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Use a calculator or computer to make a table of values of left and right Riemann sums $ L_n $ and $ R_n $ for the integral $ \displaystyle \int^2_0 e^{-x^2} \, dx $ with $ n $ = 5, 10, 50, and 100. Between what two numbers must the value of the integral lie? Can you make a similar statement for the integral $ \displaystyle \int^2_{-1} e^{-x^2} \, dx $? Explain.

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Frank Lin

Calculus 1 / AB

Chapter 5

Integrals

Section 2

The Definite Integral

Integration

Missouri State University

Harvey Mudd College

University of Nottingham

Boston College

Lectures

05:53

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.

40:35

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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Use a calculator or comput…

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Use a computer algebra sys…

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With a programmable calcul…

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Consider the following def…

So since I just wanted to use a calculator computer to find these left and right, Raymond sums for an easy go to the 5. 10 1500. Um, what I did was actually found to calculus that they have on desk most. And you can see the u R l s for the left room on some here and then for the right, Raymond, some here. So you just type those in and we'll fill these out the exact same way, and then they'll give us all the outputs that we need. Um, so let's just go ahead and first fill these out, and then we can try to figure out, um, some kind of bounds for this interval. Okay? So first I'm gonna come up here, I'm gonna start with the left. Three months, I'm I'm gonna plug and eat the negative X squared, and we were wanting to first do five. So we're going to replace in with five, and then we want to integrate from 0 to 2. Okay, What I'm gonna do is I'm just going to get rid of all of the pressure. Now. I'll keep this here. We'll need that for later on So this here gives us kind of like a graph of it. We really don't need this. Um, but it will be useful to help us answer our boundary question. So then down here is going to be our approximation. So, um, for five. For the left, three months, um, it's going to be about 1.7 for 10. Will be about 100.9 eat for 50. It'll be about 500.9017 and then for 100 will be about 0.89 189 So those are going to be our left Riemann Sums. Actually, one thing I want to kind of point out. So let me just go back to five again. Notice how all of these rectangles since we're on a decreasing interval, this left rim on some is going to be an over estimate. So when we come over here to 100 we know that this should be our upper bounds is where, um, overestimated. So, um, we eat special. Let me take a sure of that real fast, and then we can write it down on our last page. Yeah, so over here, the upper bound for this should be around 0.891895 uh, seven and then all round that up, actually to eat. Okay, so this is going to be our upper bound or are integral, and actually that more like a negative. Uh, and then that means our right re months. Um, if we were to look at this is going to be our lower bounce. Let's go ahead and plug this in. So again, we are doing, actually looks like this is flipped around a little bit. So we're going to do zero to Yeah, um, and then we want in to be five. Let me zoom into the area we're interested in. So now notice how again All these are going to be underestimating, since we are on an interval that is decreasing and the right response, um, underestimates. So we can go ahead and do five and then let's do 10. So that is going to be our approximation for the left, um, at 10. And then for 50 would be 500.862 and then for 100 it would be 1000.8722 So this is going to be our lower bound So let's go ahead and write that on this page Also. So 0.872 to 6 me to them zero point 872 to 6. So this is what we believe. This is going to be in between, um and depending on where you're around, you might get something slightly different. But for the most part, it should be in this ballpark. Yeah. Now they want us to determine if we could do something similar with E to the negative expert. But we're entering from negative 1 to 2. So let's come over here and pull this up, and then I'll do negative one here and I'll do less rectangle so you can kind of see what's happening. So we can't do exactly what we did for the left and the right response. Um, because notice here how on the left from negative 1 to 0, the function is increasing. And then to the right, the function is decreasing. So we have where some of our rectangles are going to be over estimations and others where they're going to be under estimations. And then for the right Riemann sum. If we do the same thing. Oh, let me just this back at five. You can see again. Since we're increasing and then decreasing, we have where some is an overestimation and where some is an underestimation. So we can't exactly do it the way we had it before. Um, we could possibly do something where we use left three months sums to estimate negative 1 to 0, um, and then a right Riemann sum and then just kind of use the under and over estimations and Adama. But that is exactly what I think they're looking for. So what we would say or this one is we can't necessarily do it. So can't sense e to the negative x squared on negative 1 to 2 has parts where it is both increasing and decreasing or, in other words, not mono tonic. So this is the reason why we can't exactly just use the same approach that we did for this top one again, like I was saying, you can, but you have to break it up into multiple different intervals. And I don't think that is what they were looking for.

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