a-sine-integral-by-riemann-sums-consider-the-integral-iint_0pi-2-sin-x-d-x-a-write-the-left-riemann-

Question

A sine integral by Riemann sums Consider the integral $I=\int_{0}^{\pi / 2} \sin x d x$
a. Write the left Riemann sum for $I$ with $n$ subintervals.
b. Show that $\lim _{	heta ightarrow 0} 	heta\left(\frac{\cos 	heta+\sin 	heta-1}{2(1-\cos 	heta)}ight)=1$
c. It is a fact that $\sum_{k=0}^{n-1} \sin \left(\frac{\pi k}{2 n}ight)=\frac{\cos \left(\frac{\pi}{2 n}ight)+\sin \left(\frac{\pi}{2 n}ight)-1}{2\left[1-\cos \left(\frac{\pi}{2 n}ight)ight]}$.
Use this fact and part (b) to evaluate $I$ by taking the limit of the Riemann sum as $n ightarrow \infty$.

Amit Srivastava · Accepted Answer

What does he do? Zero to buy. Gonna do? Sign off X dx. So the first part of the question say is that we need to find let human. So what is this? Red Freeman. So it is equal toe. Bring this last limit outside notice by upon to then there&#x27;s a sigma side for which we need to give a limits. Or does it good? Okay. Equal to one, two and minus one. So this end has to be able to play over here. Sign gay by a born 21 This is a left Truman B. But we need to show that limit if given by 20 and took by for yourself by this sign off by minus one. A phone do into one minus goes by. Is it going toe? Let us brood is equal. Do Once I&#x27;m taking this, it is equal toe limit. Oh, by by tending to zero limit, buy it. Ending those evil is by a born to cause off by a sign off by minus one upon one. Mindless goes by multiplying by one less course by up and one was closed by before is equal to limit off part ending to zero by about two. So this is they go to do one place caused by in due course off my breasts Sign off by minus one of on one minus go square But we know that science squared plus four squared is we&#x27;ll do one. So this value with Science Square by Liberto fired, ending to zero by a born doing Let us split this when blows goes but of born science squared by and do course by plus signed by minus one I have separate this question does able to limit off by tending to zero. So I get by a born signed by bringing one were to outside the whole in do one bless course, But this is one thing you have into limit off by attending new zero course off by minus one is one I was born signed by blows limit pending good zero signed by a boy saying to this have splendid the this this is one bodyguard. This will deployable this caused by a bone signed by players saying but won&#x27;t sign. But no, this is a weirdo. Now if you tendo zero this cause off zero is one Zito, this is 10 Indo zero exist this one. So this whole expression becomes one. This is one here, right? So this becomes too still. Limit off by attending below zero Awful. This is a little have Indal. This expression becomes one. This is one less what? And to do This limit becomes one. This is it will do a zero with a signed it was zero. So this isn&#x27;t the one. It&#x27;s on sale and we were given the left riven K zero two and minus one into sign off and k ah born do in. Is it going to do core so by a born doing Plus I know by a born two in minus one a born doing to one minor go also by a born jeweler So we need to prove But they so we need to use still So we need to use limit and to infinity to prove that this expression is one not from the previous part of the question We derive the result that the expression cause off by percent by minus one to win the one minus course by is a good one. Don&#x27;t getting is her angle is change. No, we have data. Is he going to buy upon doing so if endless, tending to infinity by standing 20 This expression left Rimal two is a griddle one.

Joseph Liao · Answer

All right, Number 100. We&#x27;re going to use, uh, Riemann sums to show that l sequel to the limit and purchasing Fendi foreign over and Oh, and tutorial. Nice. Ellen on equals Negative one support a Tolson note that in factorial and times and minus one and one is two and so on. And to use Ellen, the identity for the log function Now enough, a pose A wanna be And we want to show that else you go to the limit of and approaching effendi born over end some from one end of Ellen of K awareness L and is equal to the limit of an approaching effendi. One of her end. So from chemicals one end of Ellen of K. Over and all right, so how do we do this? Start given the l with the limited and approaching effendi of one over end Ellen of in Victoria, Monticello. Defend. I can show that this is equal to so basically fanning out the in factorial does it. Ah, to all the way down to one minus Ellen of and every comes. So we can rearrange it so that Pelin of and minus one separated using the log identity l and event. Okay, now we take we had earlier. Oh, and it&#x27;s been a purchase of funny one over and so mation firm chemicals. One toe end Ellen of K minus l it of ah, okay. This simplifies to had a purchase Infinity Pull the worn over an and you just multiply that and thank I&#x27;m just saying, um, its mission vehicles one. Okay, mine estimation Keiko&#x27;s one. And a lot of now, since these air both summations, we can fracture them together of Ellen of K ominous Allen of End. And that&#x27;s equal to the limit of and personal Cindy Warner for end. Okay, because one Keiko&#x27;s Warner&#x27;s l enough k over and And this was able to l in the beginning. In part B Part B wants us to identify the limit of some as ah, remain number for the some of the, you know grow from 0 to 1 of Elon of X and we gotta use integration by parts All right LTD and purse affinity one of her and for the some from one end of Ellen of K over and physical, too in Novo from 01 of Elon of X t X No. We evaluate the limit from the right from the left. No, this Yeah, value from the right as a approaches. Zero from the right. And a girl from eight. A one of Elon of x t x. Okay. In the end, Grove Ellen of X. It&#x27;s good X Ln X minus X This is Ah, it&#x27;s in the integration table in the back. Your book. Tomato one, this comes out to There&#x27;s a approaches. You&#x27;re from the right of on a good one minus a Ellen of a plus a put you and the one and a I guess it&#x27;s negative. One, therefore, is equal to negative one.

Brian Ketelobeter · Answer

Ladies and gentlemen, today we&#x27;re looking at 40 Problem number 47 from section 5.2, and and we&#x27;re asked to consider the function f of X here over the interval 01 and to compute or find a formula for the and three months, some and then, um, use that formula to compute the corresponding limit. So, um, to give a brief overview of what&#x27;s going on. Um, the idea is that you have, um, the interval here. So we have our interval. I, of course, in this case, I goes from 0 to 1 year. So, uh, 0 to 1 and, um, 0 to 1 and we 01 And you have the function which in this case, looks something like, um see it Maybe something like this. I mean, it&#x27;s kind of a quadratic looking beast, because I plant, um And you subdivide the interval here into partitions of length of the same length. So you petition it into n distinct partition. So, for instance, if n were, let&#x27;s say three, I would divide it into, um, I&#x27;ve divided from zero to, um 1/3, um, and from ah, 1/3 to um, from 1/3 to 2/3 and, of course, from 2/3 Shu um, 3/3. Andi, I used the 3/3 instead of one there because it helps. So it&#x27;s 3/3. Of course, that&#x27;s equal to one. But as a key is that in what we&#x27;re doing, it&#x27;s more useful to do it like that. Okay. And so we divided into intervals into intervals. Um, and each of these guys, of course, have the same length. Same with. And to denote that we do is we take We talk about the the delta here, Um, we&#x27;ll call it the Delta X K. Or, um, if we&#x27;re if we want to use maybe Delta X two, if you will. Um, Delta X two, um is equal to acts to minus x one. So, of course, X two is 2/3 X one is 1/3. So in this case, it would be equal to one, uh, divided by three. Right. Um and that&#x27;s her. So I&#x27;m just gonna move this out of the way. Hey, here. Okay, um, and now and now, um, so each of these have the delta acts to is the width of this. And now what? It also has is a the corresponding height. So we form, you know, we form a rectangle. Um, we formed the respective rectangles here. And these rectangles, of course. And they have this here corresponds to the of the West here. And this here. This part corresponds, of course, to the height and now the height. Because we&#x27;re using this end point here, the what we call the right and point the height is equal to, um it is equal to, uh it is equal to half off, um, 2/3. Which, if we go back to our function f here, of course, is X plus X squared. So this is equal to, um, you see. Ah. So can I do this? Uh, this is equal to, um Well, it&#x27;s two divided by three plus, um, or over nine for over nine. Um, which is, of course, equal to, uh, c six. Um, that should be 10. Overnight, I think. 10 over nine. Okay. And so the area of this block and we&#x27;ll refer to the block years are, um, trying. The was a block. We&#x27;ll call this block R K. So we&#x27;ll call his block. Um, the area of our two. Okay, um, which is equal to than the West Times. The height, um, is equal to Ah, one over, uh, three times. Um, pen over nine. So, times books, times 10 over NY. So it&#x27;s, of course, pan over 27. Because the 10/9 comes from this this height. And the West, of course, is 1/3. If I If I computed, I compute b proper height there. Let&#x27;s see. That&#x27;s, um, multiplying that by three. You get six over nine plus 4/9. Yeah, it&#x27;s 10 overnight. So, um, this would be my area of this block. Um, but now the thing is is in general. So in the end, when I&#x27;m going to the more general case, um, I&#x27;m going to be computing the area of R K. So now instead of are just what? Breaking it into three blocks. Well, now I&#x27;m breaking. I&#x27;m replacing two year by K, but I&#x27;m doing the exact same thing. And the wishes of the, uh, this becomes our end. So now what? I&#x27;m sorry. I&#x27;m made a little mistake there. I&#x27;m Blake breaking the the interval into N, um, sub intervals of the exact same length So, uh, this would become to over, uh and and of course, we&#x27;d have maybe, ah, instead of, um, we&#x27;d have and minus one over, and And, uh, this would be, uh, and over and in each of these beach of these guys have a length of one over end. So the delta here would become won over an instead of 1/3. Um, and the the the as the, uh the height at the height of that would be okay. So remember that the end point now is not and, I mean, not too would have to be. K is K over end K over end of which, of course, is equal to hey, over n plus k squared a squared over, um, and squared. And at this point, I really don&#x27;t want to simplify it any further. This is actually what I want is just be I don&#x27;t need to do anything more. I don&#x27;t need to simplify any further, so my, um, my area then the respective area, of course, is, though, is the with which is our end, remember? And the, um, the be height, um, which, which, as I say, is hey, over and times K over plus k squared over and squared eso if I simplify this or if I get rid of the one over an in front, I end up with this guy here. So this is the the area of beach of the individual blocks. Now, the key here is what I do next. Okay, Um, what I do next is I is remember, um, I&#x27;m not computing just the area of one of these. I&#x27;m actually computing the area off all end of them. So, in fact, what I want to do is take the some of them. So the s and here is equal to is Thea some of those areas. So it&#x27;s equal to the S. And the S end is equal to now the sum. That&#x27;s what this guy is to some all of them added together, going from of course, Ah, hey Equals to one, two, um, to end, uh, two to end of the of the area&#x27;s off be our case. So the areas of the arcades be areas off the r. K. Um, and of course, so the area they are case, of course, we know what that formula is. We have that here. So we have this respective area, we just It&#x27;s actually equal to that. So are our formula for the sum then is actually given by this guy right here. This is This is the what I would call the first formula for the area because we can actually do better than this. And that&#x27;s what we&#x27;ll do on the next page. So we&#x27;re gonna take this area now, Um, and we&#x27;re gonna do some more with it, so let&#x27;s move forward. Okay, so we know. See? Okay, so we know. Um okay, So what do we know? Okay, so we know that, um, s and worse. Um, Okay, so we know, uh, s and is equal to, um this sum. I&#x27;m sorry. Uh, I I should have Okay. I made a mistake here. Sorry. Uh, so it&#x27;s equal to this. I&#x27;m sorry. It&#x27;s equal to this. Some, um, going from, uh, hey. Equals one. Shoo in terms is okay too, And okay, so it goes from one to end off this some. Okay? And I can actually rewrite this in a simpler form. While I don&#x27;t know, simpler is not quite the right answer, but basically what I can do. It&#x27;s separate this, um, into two distinct sums. Um, has, uh, doing this Just factoring out the first and second some. So now I have the, uh oh, I I just realized something. I made a mistake on the 2nd 1 I&#x27;m sorry. I&#x27;ll fix it here in just a second once I once I get this down, Um, yeah, K equals one to, um and here. Sorry. Sorry about that. Okay. Here you go, too. Um, yeah. So this I just noticed the mistake. Sorry, but this should actually be too, because I divided by the, um, back there. Um, if we go one back, uh, write ups, that&#x27;s not the right one. Um, it should have been cube here as I multiplied by the one over and and I forgot to Don&#x27;t do it there. I just noticed that. Okay, um, right. So now this is our formula for s and okay. And, um, now, if I factor So, um, now the thing is is that Ah, if I take this, um, I take the, uh, one over and squared out, So I take one over and squared, uh, been works. I&#x27;m sorry if I divide this guy. Bye and squared Here. Okay, Bye. Okay, by, um, So I divide this guy by and squared and this guy here, I&#x27;m gonna factor out the in cube here. So I get. And to the third, um, Now what I have are I have to sums. And I should move this, uh, some over a little bit. So now what I have are two sums. Um, this is just, uh, over K, and this is over k squared. And it turns out that there&#x27;s formulas for both of these. Um, this this the 1st 1 is Beek Walter. Yeah, of course. There&#x27;s nice formulas for it. It&#x27;s equal to, um c Come on. It is. And arms and plus one, um, he divided by two. And the 2nd 1 of course, is, uh, in times and plus one times to end plus one and all of this be divided by six. Um so if we then decide that one by the squared and this one by the n cubed. Okay. Um, no, I can simplify down a little bit. Okay. Um, fat getting rid of the and in the numerator here, I&#x27;m getting rid of one of the powers of the end in the denominator. I&#x27;m similarly same. Same trick over here. I&#x27;m dropping in there. Drop one of ours here. It&#x27;s okay. Come on. Okay. It just doesn&#x27;t wanna work, huh? Don&#x27;t do what I wanted to do. I&#x27;m very sorry, folks. Yeah, I just want I want to get to this one. Change that to a square, sometimes a software. This is so terrible. Um, yeah. So, uh, okay. And of course I&#x27;m adding here. So I have some here on now. I can simplify, of course. Further. Um, just expected dividing through the first term. I get, uh, one one divided by two. Um, plus ah, us, um, one divided by two end are, uh yeah. Okay. Me? I&#x27;m doing it one divided by two end. And then, um, the other one if I used the, uh, if I If I, um, use the foil master, I get to and squared plus three and plus one, so that helps me get the proper. Um, that helps three simplify it properly and what I get. So then down here, I get, um I get 1/3 um, plus, uh, one over to end, uh, plus 1/6 and to the second. You know, if I do the same thing here instead of this funky way of doing it, then I get, um, 1/2 plus, um, plus one over to end and, um, one over to end. And this is my, um this is my s n. So this is what my s and is equal to is this guy. And then now I can, of course, simplify it further by heading the 1/2 in the is the 1/3 together to get, uh, s and equal to 36 to 6, 5/6. Plus, then, um well, one are just one over an I s one over an, um plus one over six and two the second. This is my s. And now, um, this is my ass end. So finally, what we have to do then, is used this formula too, and look at the limits. So now we have to do is we actually have to? The next step is to look at the limit. A Zen goes to infinity. So in particular than we have to take is the limit as ah c n to infinity of the s and And of course, this is equal to the limit, As in, uh, to infinity. Um, off this guy here. So we just is that we take that don&#x27;t troops. Maybe I&#x27;d bring this down here. Um, the limit is angles to infinity of this some. And we note that, of course, these two fellows goes to zero discussed zero, and that goes to zero. And so the only guy remaining is the five or six. And of course, that&#x27;s what being up. So yeah, Quest five over since as our answer. And that is the, um the the limit. That&#x27;s what we want it. So after all that work, folks, how that&#x27;s the answer, Thank you very much and have a nice thing.

Brian Ketelobeter · Answer

Gentlemen, this is a problem. Number 43 from section 5.2. And essentially, the problem is discussing reminds sums, and we want to over some interval, um, forgiven function. And we want Thio. Find a formula for the and three mon some and then take the limit. Okay, so our, um our function here today is the, um is equal to f of X R f of X equals to one minus x squared. Okay. And now, to give you a brief idea of, but we have to do our, um Let&#x27;s take a look at the graph here, so I&#x27;ll draw the, um uh, so I&#x27;m gonna draw the grab ups. Come on. Okay. So I&#x27;m gonna draw the graph of the function here. Looks like it looks like this. Um, so we&#x27;ll take the interval like this. Um, and, uh, course, this is, um are interval down here from 0 to 1. Um, and our function looks roughly like, um, this here. Okay. Um, no. So what we want to do is we break the on the end. Um, on the end, some Ah, we break the interval into end equivalent, um, sections or ah, absolve the intervals of length one over end eso. In this case, it goes something like this. So, for instance, if we&#x27;re, um, looking at, ah, the 4th 1 are made Maybe the 5th 1 here, we break this into five different subsets. So this is, um this is won over five hoops, so this is sorry about that. Okay, so we have, um um one fist. Um, this is a ah, boy. Come on. Okay. Um uh, this is two fists, two fists and so on street office and four fists that&#x27;s so on and so forth. So this is, uh this saree here is, uh, 4/5. And, um, this one in the middle is street office. And, um, then the, uh So you mean, uh and then, um, at each, uh, beach endpoint. What we&#x27;re doing is we&#x27;re forming. Um, we&#x27;re forming a, uh ah. So you you form a box here, Basically a rectangle. Um, and, uh, this here we form rectangles with this, like this, and this s so we&#x27;re using what they call the right hand endpoint for the boxes. So that means that the height, if you will so are with, of course, is the, um are with here is the 1/5 or with here, of course, is going to be the one over. And, um, and our height here then will be, uh, this This our height, if you will, is, um the, uh, ass off the height is half of in this case, it would be half up 2/5. So it would be our height would be s of, uh, to this. So when you calculate the area of this box so the air will call this box, maybe call the box? Uh, A In this case, it would be, I guess, a two. Okay, So this box here well, just name it as a too. Okay. Now a to, um Izzy s Sorry. The area. Um, yeah. Maybe I should have called it are two instead of a two. Yes, because our two would have been a better letter, I guess Are, too. So the area of our two from the area of and chu, um, is equal to Well, it is half of 2/5. Um, to this, uh, multiplied by the, uh, the with, um, which, of course, is one over. Yeah, and another way, which is kind of the ways that we&#x27;re using for this, um is the the way that they use in our the way that the reminds some set it up is that this is equal to, um this is equal to, uh yeah, that&#x27;s ah f x will sit roll call it F k um, times the delta dull Tahir uh, x k and what? Delta what this is referring to. So the delta of X k but it&#x27;s actually talking about is what is called the increment, um, lengths. So if you notice so in particular here, the increment length is the is the difference between the two end points. So in particular it&#x27;s, um it would be, in this case, two fists minus 1/5 or otherwise x k minus x k plus one or something like that. And in particular here, reusing the right hand and points. So the right hand endpoints refer to the end point on, I guess the quarter right hand side course, if we&#x27;re using the left and end point is then, um, the boxes would look a little. The rectangles would look a little different in that instead of the one we have. It would go, Uh, they&#x27;re rectangles would look like So. So the the area of the given rectangles would be a little different. In that case, obviously, they would be larger and here. But, um, the the idea, at least in these remind sums, is depending on the point to you two, you&#x27;re going to get a different areas for each some. And, ah, then the, um is the ends Riemann some is given by or these two formula that they that they use in the book looks like this. Um, it&#x27;s ah ass And, um is equal to, um, the Sigma notation, of course. Ah. Which, um, you know, looks, uh, like this guy. So, um, as, uh, yeah, says, um, Sigma of half of X. Okay, um, times the as I say the delta the dealt er excuse me, Delta X, k and crucially, here, um, the ah, end this and be in this problem. Um, were our increments are gonna be of the same length. So in each case in this problem, um, the Delta X case will, um, be equal to one over end. Okay, So are you on this problem anyhow? And other problems will be a little different, but in this problem, the increments will always be of length one over and And the, um, f x case are always found by the right end point. So, um, are are, But we have to do is first find a closed foot what they call a closed form or s in here, and then use that to calculate the limit. So, um, with this in mind and let&#x27;s begin. So, uh, first off, um, the ah for you given ends of just an arbitrary number. And, um, we get that r s in here, uh, our s and, um, again the, um Oh, the increments. So since we started, So So it starts at, um, Kay. Since we&#x27;re using this point are for Kes. Our first rectangle will go from if we replace instead of five here. If we replace this by end, Goethe one over and you over. And, um, you know, dot, dot dot This would be maybe instead of three over. And, um, this would be ah, Villa A. This would be at three over, and and then, you know, one, of course, is is is, um is an over end. Okay, so the one before it would be Ah, um ah. Let&#x27;s just say how do you want asses and minus one over and yeah, it&#x27;s getting a little tight here, Um, but and minus one over in and ah, and minus two over and all the way down to ah, and then three over and two over. And so B s. And, um So the first thing we should try and figure out is what is our, um what does our index k go over eso in this case, Kay goes from, uh, que goes from one because we&#x27;re using this side, remember? So it&#x27;s k equals toe one. Um, k equals to 12 And like I said, uh, it ends at one. So in this case, it&#x27;s K equals one, two, um, in here So it&#x27;s K equals one to end. Um, and then the f x k refers to the this right hand endpoint. So that would be, um it would be f off. It would be af off, uh, K over end and then times the length, which, of course, is, uh, one over. And so I&#x27;m gonna simplify this now just to eliminate it like this I&#x27;m gonna just write it as af of K over and, um, divided by Yes. Okay. It&#x27;s simple. Find that here. Ah. And now, in this case, of course, the function I&#x27;m using is one minus x over two. So I&#x27;m just going to replace af of axe over and buy one minus X over too. Um so let me just do that here in this form. You here. So I&#x27;m just gonna replace this by, um, this becomes, uh, one minus. Uh, Now it&#x27;s k over. And so it&#x27;s, uh, que over and square. Okay. And, uh, okay. And now we want to do, of course, is continue to simplify this now. Ah, so you&#x27;ll notice that your summing from one end. So if you&#x27;re something, um, from want and ones, then you&#x27;re going to get in there. But then you&#x27;re dividing by ends. You&#x27;re going to get one. So this is equal to, um this is equal to one and then minus um uh, is this some? That&#x27;s some again here. So it&#x27;s going to be minus whips. Uh, Minus again? Yeah. Is the, uh, um K squared? Uh, we&#x27;re gonna get troops, huh? K squared now. Provided by and square times end gives us it&#x27;s divided by n to the herd. Um, provided by, uh, case by the end square times end, which is one over our wishes and cube. And again, Now, the sum goes from K equals one. Uh, so this some goes from I want to end here. So this is our some and, um, Eagles want to end. So now we&#x27;re now we&#x27;ve calculated Are we calculate ups. Okay, so we&#x27;ve calculated this is ah, um, this is our least our first form. You. LF you Well, this is our first formula for for, uh and you&#x27;re sorry for s And so this will label as maybe, uh, f one. Okay, so we&#x27;ll call this F one. Okay, so now with f one in mind, um, now we have to move forward here. Uh, we can&#x27;t. We need to use a formula. There&#x27;s a nice formula to make this easier. So So we moved to the next stage, and, um, we want to find a formula for the end. Okay? Eso We use the formulas that, uh, this guy here, which is, um uh okay, So we used a this formula, which is actually in your book, So, um, but I&#x27;m just repeating it Here, Um, this formula here it says that the squares So the, um a equals one. Ah, to end here. Ah, wake up. Come on. Sometimes it&#x27;s gonna be so frustrating. Okay. S O okay. I&#x27;m sorry. I see what the problem is. It&#x27;s too close to the top here. So in other words, it goes from K equals one hand. I&#x27;m sorry. It&#x27;s just too close to the top. K equals one end of this guy is equal to this end and plus one time to n plus 1/6. So we use that. So then, um, from F one. So from f one, we get, um, from F one, we get that, uh, s and is an equal to one. Um, minus. Now, remember, we have. So, uh, we have this, um uh, the the F one. Oops. Uh, So the F one was this case, This some divided by n cubed. So maybe maybe what I should do is actually factor out this and cube here. Ah, see? So if I factor out the end cube, uh, you&#x27;ll see that this some here is actually the, uh, and shoot. Um, so this sum and now it&#x27;s one minus the sky. So it&#x27;s so this here we have from, uh, this This formula here gives us one minus. So, um, it&#x27;s one minus this. See, in n plus one, sometimes two and plus one. Ah. And now it&#x27;s divided by the six times the end cube from the formula week from what we had before. So this is then six thistles. Six and to the third. Okay. Ah. And now, um, what we can do is we can expand this out. Well, actually, I prefer to just cut to just cancel the end here, and, um And that would make the denominator ace, too. So instead of that, I get and squared. Um, And now what I get is that this is then equal to one. Um, minus. I&#x27;m gonna write out school. I&#x27;m gonna use the foil method to compute the numerator of that fraction there. So I get, um, to and squared, uh, plus three and plus one. And, of course, it&#x27;s now divided by, uh, six and square. So we get here, um, six and square, dividing each, uh, each fraction. Uh, I&#x27;ll be just sort of standing the given fractions here. Uh, me. And so what does this, Uh, what does this tell us? Well, we have then is ah, we can go through then. And in particular, I want to simplify this down, so you&#x27;ll notice that the end squares cancel in this 1st 1 So I lose the end squares here, and this becomes one over free in the 2nd 1 The end cancels with this one, and I get Well, it doesn&#x27;t matter here, but the final one is just 1/6 and square, so I can simplify this down further. One minus 1/3. Ah. Then gives me, uh Well, let&#x27;s let&#x27;s just move on to the next one. So So one minus 1/3 would give me 2/3 here and then plus this of their stuff. So, in other words, I get, um that my s and by s and eyes equal to, um you see, it&#x27;s equal to 2/3. Two over three. Plus, these other guys, which are, um, or I should say, minus one over, uh, to end. And then it should be minus. Sorry. Um um, minus one over to end and then, um, minus 1/6 n squared 1/6 and squared. And so this is the formula for my s end for my aunt. Uh, sorry. My ends reminds some. OK, And now what I want to do is just compute the limits. In other words. Now what I want to do is just look at the limit. Um, as an approaches infinity, um, of s and and, of course, uh, this is, uh, is equal to, um the limit is end goes to infinity of that, you know, some above, which is to over three minus one over to end. Um, minus one over six and two. The second. Ah. And now, hopefully, you know your limits Well enough to know that this these two terms here, both of these two terms are gonna go to zero. So this is going to go to zero. And this will also go to zero. And what we&#x27;re end up. What we end up with is then, um, just to third. So we end up with 2/3, and that is our answer. Um, eyes 2/3. And that&#x27;s it for Ah, this problem, folks, it&#x27;s a bit of a long, long problem. It&#x27;s, ah, number of steps, but, uh, that&#x27;s it. Uh, thank you very much, folks. And have a nice day.

Willis James · Answer

So this lesson in this lesson we use the room on some to find the area in there. That calf X plus is quiet from that one. So essentially, this is what we are finding. Okay, So the concept is that we have an interval from level 21 There&#x27;s a cab over here. No, if you divide the the area, the overall area too. So money with heaven into so many paths having the same with we would be ableto estimate that the area&#x27;s off that rectangles. Then we some all of them up. So the truck so we could have the rectangles like this. Okay, as it continues. So the length off each each with is one minus zero uba. And because you&#x27;re dividing the whole thing so many times, all right, and that is equal to one over. So what it means is that if this is there the length from 0 to 1, each of them becomes one over n. So the right part, this one of the n Then if then if this is the next division, this becomes one of another. This becomes two times one wine that is to end. This becomes three times one over N That is terrible. And and on and on until we have an over. And that is a call to what? Yeah. Okay, So what? Each off this point, we will take a sample point seeking that is the most right. The very right off each interval. So as you can see, it goes from one of the end to the end. Three of the end and all that. So it goes from key, but And for each of the cave interval, that will take the first interval. Second interval fed fourth and all that and that is the right side off the interval. Okay, so the Roman, the Romans some they were months, um is some from 1 to 10 for F or seek a this point very off evaluated. Then times the length off the interval, I&#x27;ll stand approaches in Trinity. A very huge end. Okay, so this is equal to the Yeah. Okay. So as we already know, the f off X waas equal to X plus x squared. So in place off ex you put key Seiki were there And that becomes key on n less que en or squared as we have here, then. Times one own n Because you&#x27;ve seen the length of the interval at that. So as an approaches infinity. Okay, lets go on. So now let&#x27;s deal with Onley. Does some than we come to the limit. Okay, so car. So Cuba en last key over K squared over and squared than multiplying bad. Less identify are constants one about and it&#x27;s a constant cuba in. Okay, then you also have one of and again just yeah, splitting them. Yeah, okay. Squared uba, um squared. Okay. From 1 to 10. All right, so we can take this. Came back there because the case a constant So we have vulnerable and squared then key from one to end. She then this comes back because a constant the constant is no affected by the the summation. Okay, so we can bring it back to be a multiplier. I already have one of the Ansel If one of the and squired Joyce, it&#x27;s that is this us one uber and squared. If something like that, if a joints it becomes one of the n to the path three. Okay, and you still have the submission? Oh, he squared. Okay, so having this that is K goes to one and is the heaven one Last two last three last four last five on an own class in and this is and then the last one all over to that is the sum. Okay, if you you summed them and also happen this way. We&#x27;re dealing with us some fast. Then this is heaven que codes to want em. 71 Last four last nine last 16. So yeah. So instead of 11. 12345 you have That&#x27;s the squared off them. So we have 14 that is two times two. Yeah. Three times story that is 94 times 4. 16. 5 times five is 25. Known unknown. You have, um, the last one and squared. Okay. And that ISS then and last one to end last one. All of us. Six. That iss the submission off all of them So we can substitute this into that and this into that. Okay, so it means that the whole thing one of the and squired some mention k one to end. Okay, last one of our end to the path. Three summation, K one to end off K squared would be equal to Okay, let&#x27;s do is yes. So that they would align on over and with one movie ends squared. Then in place of all this output in and last one all over to put over Sorry, the name place off the horse. And I put Onda last one to end last one. All of us six. Okay, so as you can see this to off the ends here and this one So we end up heaven one over and then one of the AIDS Council&#x27;s this one. So let&#x27;s write it appropriately. Okay, then. Yeah. And last one, two n plus one all over. So this end councils one of these. We have six and squared. Well, okay. Moment. This is what we have. I never Okay, okay. The next time we can do is to simplify. We have at this point, if you multiply the e f. And to that, you have to unscrew aired. Then you have end to this. You have n okay. And you have one multiply in that you have one multiply in one, all of us, six and squared. Okay, So what is? And last 1 to 10, then we have to end squared Last three and last one, all of our sixth and squared. Okay, so all of these is without the limit, if not taking the the limits. All right? Yeah. So, Heaven, this has an approaches in 20 K 1 for banned for a four C key. The X would be called to the limit as an approaches infinity off in class 1/2 n gig. Just splitting them. Okay. Okay. So one thing is that if the end is very huge, I didn&#x27;t want to. It will know. Make any sense, Would know would not change it. Let me put it that way. For example, if e f 200,000 and you add 12 with divided by some other number me be no change so much as you had no add one to it. Okay, so if the number is very huge, the number is very. He added a small number to it. Almost make it off. No effect. So it means that at that point, the only significant numbers or only significant path off this will be the end and the two and other down. So means that this is a quote half because at this point would ignore the one doing this and will cross out the end. Then you only have So you would know this whole paths that you have half right? Then that&#x27;s bad. Is that if n is very huge, all of this will, no matter. Adelaide to to end squared would make it as if we have no added than your thing less imaginary off three bilion than you ask. You are multiplying it by itself again. I didn&#x27;t want to it or even three billion to it will know make any it will not change is significant. Least what means that there&#x27;s a very high value would the only significant path will be to end to the power to Then that crosses. This will cross that. Ignoring this pod would have two on six. And that would be one on three. Okay, okay. So that ISS one on three probably talk about two and six. Then we can simplify it further as one on three. So here we have six. Two goes into sixth, three times 30 goes into 62 times and that is five on six. So the idea under the cab is 506 squired units. All right, Thanks for your time.

Brittany Knowlton · Answer

So this questions asking has to find a formula for the Riemann, some obtained by divine and rolled and equal stuff minerals, using the right hand and point reach ck and then taking a limit of these sums and approaches infinity in order to obtain the area underneath the curves. So first thing, any nose is exactly what is my formula for the room and sons and were given that that is equal to case able to wonder and of my function evaluated at ck times, Delta X The first thing and you find is exactly what my Delta X and CK is. So my Delta X is nearly the width of each rectangle, and I find that by subtracting my end points and dividing by my number seven rules, which in this case is and so it&#x27;s just gonna equal one over end in my CK, which is the X value of where my high school me evaluated at. That&#x27;s just going to be K times My Delta X. It&#x27;s just going to K over end. So then plugging that and see my son whenever some from K equals one to end of my function evaluated at CK so ck wass k over end. So it&#x27;s going to be three times K over and plus two times K over and squared times Delta X, which here is just one over end. So it orange to calculate what this sum is equal to. And when it just started distributing and figuring out exactly what we have inside here. Caves on end. This is just three K over end plus where I&#x27;m screwing a fraction. Both sides are Tom Bond. Get squared. Someone have to k squared over in suede. I was wonderin. Then I went to distribute that one over end. Soon I get three K over and squared plus two K squared or in cubes. Hey, it&#x27;s Natalie and start using our summation rules. I can use my son role to separate this into different to different sons, and then also, I can use my constant multiple rule. So here, rebelling from Kate that kay So this and is actually just going to be a constant years. I can take my three over and squared and two over and cute and move them to the front of my sums. So when I do that, I have three over and squared from some of k equals wonder and of K plus two over and cute comes a sum of K equals one to end que ce qu&#x27;on And in the end, we have a formula for the sun. Okay, in the sum of K squared. So this first one is just going to be three over and squared times, Um, and shins and plus one, two, this was going to shoot over and cubes times and times n plus one comes to and plus one all over six And then I just do some last simplifications. So here in animal, cancel out. So I get three over and plus one over to N and cancelled out here suddenly and squared to cancel out so much with n plus one Tim and plus one over three in suede. So last part of that question is asking us to find limits and approaches infinity of our son. So what have Lim sent Purchase and Fendi of here? Do you have two different things? Some so I can buy limit laws. Just take the limit of this first termine, then the limit of the second term. So have limit of three times and plus one over to N plus A limit s and approaches infinity. Of course, one attempts to U N plus one all over three and squared. So wanna go ahead and distribute my top? So I&#x27;m gonna have three and plus three over to n Muslim is an approaches infinity of two, one squared plus three and plus one or three and squared. He really haven&#x27;t as Anis Tane alone as enemy is approaching infinity. So since going to and bendy are lots of things that we can look at, which top or bottom has the greatest exponents? So here, if two is a great expert to his grace explode. But that was bigger than down here in this same gun. Chimp Indy, The bigger explanations in the denominator it would go to zero. But when they both have the same largest exponents on the variable so here that both Havel one here both have a too the limits just going to equal the fraction of their coefficients. So our answer here is going to be three halves plus two thirds which is equal to thirteen six

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Alternate definitions of means Consider the funct…

Problem 63 Hard Difficulty

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a. $\frac{\pi}{2 n} \sum_{k=0}^{n-1} \sin \frac{k \pi}{2 n}$
b. 1
c. $I=\int_{0}^{\pi / 2} \sin x d x=1$

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William Briggs, Lyle Cochran, Bernard Gillett

Calculus for Scientists and Engineers: Early Transcendental

Heather Z.

Kayleah T.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

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a. $\frac{\pi}{2 n} \sum_{k=0}^{n-1} \sin \frac{k \pi}{2 n}$b. 1c. $I=\int_{0}^{\pi / 2} \sin x d x=1$

Calculus for Scientists and Engineers: Early Transcendental

Heather Z.

Kayleah T.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

William Briggs, Lyle Cochran, Bernard Gillett Calculus for Scientists and Engineers: Early Transcendental

Heather Z.

Kayleah T.

Samuel H.

Michael J.

a. $\frac{\pi}{2 n} \sum_{k=0}^{n-1} \sin \frac{k \pi}{2 n}$
b. 1
c. $I=\int_{0}^{\pi / 2} \sin x d x=1$

William Briggs, Lyle Cochran, Bernard Gillett

Calculus for Scientists and Engineers: Early Transcendental