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Question

Using a Riemann Sum Determine
$$\lim _{n \rightarrow \infty} \frac{1}{n^{3}}\left(1^{2}+2^{2}+3^{2}+\cdots+n^{2}\right)$$
by using an appropriate Riemann sum.

Yaw Asomani · Accepted Answer

So when you use an appropriate Riemann some toe fight with this 11 over and cute plus one squared plus two squared, uh, post three squared buzz and squared. So, uh, you know, the function is you can see that is ah, it&#x27;s miserable Squares right. One squared, plus two squared. Plus this word so we can let every thanks b x squared. And then we have limit between zero and one. We&#x27;re trying to determine the appropriate remind some for this one. Then, uh, you know, removed. Cem is, uh, summation. I from one toe end F o c. I write change in X I in your necks. So, uh, what is the change in X change in X? It&#x27;s how it should be minus a over end. And so, what is our be Arby&#x27;s one, right? And our eight zeros. So it&#x27;s just one of red. And what is C I c. I is a plus changing X. I write a plus change in X I a 00 close. So we&#x27;re just gonna write changing eggs, genuine exits, one over end. And then I saw the Riemann some. It&#x27;s just gonna be summation. I from one toe end f off egg. Ever see? I see eyes now, uh, one of her and I rises one over and I times changing ex change in X is just one over. And so what is one ever off one over? And I means I will write c s here. I&#x27;m gonna put one over, and I So this is just estimation I from one toe end one and I squared one over in, right? And what is that? That is summation. I from one toe end. I squared right over n squared times, one over end. And that finally is gonna be this, right? That one on a different page. So summation I from one toe end, f c I changing X is so fine that this one is gonna be summation. I&#x27;m from one to end, right. I squared R and cubic because this one is one of the end times One of route times out and squared one of her and square. That is one of her in cubic and us in here. I could bring it out, right? I could bring it out. So when I bring it out that I have won over and cubic in summation, I won, and then I squid. So you see that this is exactly the same as the 1st 1 here. Can you see that? This one of her end cubic. Is this one over in Cuba here. And then this whole thing here, everything here in the parentheses is the summation of ice. Where right, so that is the appropriate relearn some. So if you have this one, this is gonna be one over n cubic. Now, what is one of her I squared is just n n plus one to end plus one over six, right? Yeah. Now that is Jezzie re months. Um, Now we have to find the limit as an approaches infinity of the three months, Um, so the limit as an approaches infinity of the Riemann. So, um, which is F o c I that It&#x27;s just the limit as an approaches Infinity of one and cubic right. And close. One two U. N. Plus one over six. You see that? So we have this one, then, uh, this end here is gonna take one of these. So this is gonna be this is gonna be a limit and approaches Infinity. Ah, endless one to U N plus one over six. And squid now have toe expand this read this current disease. So what is expansion of this one in the red might limit here? First expansion of this one is gonna be to end squared, right? Because no, this is three. And and then close one that is expansion. And then over six. And suede. This is the same as now. We&#x27;re gonna just, uh, distribute the denominator to all the numerator. So this is gonna be a limit and approaches Infinity now to n squared, divided by six n squared. That&#x27;s just one of the three, right? Then purse three and divided by six. And swear that it&#x27;s just one over to end, then plus one over six and swear. So the limit as an approaches infinity of this one, anything that has and a denominator is gonna go to zero. So finally, what you&#x27;re gonna have is just one over three. So that is the limit won over three. So one over three. Okay? I cannot Yeah,

Amrita Bhasin · Answer

okay, We know that Delta acts. In other words, changing axes. One of her on and a zero knew the formula for Delta X&#x27;s B minus over end. So Bemis here, over, and somebody we know A is zero, which gives us one of her and therefore we know the hostage equivalent to one. Therefore, what we know is that we have the integral from zero 21 of X the fourth plus ax detox integrate increase the exploited by one divide by the new exponents increase to divide by two. Our solution is seven tons.

Bobby Barnes · Answer

So they first want us to rewrite this as a Remond some s. So let&#x27;s go ahead and do that. So this would be the limit as it approaches infinity of one of her. And now that was the thing that&#x27;s changing is the power in the art of the numerator here. So it&#x27;s like 123 and just kind of keeps increasing world. The denominator always stays in. So this is gonna be the sum from I is equal to one. Or actually, I guess we would start from zero. Um because I would be the same thing to in of the square root of I over in. And now this. If we were to go ahead and pull this one of her in inside so little limit as an approach to the affinity of the some from I 020 to in off one over in hi over in notice that this is now in the form of If so, if we&#x27;re on 01 this is like B minus a over in where B is one a zero. So this is Delta X eso There&#x27;s just one my zero over in which is one or in. So that&#x27;s our Delta X Here, This is Delta X, and then this function is going to be like X I because it&#x27;s just going to be our Delta X plus R a Oh, Delta x times I like that. Um, yeah, so you could see how. Okay, well, we have one over in times I and then plus zero so that there is Route X, I and so. But this is really going to say is that this is equal to the integral from zero to one of the square root of X dx. And now we can go ahead and evaluate this using power rule truck, my pin. So remember, we can rewrite this as a one half power. So powerful says, add one to the power, so it&#x27;s gonna be x 23 house, and then we divide by the new power relapse. Then we evaluate this from 0 to 1. Um, so that would just be two thirds next to the three half evaluated from 01 So we plug in one, so be two thirds times one minus two thirds times zero. So that&#x27;s just zero. So we end up where that infinite limit ends up being equal to two thirds

Fuzail Shakir · Answer

Okay, so this is the limit As any approaches infinity of one over and multiplied by the square root of one over and plus the square root of two over and plus the square root of three over and all the way up to the square root of an end over end. So is the living is as any person infinity, discos. Zero. This goes to zero. This goes to zero. This goes to one. So therefore we have the limit as n approaches infinity off one over at times one which is just equal to zero.

Paul Teng · Answer

Okay, so for this probably want you to use Ah, the definition of a sigma definition of integral in order to calculate this some off and over and score post case where it has eliminates and approaches infinity. And they tell you that interval is from 0 to 1. So over here I have heard in the, uh, sigma location of the definition for integral. So we have as a limit as n approaches infinity of the son of f of x k time still to eggs equals to this definition definite integral that we are familiar with the f of X from A to B, where it abused Interval and can use the interval to calculate Don&#x27;t X and X. Okay, so we just have Teoh. I use the information that given us and connect the pieces of the puzzle in order to get something off the form. Something in this forms, we can easily integrate it. All right, So first in your note is that the interval is from 0 to 1 right so out represent our A to B and also help us into calculate our Delta X. So let&#x27;s go ahead and calculate Delta X or don&#x27;t. The X equals two b minus a over, and we&#x27;re being be my Z equals to one minus zero by end which goes to one over end. And to calculate x k k equals two a plus, K, Delta X, you know that a equals zero and Delta X equals to one over, and so are que XK were able to k over. And now if you look at our sigma, we don&#x27;t really see a one over and in it on neither do we see a camera and, well, that&#x27;s fine. Or we have to do is just, like, manipulated so we can get one over and in it and also a k over in. All right, So the first thing I want to do is to somehow get our one over end into this save, mom. So I&#x27;m just gonna write it down and over and squared. Plus case were and the first thing I want to do is to multiply the top and the bottom by end. Right. So in that case, we end up with and squared over and squared, plus K squared times one over end. So now, as you can see, we do have our one over end. Right? So are one over end will represent or Delta X. Now we have to find a way to get our K over and into their and no, when we can do to get Cameron is too divide the top and the bottom by one over and square. Since if we divide boats, the numerator and denominator by one over and square, uh, then we&#x27;re just incent, essentially dividing it by one. Right? So if we do that for a numerator to get into a grand score, which is one divided by whenever and square times as workers k squared, would you give us one plus case with or end squared or rather que over and squared times one over end? So this is our new form. And as you can see, we do have a k over in and we have a one over end. So if I were to rewrite the some WAAS area equals to the limit on approaches Positive infinity of the sun from K equals to 12 and oh, now we have one over one plus k over and square times, one over and and this will you cook to somehow this will you go to our definite integral a over B It&#x27;ll be of f of XK have a vax d of X. Sorry, Actually, American, we can leave the XK doesn&#x27;t really matter by looking at our pieces are one over end are one over and represents our Delta X right. So if our one over end or one of her and represents the Delta X of our formula which in turn represents the D X of our function, so are one over and represents DX and we already indicated that are intervals from A to B And finally, our third part XK XK, is okay over end and as you can see now are some. We do have a caver and which will represent our function XK. So now we have all three parts. We can go ahead and substitute them for into our definition also are definite. Integral is going to be from going to be from 01 That&#x27;s interval. We have the integral from 0 to 1 of f of X were I&#x27;ve actually goes to 1/1. Plus we can substitute kale ran with X, so get one over X squared and one over and is equal to D X. So now this is our integral And to find the anti derivative of this, we know that the anti derivative of 1/1 plus X squared is the Arc Tangent function. The our change in function right and our balance are from 0 to 1. So if we were to plug in our values for X, we get tangent, PR tendon off one. When is the Arc Tangent? What is the art? Attendant? Zero. And we know that the Ark engine of zero is zero Michael the way and the Ark tendon off one is pi over four. So that is our final answer.

Fuzail Shakir · Answer

Okay, so this is the integral from 0 to 2 off one over one plus X squared DX.

Problem 79 Hard Difficulty

Using a Riemann Sum Determine
$$\lim _{n \rightarrow \infty} \frac{1}{n^{3}}\left(1^{2}+2^{2}+3^{2}+\cdots+n^{2}\right)$$
by using an appropriate Riemann sum.

Answer

$\frac{1}{3}$

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$\frac{1}{3}$

Calculus of a Single Variable

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Grace H.

Catherine R.

Anna Marie V.

Kristen K.

Ron Larson

Calculus of a Single Variable