with-a-programmable-calculator-or-a-computer-it-is-possible-to-evaluate-the-expressions-for-the-su-2

Question

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 $

Frank Lin · Accepted Answer

So this is a computer problem. I leave that detail for you and to know how to set it up. I explain in problem night off this section. So if you want more information, please take a look at that.

Robert Daugherty · Answer

mm. So this problem, they gave us a function Y is equal to X to the 4th and roughly you should know that looks like a very steep version of a parabola From 0- one. And were asked to estimate the area. Um using rectangles, they ask us to use an equal 10, 30 50 and 100. And fortunately we are doing this with calculating tools instead of doing this manually. So if N is equal to 10, then on the interval from 0 to 1, the width of each rectangle is 1/10 and then likewise, if n is equal to 30 50 and 100, that&#x27;s where we&#x27;re going to come up with the width of these triangles, each getting smaller and smaller as we have more and more triangles. But rectangles to do our estimates for the area. So now let&#x27;s do this if I&#x27;m using right in points, so if I&#x27;m using a right in point, So think about this on the interval from 0-1, The right in point. That first endpoint is going to happen based on this width. So in the first case that first endpoint is going to be .1 in the second case, when n is 30th 30? That&#x27;s gonna be 1/30 and then 1/50 and so forth. Okay? So and then we just need to figure out that tells me the width and then evaluate the function to get the height. So we&#x27;re going to do all of this on our calculator. So we&#x27;re going to do first is let&#x27;s store this function. So X to the fourth, I&#x27;m going to store that function and just call that F of X. Yeah, for that as function. Sorry about that. Okay, X to the 4th and then move the cursor down and then store that as F of X. And I&#x27;m done. So now I can evaluate F of any any value and come up with that. So I&#x27;ve stored f now I need to create a sequence to get those right in points. Okay? So I&#x27;m gonna index with I. And where does that first endpoint start when N is equal to 10? That first endpoint is at 1/10 So the first endpoint Is going to be at .1, so that&#x27;s the first right in point. And then I&#x27;m gonna go, how far how far am I going? I&#x27;m going all the way over to one and I&#x27;m gonna increment that by .1. So that&#x27;s going to generate all the right in points for the rectangles. So I need to store this, so I&#x27;m going to store this in the variable, we&#x27;ll call it this all of our X. Values. So that gives me all of the X values I need uh in the case and it gives me 10 of those to make that work now. So that gives me the right in point of the right side of the rectangles. I know the width is 1/10. So to get all of these are gonna have to sum up. So I&#x27;m gonna have to have The width of the wrecked the triangle, the width of the rectangles, .1, so .1 times the sum. Okay, I need to add up. Well I need to evaluate F at each of those points, so F of the X values. And that will give me .25333. So that is my estimate when N is equal to 10.25333. So let&#x27;s just write that down. Yeah. Yeah .25333. Now we&#x27;re going to go back into the same thing when in is equal to 30 and so what does that change? Well when N is equal to 30 I&#x27;m going to change my sequence. Okay so let&#x27;s just go back up here and let&#x27;s just do that all over. So instead of starting at .1 is gonna be 1/30. Okay. Mhm. But that&#x27;s gonna be where the first point and then it&#x27;s also going to go to one but I&#x27;m going to increment by 1/30 and that should give me a longer list of values. Okay so now what I need is I need to estimate the area. The width of each rectangle now is 1/30. So what is 1/30? So one 30th times F at all of these X. Values Excuse me. The sum of all of those times, the sum of all of the f at each of the X. Values. Yeah .2170. So .2170. Is what I come up with their, Now let&#x27;s do the same thing with 50. So I go back to my calculator and I said okay well what would the sequence looked like if I had 50? Well the width would start the first, the first rectangle would start at 1/50 I would go to one and I would increment in increments of 1/50 And that is going to be my new list of exercise. I should have 50 of those. And then now the width is 1/50. I&#x27;m is the sum of F. Of all of the X. Values. .2101. So point to one no one. Yeah. Yeah. Yeah. Now let&#x27;s go to 100. So go back to my calculator, recall that sequence in command. That generates all of my in points. Now instead of using 50, I&#x27;m going to increment by 1/100. So that&#x27;s my first rectangle And I increments by 1 100. Those are my list of 100 x values. And now I just need to figure out well now the width of each of these is one over 100. Yeah. And times the sum of F at all of these X values. .2050. .205.0 yeah. Yeah. So what we can see here is each of these areas, as I increase the number of rectangles, the number is going down, they say, guess what you think this is? Okay. So if you already guess, you know, is it going down? Um It makes a big step when I go from 10 to 30. And then as I add, rectangles is making smaller and smaller steps. So you could guess that the exact value Is maybe .20. Okay, by the way, it&#x27;s moving there so you don&#x27;t know that you&#x27;ll find out in your subsequent section. That that&#x27;s how you figure that out. But I would make a guess. It looks like maybe this is converging to 1/5 or point to.

Robert Daugherty · Answer

Yeah. Uh huh. So problem 5 11. So we&#x27;re dealing with airy approximation using rectangles and Riemann sums. What we&#x27;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&#x27;s the same as an upper some. And if I have a rectangle and I use the right in point, that&#x27;s going to be a lower some. Okay, so what we&#x27;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&#x27;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&#x27;m always getting an upper some. So let&#x27;s just go ahead. We&#x27;re going to do let&#x27;s do all of the left, which are the upper. And then we&#x27;ll move on to all of the right to do that. So if I&#x27;m doing all of the left sums or uh Let&#x27;s do I think I&#x27;ll do maybe I&#x27;ll do the case in equal 10 and then we&#x27;ll move on from there. So in the case of in equal 10, let&#x27;s go ahead and let&#x27;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&#x27;m using 10, so the width of the inter rose one. So if I&#x27;m using 10 I&#x27;m going to have the width of being 1/10. So I&#x27;m gonna go from 2.9 And my increments is going to be .1 and I&#x27;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&#x27;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&#x27;s just do this same thing while I&#x27;ve got the number 10 there if I were to use right in points. So in the right in points I don&#x27;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&#x27;m going to start at .1 And then I&#x27;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&#x27;s just move on and do that same thing with an end value of 30. So with 30 let&#x27;s go back into our left some. So if I go back and do my left some so now I&#x27;m going to increment um Instead of by 1 10th that&#x27;s going to be 1/30. So I started zero I go up to and it&#x27;s not 00.9 It&#x27;s going to be 1 -1/30. That&#x27;s gonna be the value that I get. So this is going to give me 30 values and now I&#x27;ll just go in and say okay we&#x27;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&#x27;s when I use 1/30. Uh huh. As my width. Okay now let&#x27;s move that just to use the right side. So if I&#x27;m using the right side what changes? Well if I&#x27;m using the right side when I create that sequence that can that tells me the boundary of the rectangle. I don&#x27;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&#x27;m going to do is looking at 1/50. So if I look at 1/50 all that&#x27;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&#x27;s go to do the left some yeah Right here so I&#x27;m starting at zero But now I&#x27;m going up to 1 -1 50 and I&#x27;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&#x27;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.

Jacob Steele · Answer

Okay, So in this question, we are given a function over a specific interval. Eight were adds to calculate the right Riemann. Some given four different numbers of seven rolls. So I&#x27;ve got the answers, written it and table over here on the left. Ah, let question asked for And, um what weekend? How we need to go get a beach value as we first start off by getting Delta X, uh, which is our range divided by our number of seven rolls. And then we need funding, um, our endpoint values for each sub in Rome my okay, are XK values. So X cables, a beginning of the first sub roll, plus k times delta X, the length of each step in the role and then lastly, way to plug in. So we&#x27;re gonna get expression for that rather than an exact value. We&#x27;re going to keep ks a, um, variable there, and then we&#x27;re gonna plug that expression into the function that we were given. Um f so that&#x27;s FM XK. Ah, and we&#x27;re gonna sum that up from one to end on end while also multiplying by Delta X. So you do that and you&#x27;re gonna need your calculator to do that. Um, then you will be able to get the answers on left here in the table and impact be asked us a boat. Um, the approach, the value Reitman some as, and the number of seven year olds purchase infinity. Well, that&#x27;s even see, um, it&#x27;s it&#x27;s not changing. Our values were all 3.14159 approximately. And it actually keeps going because as I&#x27;m sure, you guys all realized the opening digits of pi. So, um, as an approaches infinity, um, the answer won&#x27;t change. The area under curve will continue to be, uh, by

Jacob Steele · Answer

All right. So this question, we are asked to estimate the area under the curve using the left green and some main point streaming some and and right Riemann, some methods. So let our function year after of X equals X squared, plus two number seven rolls of four. So the first step is gonna be to get the length of each step intervals. That&#x27;s gonna be our range between zero and two. Divided by the number seven year olds. Full r equals 0.5. So next we have to figure out thank zero. So that&#x27;s beginning us for seven role. That zero thanks. One is going to be eggs, Eero plus Delta X. So that&#x27;s 0.5 and so on and so forth to get next to experience for the next step, we&#x27;re gonna start with the left wing and some and we are going Teoh Um, yeah, I just use the form enough left even some so f of X, not plus fx. One does left next to the next three all that most blind by Delta X so ever. That&#x27;s not well. That&#x27;s the value of X squared post to when X equals zero. So That&#x27;s two radio and X. Um, if x one, these value of X squared plus two one x one equals 0.5. So that&#x27;s +212 and so on and so forth and multiply by Delta X being 0.5 and we get 5.75 And so this graph right here that&#x27;s a sketch of the left with them left Freeman Very rough sketch of what the left or even sums look like. So next we&#x27;re gonna do the midpoint calculation. Eso here, we&#x27;re gonna have to use different Um, we have to use the midpoint, stow endpoints for X values and celebrate where we doesn&#x27;t need those with a bar. So expire one is the halfway between zero and X one a k. It&#x27;s the midpoint of the first interval. So expert one is 0.25 because it&#x27;s not a zero and one is your wife. So then what we do is we plug those values onto our expression for the immediate point formula. So FX by her one, just like we did before. So that&#x27;s the value. Its value. That&#x27;s weird Post, too. When X equals 0125 So expired two is your 175 So f of X Part two is the value of X squared plus two one X equals zero points of five. And so you want to buy the that? Oh, you mad those things up By about zero point fire, you&#x27;re gonna get 6.625 And this is a little sketch with the right Riemann sums, which are a press. They are an overestimate, as we can see. So I have the midpoint reading some. So that&#x27;s gonna be our closest estimate. Lastly, you were going to do the right Riemann sons. So now we&#x27;re using just our exes again on our X bars in the formula for the right Riemann some. So we go back a beer, remember? Experiment 0.5 f of X wanted X squared plus two one X equals your heart. Five. We were golden o and multiply by Point X and we&#x27;re getting in our answer of 7.75

Joseph Lentino · Answer

so in this problem were given a function F of X is equal to the coastline of X. Now, the co sign of X is working on the interval from A to B, where a is zero and B is pi over two. Now we&#x27;re gonna split this up into multiple rectangles and the more rectangles I use, the closer my area approximation will become. So I&#x27;m gonna first simplicity&#x27;s sake I&#x27;m gonna use to I&#x27;m gonna use five and I&#x27;m gonna use 10 to get an approximation. Clearly, the more rectangles they use, the better approximation might. My answer will become and the pat you should go to see the pattern from these first examples. Now the with right, we&#x27;re gonna take pi over two and divide that by two. That&#x27;s gonna be my with now to get the heights, all of the heights, there&#x27;s only two of them. I&#x27;m gonna deal with some from 1 to 2 of the CO sign off. Now you have to be careful here. We&#x27;ve got high over two X over to, and when I enter that into my calculator, I&#x27;m going to get an approximate area of 1.3408 when I have five rectangles, all of the twos that air in the denominator here and here and here will end up becoming five. So I have pi over two, which is my and value have broken up into five pieces. Now I have five triangles and my co sign goes from Pi over two. And sorry the not next time using acts because expends up on your calculator. Um, pi over two n over five. When I approximate that value on my calculator, I&#x27;m going to get 1.1488 closer estimation to the actual area. Finally, I&#x27;m gonna take Pi over two and I&#x27;m gonna divide by 10. I&#x27;m going to get the summation from 1 to 10 co sign of pi over two n over 10. And when I plugged that in my calculator morning at 1.765 and this is

Frank Lin · Answer

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&#x27;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&#x27;re all going to be a positive height and I add up the area of all the rectangles, they&#x27;re all positive. That will approximate this area. So I&#x27;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&#x27;m gonna have 10 rectangles of with pi over 20. That will cover the interval from zero to pi or two. So we&#x27;re going to use the computer to make all of this happen. So um let&#x27;s get our calculator for all of this and let&#x27;s just figure out so the function was co sign of X. And the co sign of X. And I&#x27;m going to store that. That is my function F. Of X. I know that now F. Of zero should be one. Okay that&#x27;s what the coastline function does. Now let&#x27;s just create a sequence of numbers. And so I&#x27;m going to start at so in that first case remember I start at Pi over 20 if I&#x27;m using the right side. So pi over 20 is where I start. So let&#x27;s just do that real quick. That&#x27;s going to be sorry about that exit out of that. Um It&#x27;s going to be pie over 20 is the first right point. And then I&#x27;m going from there to hire her to. Yeah shit. Yeah And I&#x27;m doing in steps of pi over 20 So pi over 20. Yeah is how I&#x27;m stepping and I want to store all of that. I&#x27;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&#x27;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&#x27;s write that down. So 919403. That&#x27;s my approximation of the area. When I have 10 rectangles. Now let&#x27;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&#x27;m going to go over here and go back to the sequencing And I&#x27;m going to say now the width is Pi over 60. Yeah I&#x27;m going to pi over two and I&#x27;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&#x27;s go up to 50. So 50 the width is going to be pi over 100. So let&#x27;s just go back to our calculator and we&#x27;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&#x27;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&#x27;s go to 100 rectangles. The width is Pi over 200. So go back to my calculator, let&#x27;s just recall all of that command with all the typing And we&#x27;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&#x27;re saying guess where this is headed. I don&#x27;t know exactly. That&#x27;s why it&#x27;s a guess these numbers look like they&#x27;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&#x27;t know that, but just by trending by using more and more direct angles, that&#x27;s my guess is that is headed to one.

Problem

Some computer algebra systems have commands that …

Problem 10 Easy Difficulty

Answer

$R_{10}=0.919403, R_{30}=0.973592, R_{50}=0.98421$
$R_{100}=0.992125, \quad$ exact value guess: 1

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$R_{10}=0.919403, R_{30}=0.973592, R_{50}=0.98421$$R_{100}=0.992125, \quad$ exact value guess: 1

Calculus

Catherine R.

Anna Marie V.

Kayleah T.

Michael J.

Precalculus Review - Intro

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James StewartCalculus

Catherine R.

Anna Marie V.

Kayleah T.

Michael J.

$R_{10}=0.919403, R_{30}=0.973592, R_{50}=0.98421$
$R_{100}=0.992125, \quad$ exact value guess: 1

James Stewart

Calculus