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This Question:2 of 5 (0 complete) -MhisFind the area of the region in the frst quadrant bounded by the line y 3x, the line The Iolal area of the region X=4 Ine cun...

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This Question:2 of 5 (0 complete) -MhisFind the area of the region in the frst quadrant bounded by the line y 3x, the line The Iolal area of the region X=4 Ine cunve and the X-axis (lype exact answer; using radicals needed )Enter your ansi erin Ine ansi er DoXsearch6

This Question: 2 of 5 (0 complete) - Mhis Find the area of the region in the frst quadrant bounded by the line y 3x, the line The Iolal area of the region X=4 Ine cunve and the X-axis (lype exact answer; using radicals needed ) Enter your ansi erin Ine ansi er DoX search 6



Answers

Find the area of the region described in the following exercises. The region in the first quadrant bounded by $y=\frac{5}{2}-\frac{1}{x}$ and $y=x$

We want to find the area bounded by Y equals four and y equals x two, the 2/3 in the first quadrant. This area right in here, so are left bound is going to be zero. And a right bound is gonna be where these two curves intersect. It will be at four equals x two, the 2/3. So it's so for X. Well, we can raise both sides to the third power four Cube gives a 64. The threes cancel and we're left with X squared. So X equals eight. And that will be our rate bound. So are integral. Goes from 0 to 8 and we're doing for minus X to the 2/3. And if we integrate, we get four x minus 3/5 x to the 5/3 and we'll pull you in eight and get four times eight minus 3/5 eight to the 5/3 and then minus plugging in zero, we just get zero. So this leaves us with 32 minus 3/5. Eight to the 5/3 is the same as saying eight to the 1/3 raised to the fifth Power and the cube root of eight is too, and to to the fifth is 32. So what we really have is 3/5 times 32. So are areas 30 to minus 3/5 times 32 which is going to be 2/5. Time is 32 or 64 over five, and that is our bounded area.

The area bounded by two curves on a closed interval from a to B, um is determined by while the area A is equal to the integral, going from a to B off well, f of y minus g of Y D Y, where we divide these shaded region into two parts. Column A one and a two. And then we evaluate, um, each of them separately and then add them up. So for the limits of integration of a one, we have three minus y is equal to two times the square root of why So we just square here and this implies that three minus why squared is then equal to four wine. We just square both sides here to the square root here, um, and then, well, three minus y squared. Um, this is gonna be, what, nine plus y squared minus six. Y, um is equal to four. Why, we said is equal to zero, and we have y squared minus 10 Y plus nine is equal to zero. This is gonna factor as why minus one times Y minus nine is equal to zero. Okay, so now we have, um Well, here we just take y 01 because Y is equal to nine. Um, doesn't satisfy here, So we take y being equal toe one. Okay. And here we have a equal 01 and b equals one. And we have f of y is equal to two times the square of y and g of y is equal to zero. Therefore, we have a sub one is equal to the integral, going from zero toe, one off, two times the square root of Why de y Okay, if we evaluate this, um, this is gonna be equal to Well, we have to y to the three halves divided by three halves, evaluating from 0 to 1. So we get four thirds and then just times while one minus zero. So this is equal to four thirds. And then for the limits of integration of a sub to I'm second area here. Well, we have why minus one squared is equal. Thio three minus. Why? So that implies that why squared? Um, well minus two y plus one is equal to three minus y. So you confess is equal to zero and we have y squared minus Y minus two is equal to zero So we factor this. This implies that we have why, minus two times why plus one equal to zero. And here we take that y is equal to two because you have y being greater to zero. So one doesn't satisfy here. Okay, so again we take a equal to one, be equal to two, and f y is equal to three minus y and g f y equal to y minus one squared. And that therefore we have a sub to equal to the integral, going from 1 to 2 off well of three minus y minus. Why? Minus one? Whoops. Why? Minus one quantity squared de y. Okay, so we evaluate this and we get this is equal to get the integral from 1 to 2 of just, uh, two plus. Why minus y squared de y. So therefore, this is going to be equal to while we get two y plus y squared over two minus. Why so on stretches down here. We get to why. Plus why squared over two minus y cube over three. And then we are evaluating this from 1 to 2. Okay, so we first plug in to instruct off. We give you plug in one. So we get four plus two minus a thirds and then minus to plus one half minus one third. So that gives us four minus one half minus seven thirds, which gives us 21 minus 14/6. This is equal to 76 Okay, so therefore, the total shaded area is just equal to a one plus a two. That's just equal to four thirds plus 7/6. So that's just equal to come. Denominator. Here it's gonna be equal to eight plus 7/6, which is equal to 15/6. So therefore, um yes, the solution here, the area bounded by these curves is 15 6. All right, take care.

So we want to find the area of this region that's pictured here. So we're given several functions of X in terms of why, and you see how maybe we can sort of separate this into two different regions that are bounded by two curves. So if I draw this horizontal line, X is equal to sorry. This horizontal line Y is equal to one we see below the line. The region is bounded to the left by just the Y axis that's X equals zero and to the right, by this curb, excess to square roots of why and above that our upper bound is three minus wise, equal X, and are lower bound his exes. Why minus one squared. So the area we can write down as being the integral first from 0 to 1 with respect to why of to square roots of why minus zero. Okay, because the Y axis is the graph of X equals zero. All right, so then yes, sir, than d y here. And then we would just have the integral from 1 to 2. Now what's our upper function is three minus y and then our lower function is this one red. Okay, so these two in a girl's add that together gives you the total area of the shaded regions. Now we won't evaluate thes because it should be straightforward. You just expand here on the right hand side. You're just integrating polynomial. So to find the anti derivatives, just reverse the power rule. If you're confused about how to actually integrate thes these two look back to some previous videos where we work through all the steps and full detail.

In this problem. We're looking from the area bounded by y equals one plus rid X y equals two of a rude X Y equals X over four and the y axis. So I've done the graph, You're ready and you'll notice that I've actually put another line right down. Why equals one? That's because, ah, you know, we've got multiple curves here. Normally we're just used to dealing with too. But because we've got three of them, we're gonna have to separate this into a couple of different regions. Uh, for the first region, we are going to take the area from 0 to 1 of one plus rejects and minus the red curve. And for the second region from 1 to 4, we're going to take the area using the black curve minus the red curve. So we'll have to add up the two areas so area is equal to the integral from zero toe once of the blue curve minus the red curves. So one plus rude x minus X over for the ex, plus, the second area is from 1 to 4. Black curve would be to overrule Dex minus X over for the ex. Luckily, for us these air not difficult expressions to take the anti derivative for so we can go straight to it. That will be X. Ah, that's X to the power of 1/2. So we know we want extra power three over two and then taking into account the coefficient, that'll be two over three X to the three over two. And that's one over four X. So that's gonna be minus one over eight X squared, evaluated from 0 to 1, and then we gotta add it to anti derivative for the next one. So that's two times X to the negative 1/2 which makes it for X to the power of 1/2 minus won over eight X squared, evaluated from 1 to 4. So we've got a little bit to evaluate here, putting one into our first bracket that will be one plus two over three minus one over eight. I'm gonna skip putting zero in there because that old zero every time add it to the next one. Putting four in there, it's gonna give us, um, eat minus two and then putting one in there will give us ah four minus one of rape S O. If we group the first racket together as one fraction, that will be 37 over 24. And if we group everything in the second fraction that that's a being 17 over eight, giving them a common denominator and simplifying will actually give us the area of uh 11 over three.


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