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With a programmable calculator (or a computer), it is possible to evaluate the expressions for the sums of areas of approximating rectangles, even for large values of $ n $, using looping. (On a TI use the Is> command or a For-EndFor loop, on a Casio use Isz, on an HP or in BASIC use a FOR-NEXT loop.) Compute the sum of the areas of approximating rectangles using equal subintervals and right endpoints for $ n $ = 10, 30, 50, and 100. Then guess the value of the exact area.
The region under $ y = \cos x $ from 0 to $ \pi/2 $
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Calculus 1 / AB
Areas and Distances
Missouri State University
University of Nottingham
Idaho State University
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.
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.
With a programmable calcul…
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Yeah. So problem number 10. We have a function F of X. Cosine of X on the interval from zero to pi over two and they want us to approximate the area. Um sampling it with the computer using 10, 30 50 and 100 rectangles. So real rough, you know that when you graph the co sign function from zero two, pi over to co sign function starts at the value of one and it's gonna end at zero. So what you see this curve doing is something like this is an important thing to note is this curve lies entirely above the X axis. So when I create the rectangles, they're all going to be a positive height and I add up the area of all the rectangles, they're all positive. That will approximate this area. So I'm going to use my computer to do this. That the key is. What is the interval, the length, the width of this interval here from here to here is Pi over two. Okay, so when I divide that by 10 so if we just take a look real quick, what are going to be the widths of our rectangles? Okay When I have 10 rectangles, the width of each rectangle Is Pi over 20. When I have 30 rectangles, the width of each rectangle is pi over 60. When I have 50 rectangles, each rectangle is pi over 100. And at 100 rectangles each rectangle is pi over 200 in width. Okay so I'm gonna have 10 rectangles of with pi over 20. That will cover the interval from zero to pi or two. So we're going to use the computer to make all of this happen. So um let's get our calculator for all of this and let's just figure out so the function was co sign of X. And the co sign of X. And I'm going to store that. That is my function F. Of X. I know that now F. Of zero should be one. Okay that's what the coastline function does. Now let's just create a sequence of numbers. And so I'm going to start at so in that first case remember I start at Pi over 20 if I'm using the right side. So pi over 20 is where I start. So let's just do that real quick. That's going to be sorry about that exit out of that. Um It's going to be pie over 20 is the first right point. And then I'm going from there to hire her to. Yeah shit. Yeah And I'm doing in steps of pi over 20 So pi over 20. Yeah is how I'm stepping and I want to store all of that. I'm going to call those my X. Values. So that should create for me a list of 10 Right in points of length. Power over 20. So remember the width is what Pi over 20 in this case the pie over 20 is the width. So now to figure out the area it's going to be So pi divided by 20 times the sum of all of those X. Values. So F. The X. Values. Yeah. 0919403. So that 1st 1919403. Let's write that down. So 919403. That's my approximation of the area. When I have 10 rectangles. Now let's just go back and it should be a little bit quicker now when I do this with 30 rectangles. So now the width is Pi over 60. So I'm going to go over here and go back to the sequencing And I'm going to say now the width is Pi over 60. Yeah I'm going to pi over two and I'm going to increment and pi over 60. Those are my new X. Values. and now the width of each rectangle is Pi over 60. So I need to figure out what is pie divided by 60 times the sum of all of those X. Values. 973592 973592. Mhm. Yeah. Yeah. Yeah. Okay let's go up to 50. So 50 the width is going to be pi over 100. So let's just go back to our calculator and we'll just redo that sequence command. Now what changes is the width is pi over 100. Mhm. So that gives me 100 values each of with pi over 100. And now I'll just come in and The width is Pi over 100, Sum up all of those values. 984-1 is the approximation here, so .984-1. Yeah. Yeah. Mhm. Yeah. And now let's go to 100 rectangles. The width is Pi over 200. So go back to my calculator, let's just recall all of that command with all the typing And we're going to change this the pi over 200 as the width and the implement. And then go back to where we did that some. And now just change this from pi over 100 pirates 200 99215 so .99215 mm. Now they're saying guess where this is headed. I don't know exactly. That's why it's a guess these numbers look like they're getting closer and closer to one. So my guess is my guess is that the exact area? Yeah. Yeah. Is one square unit. Okay. So I don't know that, but just by trending by using more and more direct angles, that's my guess is that is headed to one.
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