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Use a calculator or computer to make a table of …

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Problem 14 Easy Difficulty

With a programmable calculator or computer (see the instructions for Exercise 5.1.9), compute the left and right Riemann sums for the function $ f(x) = x/(x + 1) $ on the interval $ [0, 2] $ with $ n = 100 $. Explain why these estimates show that $$ 0.8946 < \int^2_0 \frac{x}{x + 1} \, dx < 0.9081 $$


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00:03

Frank Lin

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Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 2

The Definite Integral

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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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Video Transcript

all right, so let's go ahead and start doing this problem. So the question basically says You have the integral off X over expose one. It's definite. It's asking you to. You gravitated from 0 to 2, and it's a little bit trickier than usual. It wants you to basically calculate this numerically using a programmable calculator or a computer, and show that if you use the remand sums, the the lower bound is going to be 0.8946 and the upper bound is going to be 0.9081 using the left hand rule in the right hand rule. Okay, so first, I am just going to go through the idea of what we're going to do. And I will use Excel in order to show you that we actually do get these numbers and the idea of why it's going to be between those two numbers. So I'm going to start with that part first. Now you are going to be able to prove that this if I call it why, why Prime is going to be a positive number on the interval from 0 to 2, why double prime is going to be a negative number in the same interval. So this graph that I drew right here is basically, um the graph of X over X plus one from 0 to 2. Okay. And that portion, the algebraic calculation. I will let you do that. However, the idea is this. If I use the left hand roll and let's see, I split it into five equal basis. When I calculate the rectangles, the first height is going to be the left endpoint. So it's gonna be that rectangle off course. The height is zero. So there is no area. The next rectangle is going to look like this. The next one like that make something like this and there's one more. As you can see, the integral here is the area under the curve. But we have a little bit of a gap here, so we know that this is going to be the lower estimate. So if we're able to show that this is indeed the value of the Riemann sums with 100 rectangles, that shows that the integral is going to be greater than this number. Okay? Similarly, if I use the right handle, the first rectangle will start right here with this hype. And the next one is gonna be here, the next one there, the next one there and the next one there. So it is very clear that it's going to be an overestimate, because this portion is above the area off area of the curve. So we know that if we can calculate that this is precisely the value of the remand, some with 100 rectangles, it is going to be above this value right here. So that's the idea off how we show this. Okay, now all we need to do is to actually calculate it. So I will show you how I do that using excel. Okay, I actually already did the calculation, so let's go through it anyway. Stock. So I know that the Delta X, the with off the rectangle, is going to be 0.2 because the distance from 0 to 2 is too. And there are 100 rectangles. So this is where you can see the calculation right there. Now, each off the positions using the left hand. Rule it. The first number starts at zero. To get the next number, I add 0.2 to get the number. After that, I add 0.2 again. So how many times a My adding 0.2 It's simple. It's actually the number of times that I see the indices right here, um, in the X minus one. So it is going to be I am adding zero times here I am, adding, once there I am adding three times there, so the calculation would be this number minus one. So it's basically one minus in, uh, index minus one times 0.2 I'm not going to do any of the fancy absolute reference and stuff because this is a simple calculation. So let's just do that a quick and then we get all of the numbers. One way to check that we actually did the right thing is to make sure that the very last number has to be close to two. And it is. In fact, it's 0.2 before, too. So this is precisely what we're looking for. Okay, Next, we want to find the height off the rectangle, so it's f of the input right here. So we're simply going to dio this number divided by this plus one because the function was X over X plus one. Hey, of course, that value is going to be zero little quick. And these are the numerically approximated value off X over X plus one at each of these given inputs. Let's see. Well, I cannot do it in my head, but apparently the very last value is going to be approximately 0.66 What's to over two plus one? That's 0.6 bar. So, yes, it makes sense that the heights are going to be that much. Now we actually calculate the area F of X times, Delta X. So this is going to be this value times point to cannot use parentheses. I need to use a star table. Okay, go. And we got all of the areas. Now let's calculate what the total area is going to be. That is going to be simply the some off each of these area. There's 100 of them, as you can see right here. Did I get 0.894691 precisely what we were looking for. So if I go back to here, we definitely got that number. We can do the same thing with the right hand roll. Now, instead, off using zero as my input. Let me just erase this one. What's gonna happen here is the first height actually starts on the right hand. So, um, the first height is the first exposition is going to be 0.2 instead of at zero. Okay. What's the next height? You add 0.2 to that one. What's the next height? We add point on 10.0.2 to that one. So what ISS The relationship here between K and X stars. Okay, well, it's simply going to be how many times a my adding 10.2 the index many times. So it's simply eagle, too. This number Times too. If I do this, I feel everything out. And this is what's awesome about Excel. I already did these calculations based on the same idea, but the input is just going to be changed, and all of the calculations are already done for me. So I know that all of these are gonna be right. And according to this number, 0.9080 we get that. Oh, it's just slightly off. But, uh, I don't think this number is right, because I am very confident with what Exhale just did for me. Anyhow, I am going to decide to say I am right. Okay, anyhow, at least that is good enough for me. And that's how you numerically calculate what the left hand role in the right hand rules are going to be. I already explained why it's going to be between those two values, and that's how you answer these type of questions.

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Video Thumbnail

05:53

Integrals - Intro

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.

Video Thumbnail

40:35

Area Under Curves - Overview

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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