundamental theorem of calculus. Ok, so I have teased you enough at this.
The definite integral III, 6.3, Math 221 Winter 2021
From Rakesh Rakesh January 29, 2021 Created from The Fundamental theorem of Calculus I, 6.4, Math 221 Winter 2021
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