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Prove Formulas 9.

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$\bar{y}=\frac{M_{x}}{m}=\frac{M_{x}}{\rho A}=\frac{1}{A} \int_{a}^{b} \frac{1}{2}\left[f(x)^{2}-g(x)^{2}\right] d x$

Calculus 2 / BC

Chapter 8

Further Applications of Integration

Section 3

Applications to Physics and Engineering

Applications of Integration

Baylor University

University of Nottingham

Idaho State University

Lectures

in this video, we're asked to prove the theorem of the moments basically, which is given the coordinates of given a region, the X and y coordinates of that central are given in the formulas above. Yeah. Uh huh. So how do we do this? Let's first let's consider a line. Let's consider just a graph of any arbitrary function. I'll just call that one F. Yeah. And we'll call this one here. Gene wow. And we're going to assume that it's bounded in between some two points here. Just call them A and B respectively. Yeah, this will just make it easier for us to do this later on. So suppose we have now if we're going to consider the areas between these two curves, let's consider that we have to deal with some moments. Suppose we have these four because remember there are moments for each curve. So and there will be both an accent one moment. Mhm. So obviously because there are four moments, we can say that between these shared curves, we can say that there are two moments for the total. Let's just say that they're F G okay for X and Y respectively. So obviously this has to be true by the fundamental theorem of movements. Uh huh. M F G. Why? Just like that? Yeah. Yeah. And we can always arrange the moments in this following way and they should actually not be F G. They should just be F in the moments of both of these curves. We can rearrange this we can say M F G in general is going to be an F minus MG. So M f x minus M F Y and M f g Y is equal to or actually not fy again G y G x G X there and this would be equal to M f y minus M G Y. Now we need to also remember the definition of the century. I remember that the X coordinate is always going to be dear the Y moment in general over the mass and the white quarter of the century, it is going to be the X moment over the mass. In our case our moments are both M F G. So what that means is we will have expire will be defined as mm F G. Why why over em which is the same as M of F y minus M G Y. The whole thing over em or both of these things over em. Because we can split the fraction and the same with the Y bar and and lowercase M here denotes a total mess. Yeah. Okay. And we remember that the remember from the definition of the moments in general. Yeah. Oh yes. The moment, the x moment would be equal to 1/2 A row times integral from 1/2. Sorry, let's fix that. The integral from a to B. Of one half times f squared of X. And the Y moment would be equal to rho times integral from A to B us X times F. And and again the excellent wine moments are the moments the F. On regarding the F. G. Access and also the mass in general is as above just like that. Now of course this F is just the union of two functions as shown there. So substituting all of this in, we're going to get that X bar will be equal to rho integral from A to B. X. S right minus rho times integral from A to B X. G of X. All of this will divided by rho integral from A to B. F minus G. All this over rho integral from A to B. Like that. And obviously all of the rows will cancel, wow, mm hmm. Time the same thing with the Y coordinates we have the following. Yeah. And again, all of the rows will cancel. Oh and recalling that the mess is just the bottom. We've arrived at our formula. That's how it's done.

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