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Problem [20] Suppose that f € L'(R) prove thatIf(oz) f(z) du(z) - 0 as 6 > 1Here and in what follows / is the standard Lebesque measure....

Question

Problem [20] Suppose that f € L'(R) prove thatIf(oz) f(z) du(z) - 0 as 6 > 1Here and in what follows / is the standard Lebesque measure.

Problem [20] Suppose that f € L'(R) prove that If(oz) f(z) du(z) - 0 as 6 > 1 Here and in what follows / is the standard Lebesque measure.



Answers

Compute the left and right Riemann sums- $-L_{6}$ and $R_{6}$ , respectively- for $f(x)=(3-|3-x|) \quad$ on $\quad[0,6]$ Compute their average value and compare it with the area under the graph of $f .$

Okay. This problem, we're gonna compare the left endpoint some the left, Riemann, some to the right. Riemann some, um, over six sub intervals for this function On the interval from 0 to 6. It's right that appear 02 six. Okay, I've already got a couple pictures drawn over here to help us understand this. Uh, the left some is. Ah, First of all, we're gonna gonna figure out, you know where exactly we have to divide up these rectangles, But it's pretty easy to see if we do the entire interval. And we divided up six ways that each sub interval is gonna have a width of one. You can see that here. That's why I have this, uh, the X axis stepped off this way, cause if we just add one to each X value will get the next X value. Um, and it's the same way with the right, um, in point. Some also. So to execute the left Riemann some for six sub intervals. We just want to we just want to find the area of all six of these rectangles. So it's like we have a rectangle here here, and we actually have one here that you can't really see because the left hand side of it has a ah, a function value, a y value of zero. So if we find a, uh, area of a rectangle with a Y value of zero height of zero in that area is just gonna be zero. So that's, er, that's gonna be our first rectangle on the left hand, Riemann some it we they all have a width of one. No actual. Go ahead and use the distributive property there, a width of one. And then we just need to find the height of each of these rectangles. So we know our first heights gonna be zero. Our next tight is gonna be whatever the function value is at one. Okay, at this X value. If we plug that X value into the function, then we'll get this value right here, which is the height of this rectangle on the left hand side. So if we plug a one in here, we're going to get one minus three is negative, too. And negative two squared is gonna be four and nine minus four is gonna be five. So it's gonna be Route five. Okay, the height of my third rectangle. This rectangle here is gonna be the function value at two. Well, give me this point. So if we evaluate this function at 22 minus three is negative. One negative one scored. His 19 minus one is eight. So this is route eight, and we're gonna continue right on down the line when you get this next function value at three. Um, for the fourth rectangle. So that one is gonna be three miles, 309 minus zero is nine. And the Route nine is gonna be three. That we suspect this picture has some symmetry. Um, about X equals three. So go ahead and see if we get the same when we evaluate this four. Um, the function value 44 mice, three is 11 squared is 19 minus one is eight. So Route eight. Okay, so when we evaluate the next one, actually, the final one at five. We're gonna expect it to be Route five. Okay, Now, each of these are the heights of the rectangles with six rectangles, and we're gonna move. The width of every rectangle is one. So if we just add these up and multiply it by one. We're going to get well, just approximate this. Actually, it's gonna be in here. Rational number 13.1 three is our area of those rectangles figured with the left hand. Riemann some. Now, let's figure out the right hand, Riemann Some It will be able to shortcut this a little because we've already evaluated all these function values at the at the X values we're gonna need just require a little bit of shifting. Um, the width is still gonna be one for every rectangle. But now, the first height we're gonna want to use for this first rectangle This is the key thing to understand. We want to use the right hand height. So that is the function value right there on the curve, evaluated at X equals one. And that corresponds to Route five up here. So we're just shifting all of these. Why values over one and we're just not going to use the 1st 1 We're not going to use the left hand one. So the next one it's gonna be Route eight in the next one. It's gonna be three. Then the next one will be Route eight and the next one will be Route five. Okay, but that's only five function value. So that's only the 1st 5 rectangles. We still need this sixth rectangle here, which will evaluate on the right hand side at X equal six. We suspect it's gonna be zero from the picture. Six minus three is 33 scored. His 99 minus nine is zero. So that is zero. Now, with some observation up here, these air the same six numbers that we have down here. It's just they've all been shifted over, and the zero is now in a different spot. But this has to give us the same approximation. Uh, 13 point one three. So these Ah, the left hand Riemann some and the right hand Riemann. Some are equal here, and it's because of the symmetry of the picture. So we have figured out both of those and discovered their equal

So for this problem, we're going to be competing. The left and right hand remains, um, for our function FX. And so our first of seven, this problem is going to be computing the er sub intervals for our remains. And so we know that we're computing are some over the interval from 0 to 6. So to actually the, uh, Delta X or the width of each sub interval, we're just going to subtract the right and point, which is six minus the left endpoint, which is zero, and divide that by the number of sub intervals. So we're finding l six and our six, so in this case, they will both be six, and our step size is going to be one. So that means that in order to pick our X values, we're just going to start at zero. Because that was the left. I left him End of the sound of the interval that we're finding that you're even some lover and we're just going to increment this by one all the way up until we get to the right and plate. So our X values will be just be 012345 and six. And our next step is going to be calculating FX at each of these values. So, for instance, F zero is just going to be the square root of nine minus three squared, which is heard in nine months. Nine, which is just syrup. And we can calculate each of the other FX values at our corresponding X points. And we do that. We will get the square to five Skirt of Eight three squared of eight this 45 0 And so from here we're going to be able to calculate or left and right hand remained Sum's so we could go and start by finding our left remains. Um l six in the way that we do this is we're going to multiply our Delta X or our step size by the sum of the 1st 6 terms the 1st 6 y values that we found. So this means zero close escort five plus a skirt of eight plus three close scored eight plus is fortified, and this is going to end up adding up Thio about 13.129 and we're also going to be able to use the same set of information to find our right hand remained. So So again, we multiply our step size still two eggs, which is one. But this time we're going to multiply it by the last six terms. Oh, that we found that correspond to our ex cornets. So that means that we are going to start at square 25 this time, Um, and again, add up all of the terms that lied to the right of it or the next six terms, that lying to the right of it. And when we do that and calculate our agreement so we can see that are right, And rheumatism is also going to be equal to 13.129 So in this case, are left hand remains, um, in right hand remains, um, are going to be

The question We have improved their you're years too many that that thing g practiced by the preferences. That just kind of no, you cannot have. And that is for my life. Uh, Mexican? What? It's the collected over the region. Well called that or yes, it's help with him. That speech solution as a that just Children is Yes, yes. Yeah. Look, send you I would in Yeah s Is it lost? So based tracking is division Region inside. That's the basement site there. So simply that will instantly and stay there. See? Yes, these the tents. Uh, okay, 18. Thanks. I think to see the city. Mhm. So we used air that fact. Good. Chess off early off the function F S c. So you say this the idea. But on big breasted Yes, this is the answer. Get

Hello there. So here we have. The following limit is that involves an absolute value off X minus six. So let's evaluate this limit. This is almost trivial. This is just the absolute value of zero that is equals to zero. And this is or did I mean here. Then we need to use the EBS Lindell definition that say that for every epsilon greater than zero, the exist a delta greater than zero such that such that these difference that is going to be X 96 is greater than zero on a less them. This helps this delta. These implies that the difference between the function and the limit that in this case is just zebra is less than the absolute here. So what happened here is that we go to the limited zero, so we just have the absolute value of the function which, conveniently, in this case, is just the absolute value of the absolute value off. Okay, 96 on these should be less than done. Absolutely. But here's what we want. Actually, this is just the absolute value off X men. Six is less than Absalom. And here we have what we want. So We want something similar to this in order to find that delta that we need For every absolute on here, we have find it is just absolutely so. The Delta. He's just absolutely


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