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Some computer algebra systems have commands that will draw approximating rectangles and evaluate the sums of their areas, at least if $ x_{i}^{*} $ is a left or right endpoint. (For instance, in Maple use leftbox, rightbox, leftsum, and rightsum.)

(a) If $ f(x) = 1/(x^2 + 1) $, $ 0 \le x \le 1 $, find the left and right sums for $ n $ = 10, 30, and 50.

(b) Illustrate by graphing the rectangles in part (a).

(c) Show that the exact area under $ f $ lies between 0.780 and 0.791.

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a. Specifically, $R_{10} \approx 0.7600$ $R_{30} \approx 0.7770,$ and $R_{50} \approx 0.7804$b. GRAPHc. 0.780 and 0.791

00:15

Frank Lin

Calculus 1 / AB

Chapter 5

Integrals

Section 1

Areas and Distances

Integration

Oregon State University

University of Nottingham

Boston College

Lectures

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In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

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In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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Yeah. Uh huh. So problem 5 11. So we're dealing with airy approximation using rectangles and Riemann sums. What we're asking here to do is to find the left and the right some for this function F of X is one of the X square plus one for different numbers of rectangles. So if we look at this a couple of things, we know about this function. If I look at the first derivative of this function that is going to be minus two X over X squared plus one squared, and on the interval From 0 to 1, this is always negative. So the first derivative is always going to be um less than or equal to zero. That means this function is always decreasing. So if a function is always decreasing and I draw a rectangle, if I draw a left rectangle, that's the same as an upper some. And if I have a rectangle and I use the right in point, that's going to be a lower some. Okay, so what we're asked to find is using left and right. So the left is going to be an upper some, and the right is going to be a lower some in this particular case because of the decreasing function. Now, if we take a look at this graphically, well, see here's one over X squared plus one. If I use 10 rectangles and I use the left some. So you see, if I use the left side, I'm always getting an upper some. So let's just go ahead. We're going to do let's do all of the left, which are the upper. And then we'll move on to all of the right to do that. So if I'm doing all of the left sums or uh Let's do I think I'll do maybe I'll do the case in equal 10 and then we'll move on from there. So in the case of in equal 10, let's go ahead and let's do that real quickly. So for this I want to do the function is one over X squared plus one. And I want to use 10 rectangles and doing the left some. So my function is one over X squared this one. And that is going to be stored or just call this F. Of X. So that is my value of F. Of X. And then now I want to do is create the left sides of each of those rectangles. So to create those numbers uh for the left side you start with zero. If I if I'm using 10, so the width of the inter rose one. So if I'm using 10 I'm going to have the width of being 1/10. So I'm gonna go from 2.9 And my increments is going to be .1 and I'm going to store this as all of my ex values. So these are the left side of those rectangles. These are my X. Values. And so I had 10 X. Values which are the left sides of those rectangles. Now what I need is just to evaluate the some. So the width of each rectangle is 100.1 and then the height of each rectangle is going to be the function evaluated. So it's going to be the sum of all of these X. Values. So 809981. So that is the value 809981. So if we look at that 0.8 mhm. Oh What was that number again? It was 809981. So the width of each rectangle here was 1/10. Okay now let's just do this same thing while I've got the number 10 there if I were to use right in points. So in the right in points I don't start at zero I start at 00.1. So what I would do there is I would go back and say okay this sequence that I just created Instead of starting at zero I'm going to start at .1 And then I'm gonna end at the value of one and those are going to be my new X. Values. And now when I look at the somme yeah I get 759981 so seven 759981. Now this is what I expect remember this was an upper some this was a lower some the function is above the X axis. I expected to be lower. Now let's just move on and do that same thing with an end value of 30. So with 30 let's go back into our left some. So if I go back and do my left some so now I'm going to increment um Instead of by 1 10th that's going to be 1/30. So I started zero I go up to and it's not 00.9 It's going to be 1 -1/30. That's gonna be the value that I get. So this is going to give me 30 values and now I'll just go in and say okay we're not. What I need to do is to say okay one 30th is now the width times the sum of all of those X. Values. Yeah So 793685. So 793793685. And so that's when I use 1/30. Uh huh. As my width. Okay now let's move that just to use the right side. So if I'm using the right side what changes? Well if I'm using the right side when I create that sequence that can that tells me the boundary of the rectangle. I don't start at zero. In that case I start at 1/30 And I go all the way over to one and a width of 1/30. So I create all of these And and then I go and just do the same thing that I did a moment ago which was 1:30 at the time as that. Some and I get 777019 so 777019 Yeah. And again the upper some was bigger than the lower some. And then the last one I'm going to do is looking at 1/50. So if I look at 1/50 all that's changing is the spacing of the rectangles. The rectangles are skinnier and I have more of them. Okay so if I look at what changes there let's go to do the left some yeah Right here so I'm starting at zero But now I'm going up to 1 -1 50 and I'm spacing myself by 1 50. So I create 50 values of the left side of the rectangles. And then I go in and say okay well now what is the width width is 1/50 times the sum half of all the X values. 790381. So 790381. Yeah 79 oh 381 And now let's do that same thing on the right side. So if I go to the right side what changes there Is I no longer start at zero. My first rectangle on the right side is going to be the width of the rectangle which is 1/50. Go all the way to one. I increment myself by 1 50th And then I would say okay then you have won over 50 times the sum of f of all of the X values. 780381. So 780381. So 780381. Okay now a little bit of thought about this. So these are your upper sums right here. These are your lower sums. Okay so no matter how many rectangles you get, the actual value is going to have to be in between these two numbers. Okay so the tighter I get it, I know that the actual value of the area has to exist somewhere in between these two numbers because the actual value has to be bigger than the lower some estimate, the best, lower some estimate, and it has to be smaller than the best upper some estimate, and it has to fall somewhere in between there.

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