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QuestionSet up the definite integral that gives the area bounded by the curves y-2x and y-x2J6 (x+2xldxJ"(x_2xldx(2x -xldx(xz 2x)dx"(2x- x?Jdx...

Question

QuestionSet up the definite integral that gives the area bounded by the curves y-2x and y-x2J6 (x+2xldxJ"(x_2xldx(2x -xldx(xz 2x)dx"(2x- x?Jdx

Question Set up the definite integral that gives the area bounded by the curves y-2x and y-x2 J6 (x+2xldx J"(x_2xldx (2x -xldx (xz 2x)dx "(2x- x?Jdx



Answers

Use a double integral to compute the area of the region bounded by the curves. $$y=x^{2}, y=x+2$$

Okay, So to find a area, uh, the region bounded by these two functions. So when you to set these two functions eat well so that we can find this intersection point. However, we have this function in terms of acts in this function in terms of why so we either need to solve this one for ATS are sold. This one for why, in this case, I'm gonna solve this for why? So this functions also gonna be Why equal to the square of X? Okay, so then with this now I can go ahead and set thes two equal and on the intersection point. So with that, I have X squared is equal to the square root of X. Then if I square both sides and get X to the fourth is equal to X. Now if I bring this extra left hand side so I have X to the fourth minus X is equal to zero and here I can go ahead and factor out an X. So I'll get extra third minus one is equal to see your out. Okay, so this will give me two results, will have x equal to zero. And if I have X cubed minus one equal to zero. Then, in this case, X huge equal toe one. So this will say that X is going to be ableto one. So this means that, um, these two functions intersect at X equals zero and X equal toe one. So now with this I'm going to do, I'm gonna use vertical cross sections. I'm gonna do the top most function, which is a squared of X minus the lower most function here, which is gonna be X squared. So using vertical cross sections. So with that, I'm gonna do The area is equal to the inner girl of the top most, which is gonna be the square root of X minus X squared. Since this is the lower and then times, DX and X is gonna be from 0 to 1 case of them From here, we can just go ahead and I'm gonna rewrite the square root of X first. So that's gonna be you could rewrite the squared of X as X to the 1/2. Then this is minus X squared times, DX. Then we can go ahead here. I just find the anti derivative of each term, so the anti derivative of X to the 1/2 is gonna be 2/3 So 2/3 X to the three halfs and then minus the anti derivative of X squares on the X, cubed over three. And then we're gonna evaluate this from 0 to 1. So I'm gonna go ahead and plug in one, and then we're gonna plug in zero, and then we're going to subtract our results. So with that, we have to we have to over three times one to the three halfs minus one cube or three. And we're going to some shopping and I want to plug in his euros. We have to thirds time zero to the three halves, minus zero. Cute over three. So with that, we get a result of 1/3 or 0.333 and so forth. So that'll be the area of the shaded region

Okay, So to do this, I need to find the intersection of these two functions, so we're gonna go ahead and set the two functions equal to each other. So with that, we have X cubed is equal to X squared. And from here, I'm gonna bring everything to one side. So have X cubed minus X squared is equal to zero if I factor out an X squared, so that'll leave me with X minus one. So this will give me the X intersections. X equals zero and acts equal to one. So this says a zero they intersect and the X equal to one to this intersection right here. So then I can go ahead, and I'm gonna use horizontal cross sections. So I'm going to do the right most function minus the left. Most of the right most is gonna be X squared. And the left of us is gonna be execute so I can go ahead and find the area he's in the inner Groll of, So I'm gonna do you the right most, which is gonna be X squared and then minus the left most, which is gonna be x cubed, then times DX and so acts is going to be from 0 to 1. So then here we just go ahead and integrate finding the anti derivative. So the anti derivative of and X of X squared is gonna be X cubed over three. And then the anti derivative Oh, excuse is going to be extra fourth over four, and then we're going to evaluate this from 0 to 1. Okay, so we're gonna go ahead and plug in one. First, we have one cubed over three, minus one to the fourth over four. Then we're going to subtract that from. So now we go ahead and plug in zeros. We have zero huge over three minus euro to the fourth over four. And with that, we'll get a result of So it's a very small region that's shaded, So we should expect a very small answer. So our answer will be 1 12 So that's gonna be the area of the the region bounded by these two functions.

This problem focuses on finding the area of regions bounded by the curves. So our first function given is F of X equals X squared. And then our second equation given is f of X equals X. Because we're not given to balance that we have to determine Are balanced by where the two graphs intersect. We can do this just by equaling the two equations equal to each other or setting the two equations equal each other, X squared equals X. To do this, we have to subtract X both sides and then factor out it's this way we do not forget a factor. So we can determine that to grass cross at X equals zero and X equals one. So are bound for the integral are going to be 0-1. The next step is just to determine where or which graph mhm Is on top of the other two set up the integral correctly. So um uh an X squared. So we could say this is one, this is one. So both are going to intersect these two points right here, we change the color of this. These two points right here, the parabola X squared is gonna go wait this continue up, wow Y equals X. It's gonna go in a straight line like this. We're gonna find this area right here. So we can tell by this craft that Y equals X is on top of the X squared equation. So can set up X minus X squared D. X. Now if we solve this equation out or there's been a role, Our final answer should be won over six. Yeah.


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