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Greninjack Dan

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

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Lowie Tambis

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Felicia Sanders

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Okay, So recall that in couch one if we had a continuous function defined on a closed interval, a B so continuous in continuity is just kind of ah, technical assumption that makes everything simpler. And I mean most functions. We're going to deal with our continuous. So it's not a tremendous hindrance toe what we're going to be doing. But if we have such a situation and we look at, say, the graph of ffx on disclosed Interval A B and we look at the area bounded between the function in the Y axis between X equals A and X equals B well, that's given by this definite integral the integral from A to B ffx the x okay and you spend a good chunk of calculus one talking about how to find these integral is using the fundamental theorem of calculus. But we want to zoom in on the geometric interpretation that this definite integral it really is just the area under the curve. And now really, it's the signed area under the curve because let's say the graph of effects goes below the X axis, like over here. Well, this area that's below the X axis is considered to be negative. Okay, So we can sort of just imagine that if we had a function that was always positive, then we really can think about this. Integral is being a meaningful positive area, okay. And then just sort of adjust when we have, say, a negative area like this. So for functions of two variables, we want to have kind of a similar geometric idea to area under a curve. But now, if you think about the graph of a function of two variables well underneath the surface, that is actually the graph of the function. You have a volume, and now you're not gonna look at a volume above an interval, but actually a volume above or bounded above a region in the X Y plane. Okay, so let's talk about that a little bit. So we have a function S o Z will be some function of X and y, and we're going to again assume that this function is continuous to make everything clean. Don't worry so much about that. We're also going to assume that this function is positive words more specifically, not negative. So it's greater than or equal to zero just to kind of simplify this geometric interpretation. So let's say that f is defined on a region are that some subset of the X Y plane are too. And we want our to be just like our closed Interval A B. We want our to be closed and bounded. So our function f is positive on this region are it's continuous on r and r is closed. Inbounded. So this is a similar set up we saw when we did optimization of a function of a continuous function on a closed inbounded region. Okay, so what we have is the volume bounded between the region are and the graph or the surface of ethics. Why, yes, given by Well, this is just symbolic. Okay, so I'm going to write to integral signs to signify that in the domain, there's actually two variables I'm gonna put this are underneath. So this is like the double integral over this region are of the function. And then I'm gonna put d A. And now what is D. A. D. A. Is just a small area region. And so the notation here is actually kind of suggestive of what we're going to do to kind of compute this volume. And now I haven't really done anything. All I've said is that, you know, given a function defined on a region closed, unbounded. This volume bounded between the region are and the graph of F is just symbolically notated by this double integral. Okay, so, you know, we want to think a little bit about how we would actually compute something like this, Similar to how we did a remind some for a definite integral of one variable. All right, so let's draw the X Y plane and just think a little bit about how to tackle this volume question or how to actually, you know, compute the volume that we were, uh, denoting by that double integral for a function of two variables. Okay, so here's the X Y plane and recall that our function is going to be defined on some region in the X Y plane that's gonna be closed and bounded. So I'm just going to kind of draw it like that. Yeah. And so the simple thing to do here is very similar to what you do for a definite in a girl in Calcutta one. But instead of breaking up the interval into really small pieces. We need to break up this region into really small pieces. So what I'm gonna do is just kind of cut this up into little tiny pieces. But I need to do it in both dimensions. So you see that if I cut up this region into a little just think about these squares. Each one of these little tiny squares has an area d A. And I can think about this D A as being just a little small DX in this direction times a little small d y in the vertical direction. That's the area of this tiny, tiny little square right here that I've broken up. Okay, but if I think about the graph of the function So if I look at this one little chunk of the function, what I'm drawing here is a rectangular prism that's going up to the height of the function. So this is X, and why than this height in the Z direction Is the function evaluated at X and y? And so this little small little volume is just d a. Just the little area of that small square times its height in the Z direction. So this little piece of volume is just going to be, well, the height f of X y times d a. Like that. But what I want to do is add up all of these little pieces of volume. These little rectangular prisms, that air sort of filling out the volume. So what I do is just say okay, Well, I wanna add them up using this double integral. And so what do I mean by this double integral? Well, I really mean a double remond, some where I'm summing over sort of my partition and the extraction in my partition in the Y direction. And so again, the way I would really compute this is by taking all of these function values and multiplying by these d a s or these d x times d wise and then adding all of them up. So that's sort of the naive way to do that. But what we're going to see is that there's actually a better way to actually compute this double integral, or this volume bounded between the graph of a function over a region, and that's using the idea of an ID aerated in a girl. And like we've seen throughout this class, what we're really going to do is we want to instead of recreating the wheel. You know, with this double room on some and kind of doing the naive idea of just breaking up the volume into small pieces. We want to recycle the theory and the ideas from Cal Kwan that we've already developed. And that's what I'd aerated inter girls are going to allow us to dio. Okay, so recalled that are set up is we have a function of two variables and we're assuming it's positive or non negative. We're assuming that it's continuous, and it's defined on a closed and bounded region are. And so the idea is that we can think about are in the X Y plane as being well either bounded between two functions of X or two functions of why, and so I'm going to think about this region, and this is a slight simplification, but it actually is pretty general. So if we have a region, are then I actually want to think about this as being bounded between two functions of X and mind. So here's X equals Hey, here is X equals B. And so this top function. Let's call this G one of X and this bottom function. Let's call it G two of X. Okay, And then I want to think about my region is being this area bounded between these two functions of X. Okay, so this is really what I'm calling are right here this region in most regions I could describe this way I can describe them as sort of being bounded between a one function of X and another function of X, or maybe one function of why and another function of why. But that's really going to be the same idea. So kind of, without loss of generality will assume that our is bounded between two functions of X like this. And here is Thea Dia say that I fix a number X between A and B. So here's my number X So fix x so for fixed x. Yeah, ethics. Why is a function of why? And actually it's a continuous function of why, and it's also defined on the closed interval. Well, so I'm sort of cutting my function right here for a fixed X, just fixing an X value there and is going from well, G two of X all the way to G one of X And again, this is for a constant X. This is very much similar to what we do with partial derivatives. Were just fixing an X value. And we're treating X is a constant and letting why vary from wherever it varies? Well, it goes from whatever this number G two of X all the way to G one of X But just think about the cock. One idea. I can now find the area underneath the graph of my function f x y along this small little line right here. So I can just think, OK, well, I can integrate from G two of X for a fixed X to G one of X of my function. And of course, it's changing with why so I d y But this is going to be true for every single ex between A and B. So this itself actually gives me a function of X from A to B. So this is really amazing. So you should see the analogy between doing this and doing partial derivatives because with partial derivatives, we also fix the variable and just thought of the other variable is changing here we're fixing an X in thinking about why changing and adding up the area underneath that curve. This is sometimes called partial integration, sort of for obvious reasons reasons because it's analogous to partial differentiation. Okay, so I'm fixing an X value on then sort of getting this function of X which recall this function is continuous, so an integral to find function will be continuous. This is a closed interval. So what does that mean? That means I can actually integrate this function capital ffx so I can integrate from a to B of capital ffx dx. But what is this giving me? Remember that the output here. So I pick an X value and I'm getting an area underneath this curve. But that means if I add up all of these X values, I'm adding up all of these areas between A and B and what that's going to do is give me the volume under the region. So this is actually equal to the volume that we want in this volume that we want we found by sort of four first integrating with respect to why and then integrating with respect to X. So if we write this in the following way, we have the firstly Why then the X So this sort of expression, now that we have an inner integral and then outer integral, this is called an ID aerated and a girl in the way it works. When you see an ID aerated, integral like this is you start with the innermost integral and you treat X is a constant. So X is in the outer inner girl. Some entry X is a constant. I find an anti derivative of the inside evaluate from this function of X up to this function of X Then I deal with the outer integral. Then I take the anti derivative with respect to X and evaluated being a limits like that. Okay, so also, a region could be bounded between two functions of why so it could look something like this is Well, so maybe this is over here. This function is say F one of why, in this function over here is if two of why and everything is gonna work exactly the same way, except we're actually going to do the integral with respect to X first and then why we're still going to cut this up for fixed values of why we're going to cut this up. You see that for why exes going from this function of why did this function of why? And we're going to do everything exactly the same way, So just sort of to reiterate that this is my function are and we have our function f of x y. Then the area bounded between this region and the graph of f of X y, which were denoting by this really is just equal to well, say that this is B and this is a well, now we're going from A to B for why? And then the inner ID aerated in her girl is this hath one of why f two of why f of X y yeah, but now notice that we're doing x first. So we're integrating the function, assuming why is the constant with respect to X evaluating at these functions of why and then taking the anti derivative with respect to why and evaluating at A and B so it's exactly the same. But sometimes regions are easier described as two functions of why Sometimes they're easier to describes two functions of X. So you just sort of have to make that choice when you look at the region. Okay, So the situation is even easier when our region are is a rectangle because then we're actually free to integrate in any order we want. So if we take our integral of our region f of X Y d A. Well, we can actually set this up as an ID aerated integral, so we can go a to B C to D. So we're gonna integrating with respect to why first. So it's gonna be d Y d x so in the inner integral. Why, for a fixed X Y is going from C to D because this is a rectangle and then X is going from A to B. But this is the same thing as first fixing. Why value between C and D and then saying Okay, well a where X is just bearing between a and B. So let me add of those values first, these will be exactly the same. And this result is sometimes called to Beanies era kind of fun to say. And it's the fact that for ida rated inter girls. We can actually switch the order and now you say, Well, why would I ever want to do that? Well, sometimes an integral might be impossible to integrate first with respect to X, and you have to integrate, versed with respect to why or vice versa. And so this is an extremely useful The're, um it's a sort of creatively be able to solvent girls that maybe you wouldn't be able to solve otherwise without kind of switching the order of integration in that way. All right, so the last thing we want to talk about in this lecture is a couple of applications of the double integral. So again, let's say we have a region are and a function that's continuous on our our were again going to assume is closed unbounded F full. Assume it's positive. And so the first thing we could do is talk about the area of are Well, this is actually really easy. Okay, so the air, let's say if I wanted to find the area of the region are I can actually write. This is a double integral. I could just integrate over the region. The function one and you say Wait a second. But I thought this was a volume. So this should be the volume above the region are and bounded below the function one. But if you think about it, that volume is going to equal the area to say I have a region like this, we'll just say it's rectangular and I want to find its area Well, the area is going to be equal to the volume of the region above the area that has a height one above its of this height. Here is one and this is our region are Then the volume here is going to equal the area of the region. So that's the idea here. So if we integrate dysfunction one numerically, we actually are giving the area of the region. Okay, so the next thing we can talk about now that we know, you know, a quick and easy way to find the area of the region is the average value of a function. So this is something that you talked about in Calculon to, and the idea is pretty simple. So the average value of F over the region are, and we'll have the same kind of assumptions on F and R F is going to be continuous and not negative, and our will be closed and mounted. Well, all you do is just take the double integral of F over our and divide by the area. So sometimes you'll see for average value, there's different notations, but you can say like F bar. So it's like the average of F is just the integral over our death and then divided by the area. Now notice here for the area. I'm not explicitly writing this one, okay? I'm just writing, uh, just the d A. But of course, this is one times d. A. So this really is just the integral. So I'm adding up all of the function values and then dividing by the area of the region. It is kind of a very natural definition for the average value of a function over a region, and this has a lot of really useful applications. Things like finding the center of mass. The center mass is essentially the average value of the X coordinates. Uh, Y center mass is going to be the average value. The Y coordinates, etcetera.

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