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Lowie Tambis

If $f(x)=x+\sqrt{2-x}$ and $g(u)=u+\sqrt{2-u},$ is it true that $f=g ?$

00:50

Masoumeh Amirshekari

01:09

Felicia Sanders

Simon Exley

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Okay. So as I said, we're going to turn our attention now to finding the area underneath. Curse. Okay, so the set up is pretty simple. We have a function at the X defines on a closed interval. Okay. And so again, sometimes we use open intervals here. We're really going to use a closed interval just for some technical reasons. You don't really need to think too much about it. But the idea is that here is our function that we can plot. So maybe here's a and here's B So our function look something like that. And then the question is really simple. In between X equals A in X equals B is a bounded area. So the area is bounded between the graph of the function and the X axis. Okay, so this is just some area, and our goal is just to figure out what that area is. Now that's going to take a little bit of time to do. But the first thing I want to mention is that if this area is a recognizable geometric shape, good than we can apply known formula. Okay, so let's just review a couple situations that could come up. So for a triangle if we recognize this region. So if we have, you know, some special function were actually this region just consists of triangles, like if it's a line or the absolute value function, something like that, we can just use the fact that the area of a triangle is one half base times height. And then if we have maybe a circle course, we just have pyre squared. So somehow, if the graph of a function is consist of, uh, pieces of circles, sort of sectors of circles, then we can just use the area of a circle and just divided up appropriately. And you know, there's some other situations do. So if it's a rectangle, or maybe somehow it's it consists of rectangles. We conform it by just kind of Hastings and rectangles together. Of course, the area of a rectangle is length, times, width and etcetera. I'm not going to sit here and list out all the possible geometric formulas that we may run into, but the point is, is that in some special cases, we actually can already solve this problem. If we recognize the graph of the function is being the boundary of some geometric shape or some union of geometric shapes. Then we could just use are known geometric formulas. Where it really gets interesting is when our function does not fall into one of these categories. Okay, so here's a picture that's going to illustrate kind of a first attempt to solve the general problem of finding the area under a curve. Now here the function h of X is equal to X square. Okay, so this blue line is the function X square, and we're interested, and this area bounded between X equals zero and X equals one. And then the graph of X squared in the X axis. Okay, so it's almost a triangular region. But you see that the top side right here is actually curved. And so the initial idea or the kind of naive thing to do is approximate the area by rectangles. Okay, So notice here. These rectangles, I've sort of just broken up the region into for pieces or four parts. I've broken up the interval 01 into four parts. So another way to say this is that I have partitioned the intervals you're with one into for equal parts. Okay, so we have the interval from zero. 2.25 The interval from 0.252 point five, the interval from 0.5 0.75 And then the interval from 0.75 till one. Okay. And so each one of these parts of my partition. So this word partition simply means that I'm breaking it up. And the union of all of these intervals is the whole interval on their intersection. Eyes empty there. Paralyzed intersection is empty. Okay. And so each one of these parts of my partition gives the base of one of these rectangles. And now I used equal parts so that actually each one of these sub intervals has the same width. So they're each of just length one quarter or 10.25 So then now I just need the height of each rectangle. So the bases 0.25 here and here and here and here. But what about the height? A beach rectangle? Well, here I actually have to make a choice. I have to pick how high I want the rectangle to go up. And here I've actually specified the right in point of each side. Venerable so 0.25 0.5 point 75 and one to be the height. So the height here is f of one. The height here is F of 0.75 The height here is F of 0.5 and then the height here is F of 0.25 Okay, but I could have just as easily used the left endpoint. So for the height of the rectangle, we can use the right end point, the left endpoint, the midpoint. So the midpoint would be like this point here that would give me a different height. Or I could really use any point I want in the sub interval right here to give the height of directing. And it turns out that it's not really going to matter in the long term. All that's really going to do now is that sometimes in specific cases, using the right in point, using the left endpoint or using the midpoint might give a more accurate approximation, or it might given overestimate or underestimate, just depending on you know what the shape of the function is. Okay, but the point is here to approximate this area, we're gonna break it up into rectangles and of course, here I broke it up into four rectangles. But I could have broke it up into more right. I could have broken up into 100 rectangles. And of course, the idea is, the more rectangles I use, the more accurate the approximation will be. So let's just see, actually what this approximate area ISS. So let's just think we can make a choice, right? So we can either use the right in points, three left in points or the mid points. So let's see how that goes. Okay, so remember, my goal is to approximate the area between the X axis. The graph of y equals X squared X equals zero and X equals one. Okay, so that region we had before now I can use right in points which all denote by the capital are so And then I'll put a subscript of n So this is going to be in rectangles, equally spaced rectangles. That's important with right endpoints for the height. So I'm going to evaluate the function at the right end point to give the height of the rectangle. I can also use what I call Elsa, then which will be in rectangles with left in points again. For all of these, they're going to be equally spaced. And then I can also use what I'll call Capital M seven, which is in rectangles with mid points. So, of course, when I say right in points left in points mid points, that's the point. In the sub interval that I'm using, I'm gonna evaluate the function there to give the height. Okay, So for our problem, let's just think about what is our four. Okay, so this is going to be, well, it's just gonna be I'm just gonna add up the area of each of those rectangles and the height I'm going to use the right endpoints. So the width of each one is your 10.25 Okay. And then the height of the first rectangle, I'm going to evaluate the function X squared at the right in point, which is 0.25 squared, and then the second rectangle has with 0.25 they all have with 0.25 But now I'm going toe, evaluate the height of the rectangle at okay. The function evaluated at the right in point of the seconds of interval and then I have the third rectangle and then the last rectangle. Okay, so this is just rectangle one rectangle to rectangle three rectangle four in that last picture that we saw. So that was an exit. That was an illustration of our four. And so what did we get here? Well, you just do it and you get zero point for six, eight, seven, five. Okay, so that is the approximation using in equally spaced rectangles with right and points for the height of the area between the X axis y equals X squared X equals zero and X equals one. So we can also look a l for and now l four is going to be almost exactly the same thing. But instead of evaluating the function Y equals X squared at the right end, points were going to evaluated at the left end points, but the width is going to be the same. But now, instead of 0.25 the right end point of the first rectangle, I'm going to use the left endpoint, which is 00 squared. And then now the left and point of the second rectangle is 0.0.25 the left endpoint of the third rectangle is 0.5. And then finally, the left endpoint of the fourth rectangle is 0.75 and we compute this and we get zero 0.2 one 875 Okay, so we get a lower estimate of the area. And so actually, from the picture we can see that are four is going to be an overestimate, right? The rectangles hangover. Whereas l four is going to be an underestimate because if you draw in the rectangles using the left endpoint, you'll be underneath the area. So typically a better approximation is to use the mid points because it will somehow average using the right and point in the left endpoint. So it will give a better approximation. Typically, yeah. So using the midpoint again, the ba width of each of the trump of the rectangles 0.25 that's not going to change. But now we're going to evaluate the function X squared at the midpoint. So the midpoint of the first So the interval is 0.1 to 5. So I got that just by averaging 0.25 the left and right in points, so that right in the middle is 0.1 to 5 and then for the second sub interval between point to 5.5, I do the same thing. I just find the midpoint that's going to be 0.375 And then for the third Cem interval, we'll have zero point 6 to 5 and then finally we'll have point 875 okay? And then what do we get when we compute this, we get zero point three 28 one to five. So you see that the approximation using the midpoint of the interval is somehow in between the right using the right in points which we said is an overestimate and using the left endpoints, which is an underestimate. And in fact, what we'll see in just a second is that the actual area underneath the curve of y equals X squared between zero and one is one third. Okay, so see, noticed that m four using the midpoint is pretty close to one third. So it really is a pretty good approximation, whereas using the right and left in points are pretty far off. Okay, but that's really just a case of using relatively few rectangles. All right, so we're interested in getting better and better approximations. So to get a better approximation, what do we do? Well, we want to use more rectangles, and I'll say that as we use more and more rectangles, it's going to become irrelevant whether we use the left endpoint, the right in point or the midpoint. And so, just to be consistent, let's just used RM. Okay, So the same question we want to find the area bounded between the function. Why it goes X squared in the X axis between X equals zero and X equals one. Okay, okay, so let's just write down what are subject Ennis in general. So what do we do? Well, we're gonna use in rectangles using in rectangles and of course, these air going to be equally spaced. Otherwise, things get unnecessarily complicated. So using in rectangles of width Well, what's the width of each rectangle going to be? Well, this is just kind of a geometry question. If I want a partition in interval into in equal parts, well, I'm just going to take the width of the total interval. So the width of this interval is one. And divide it literally just divided bythe number of part someone. Okay, so in rectangles of with one over n and then height. Well, what is the height of each rectangle going to be? Well, I need to think about what are the right in points of each so miserable? Well, if I just start breaking it up here zero all the way, toe one. Well, I'm gonna have one over in, and then my next right in point is going to be too over in. And then my next and point is going to be three over in all the way. My second, the last one is going to be in minus one over in. And then my last one is one which I can think about is in over in. Okay, so the height is going to be I over in squared. OK, so I'm just gonna pick one of these right in points for the eighth rectangle. Okay, so where I is between, you know, it's starting at once between one and in. We have our spin equals. Well, we're gonna have one over in which is going to be the width of each rectangle, and so I can actually just factor out that one of her in. So that's just each term is gonna have a factor of one ever end. That's the base of each rectangle. And then the high to be tricked angle is gonna be I squared over in square, but that means each each height has a factor of one over in squared, so I can factor that out. So I have one over in cubed and then I just have one squared plus two squared plus three squared plus all the way up to in squared. Okay, so all I did, you know, really, If I expand all this out what I'm saying this is one over In times one over in squared, plus one over n times two squared over and squared, plus one over in times three squared over in squared. Okay, that's the same thing. I just did a little bit of simplification, so I just factored out all of the common terms. Each term has a factor of one over in cute and again, this is just base times height, base, times, height, base, times height for each of the rectangles. I'm just using the right in point to evaluate the function to get the height. Well, it turns out, so we can show which we will show later. I promise that one squared plus two squared, plus three squared plus all the way up to and squared actually has a closed formula. So what it is is, in times in plus one times two in plus wine all over six. Okay, so actually, it's not that hard, at least for this simple function. Y equals X squared to give a nice formula for the approximation of approximating the area, using in rectangles, equally spaced of with one of her end and then height given by the function evaluated at the right end point, which is just I over in that quantity squared. But this is really, really nice, because what we're going to do next is think about what happens when N goes to infinity. Or, in other words, when we use mawr and more and more and more rectangles. Alrighty. So we're going to use a lot of rectangles to approximate and eventually hopefully find the exact value of the area between X equals zero and X equals one bounded between Why equals X squared and the X axis. Okay, so recall in what we're doing is we're breaking this up into a bunch of really thin rectangles like this. And now, just to be consistent, we're using the right in point to give the height of these each of these rectangles. And you see that as we use more and more rectangles So this is in rectangles. We're in here. I just want to think about is a very large number. So the wit is one over end the height of the ice rectangle. Is I over in squared? Okay, so we actually know what this area is of all of these rectangles added together. So our sub in, as we just saw is one over in cubed times in times in plus one times, two in plus one, all over six. Okay, so that's the area of the sum of all these rectangles which, as you can see, is getting very, very close to the actual area underneath the curve of why equals X squared. And so we expect that the area is the limit is in, goes to infinity. Okay, So, basically, as we use in arbitrary amount of rectangles, so essentially infinitely many rectangles. Now each one of those rectangles will be infinitely If Intesa Mally skinny Infanta zimmel is just kind of the opposite of infinite, just very, very small. So we expect the area is the limit is in, goes to infinity of our cement and that's actually true. Okay, so we have not justified that. But I think the justification, or at least the intuition, is clear as we use more and more and more rectangles that they're getting skinnier and skinnier and skinnier, they're better approximating the area in each little kind of small slice, if you will, of the actual area. Okay. And now it's just kind of a pretty elementary exercise to take the limit of dysfunction is it goes to infinity. Notice that this is a rational function, right? So if we factor out the highest factor event on top and the highest factor event on the bottom, then we're gonna have basically something like to in cute over six in cubed times, one plus some stuff that's going to zero over one, plus some stuff that's going to zero. So the interesting thing is just kind of the leading terms, or, if you will, the limit should be approaching the ratio of the leading coefficients, right? In other words, we have a horizontal ascent. Oh, at one third. So that means the limit is in, goes to infinity is approaching one third. And so, in fact, one third is through the area. Underneath the graph of y equals X squared between X equals zero and X equals one. So that's the idea behind actually computing exactly the area underneath the graph of a function. You just break it up into small pieces and small rectangles and in the limit, it's not really gonna matter. Whether you use right and points left in points from mid points doesn't matter. But you take the limit and it should be approaching the area underneath the curve. Okay, so let's talk about the general set up. So the area under don't say it this way. The area bounded between a function ffx and the Y axis on some closed interval A to B. And so again, the picture just looks like this. That's what we're looking for right there. The area bounded between ffx and the Y axis. Now we're going to make a technical assumption about ffx that it is continuous. And now if it's continuous, this will all make sense on, you know, kind of Ah, very technical level. So the area bounded between this continuous function ffx in the Y axis on this closed interval A B, that's our goal. Well, the first thing to dio is partition a B into in equal sub intervals each of length. And here I'll just use notation. I'll say Delta X is the width of each of those small serve intervals which will be the width of each rectangle should be the length of the interval. So B minus a over in. And so what we have are a collection, a set of intervals that looks like this x of I minus one t x I So here I is going to go from one thio end. So this might just collection of some intervals. And now what is X I so x of I is going to be a so the left endpoint of the interval, Plus I times Delta X okay. And you can verify that this is the case, you know, so each you know. So I'm starting at a and then each part. I'm just gonna add Delta X Okay. So if I want kind of the the AIF right in point in this case, I'm just going to add built X I times. Okay. Not too hard to see. Okay. And then what's next? So we want Thio evaluate F MX hat the ice right in point. So the either rectangle has area. Well, it's gonna have area What? Delta X base times the height, which the height is f of a plus Delta x times I Okay, now this is for are seven. What about if we wanted Elsa Van? Well, for Elsa been, We just want to you evaluate not at the eighth, right in point, but the left right in point. So that's X by x of I minus one. So that's going to be a plus. Delta x Times I minus one. This is for else Ammen and then for the midpoint. If we wanted to do that, it's just delta x times f of and I'll just write it this way. Whatever x of I minus one is plus X by over to. So it's the average of the left endpoint in the right end point. Okay, so we're saying right in point and if we're just going for the exact value and taking a limit, it's really not going to matter. So we might as well just use the right in point. But we can also just do approximations by the left endpoint in the mid point. So the final step to find exact area is just going to be dealt X times f of A plus doctor X plus f of a plus to Delta X plus 2 to 10 plus f of being. So this is our seven, but we're going to take the limit is in, goes to infinity. Okay, so that's gonna be the exact area. Now we would get the same thing if we did the left endpoint or the midpoint, but it doesn't matter. Okay, so that's the general set up. You have a continuous function. You want the area, your partition, the interval. You're interested in into ends of intervals, each of length Delta X, the right and point of each one is a plus. I time still two x and then we're just going to evaluate ffx of the ice right in point to get the height and then to get the exact area. We just take The limit of the sum of all the rectangles is and goes to infinity. So that's the idea. Pretty simple, pretty naive. Again, We're just using this machinery that we've built up at the beginning of the course about limits to have really given answer to a question that's that we would not be able to really solve otherwise.

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