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Find the area enclosed by the curves y = V5S-xy=x+ land the X-axis...

Question

Find the area enclosed by the curves y = V5S-xy=x+ land the X-axis

Find the area enclosed by the curves y = V5S-xy=x+ land the X-axis



Answers

Find the area of the region between the curves. $y=x\left(x^{2}-1\right)$ and the $x$ -axis

All right. Now sketch the region enclosed by those two curves. So this is our XY plane. And the first function is why he goes to X Cube. So you should look like this. This is our by equals two x Q And why he goes to X. Okay, it's a street line. This is why it goes x in this those two eco region since everything is symmetric, uh, with respect or region. So those two are exactly the same. Um, they have the same area, and now we need to find this shaded region area. Um, so they will be, um yeah, we only need to calculate one side. So wait time and then times two. So it will be two times. Um, this shaded region. Yeah. Um, and, uh, I want to calculate this region by taking integral with respect to X. Since everything is represented bags All right. Right. The next step, we need to find the boundary. Correct. Clearly, X goes from zero to this point this point x two all right and x two should satisfy the those two functions, which is X square equals two x, so x two should satisfy this equation. Yeah. Since extra is not zero, we can cancel one X Then we got X squared equals one, which gives us X two eco swat. Since X two is part, so are integral. That goes from zero to one. Okay, um, so then the thing inside the integral is the upper curve, minus the lower curve. So our upper curve here is a straight line X minus glower, curve xq. Yeah. Then let's find anti duality of that. That will be up half black Square, minus fourth. Thanks for you had a boundary one and zero. So Well, X equals one we plug in. This will give us one half minus or quarter, which is a quarter times two is about half right minus, we know, actually called zero. This term, zero times two is also zero. So it's like one half minus zero. Our answer will be just 7.5

Uh, okay. Sketch the region closed by the given curves and finance area. So first, let's draw those two curves. This is our XY plane. And, uh, the first function is X equals. So why to the power floor? So it looks like a problem, but it grows much faster. So it opens to the right like this. This will be my ex seacoast y to the power four. Oh, and then the second term here, Um, why he goes to the square root of two minus x. Yeah, this we know. Oh, yeah. Um, it can be represented as if we take the square. This is just a wide square equals two to minus X. We move around things. This is equivalent with X equals two two. Wine is Y square. What? Let's score. Yeah, and, uh, be careful here. We also need this indication why is greater or equal to zero? Because you know, the square root my equal to the square root of something, So why cannot be negative? Yeah. Okay. And this is also a problem. He opened to the left on a start at two. And open to the left. Okay. They will be enclosed area will be here. Okay. And here is to Here is General Oh, uh, and as we can see, um, since we reformat the second curve right now, everything is represented by why. So if we want to find an area and take the integral with respect to why, instead of facts and and the boundaries at this point in this point, So if I do know this s by one Yes. S y two how to find them? It's just, um Oh, be careful here. The enclosed region is actually now the entire thing, because we could this extra condition here when we reformulate the second curve, why should be no negative. So, um, this enclosed area is only this half of them is here. Here is empty. Okay. Um, therefore, we don't have this by one. We only have this fight to on our right. One will be zero. It goes from zero to white too. Okay. Mm hmm. So now we know. Mm. Our boundary by one will be terrible. And why to surely satisfy in a way to is positive. So what? You should satisfy two minus. Why square the second curve and the first curve echoes too wide. The power of four. Okay. Mm. Uh, as we can see, if I plug in by coast to one. This equation hopes so you can say are way too. Must be one. Okay. Or you can do the factoring Then you can solve for this, uh, um, polynomial equation, you'll find your roots, and the positive route will be our right to, and it has to be one. Um, So the boundary will go from 0 to 1. Um, seeing inside the integral is the record minus left, Curve like curve. Here is just two minus. Why? Square minus the left. Her left her is, uh, right to the power of four. An anti duty of that is to why mine is one third like cube, minus month. Faith, my faith and inveterate one and zero and receiving Michael 20 this whole thing you go to zero. So we only care when y you go to one plug in white. Go to one. We got two minus one third, minus one faith. So it will be on two minus one third. My nurse on fifth. So this this is, uh, 8/15, right? So it will be 30/15 miners, eight or 15. Oh, So this whole thing will be, um, 24. 22 over 15. Mm, 15. Okay.

Oh. Mhm. And this question we're asked to find the area between three curves and one of the curves is actually going to be a starting boundary Now because I have one function in terms as X as a function of Y. That means that I would like to transform all of my function so that I have X as a function of why I did that for the Y equals description of two minus X by squaring both sides and then adding the X and subtracting the Y squared. So hence we have X equals two minus y squared. Now we need to find the intersection points in order to find the intersection points. We need to equate our two equations namely to minus Y squared Equals Y to the 4th power. And by getting everything over to the left to the right hand side, in this case we'll get that zero equals y to the fourth plus y squared minus two. Here, I'm going to make a substitution. Let's let let's let for example you be equal to y squared so that we only have to deal with a quadratic equation. So we'll have you squared plus u minus two equaling zero. Which when we factor this we're going to have this be equal to you. Yeah plus two U -1. Which means that you will have to be equal to negative two or you will have to be equal to one. But that's not what we're looking for. Remember that you is why squared that means that why I swear it has to equal negative too. Or why squared has to equal one and we can reject the negative too. So that means that our other intersection point is why equals one. Now from the graph, since it asked us to graph the region, we know which function is going to be on the right hand side, in which function will be on the left hand side. We know that the area as in terms of with X as a function of Y is over a specific region of integration. We'll have this be equal to the right minus the left, the Y. Yeah. And from our graph we have the wow two minus y squared is on the right hand side, so be integrating from 0 to 1 will have to minus y squared. Yeah. And obviously the function on the left is why to the 4th power -Y to the 4th power. And integrating this is very straightforward because all we need to do is just use the power rule still have to I minus two, Why cubed -1? Over 5th, 1/5 White to the 5th. And all of this will be evaluated from 0-1. And the good thing is that the zero will just disappear Because zero and a polynomial is just zero. Hence when I do this, oh, We're going to have two times 1 We'll have two times 1 -1, 3rd times one cubed, minus 1/5 times one. Which is just going to be equal to when you do all of the multiplication and simplification, you will get that. This is equal to 22/15. Yeah, that's how you do this question. Don't get a lot like.

Uh, not sure where they wrote the problem this way. Um, but y equals X squared. Looks like this to probable. I think everybody knows that if you're in, if you're in countless, um, and then the other function X equals Y minus two. Um, I write as why equals X and you have to add the two over, Um, so, like, shift up to and the soap is up 1/1 line like this. Now, here's the issue is we're definitely find in the area between these two, but we don't know those bounds like I don't know this x value in this X value question mark Question Mark. So you find that by setting them equal to each other, X square is equal to X plus two expo area and subtract X. I'm gonna subtract two over. I can look at this as a quadratic that a factors A to, um one into that had to be native one. Well, that works of it's plus one minus two. So I get X plus one next minus two, which means my bounds and this should make sense is negative one and positive two. So if you plugged these and you'll get the same. Why values A one plus two is one. They won't square this one. Two plus two is 42 squared is for okay, so don't checks out. So those are my bounds from native wanted to. My upper function is the X plus to function and you need to subtract off the lower function, which is the X squared. And so you find area The X man From here, it's just the anti derivative. So the anti derivative of X, you add one to your exponents and then you divide by that new exponents plus two x again, you had one to your exponents and divide by your new exponents. You can double check the derivative of this needs equal this so and it does You just believe me from negative 12 tubes. And now we need to plug in your bounds. So, you know, plug in to when you square it becomes four. Two times two is four. I'm plugging into, by the way to Cuba's eight thirds and a minus. When you plug in negative one into all these negative one. Once you squares positive two times able in this minus two and then a one cubit is still negative. One so changed that to be close one third. Now, from here, I would go to a calculator cause you're gonna get some fractions and just you can go with me as I'm this left wipe in parentheses here is 13th. Um, and when I get on this right from parentheses is negative. 76 And so going back to the 13th minus negative 76 you get an answer of nine house or 4.5. I like fractions.


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