find-the-areas-bounded-by-the-indicated-curves-yx2-y2-x-y1-3

Question

Find the areas bounded by the indicated curves.
$$y=x^{2}, y=2-x, y=1$$

Tyler Moulton · Accepted Answer

uh huh. We want to find the area founded by the curves, Y equals x square Y equals two minus X and Y equals one. This question is challenging understanding of how to apply integration for the purpose of areas derivation so it&#x27;s not appropriate definitions for the area between two curves for vertical components it is A equals integral A to B Y tu minus one. The X for horizontal components in a equals integral C two D x two minus X 20 Y. That&#x27;s if we sketch these curves out and identify the appropriate bounds and functions to integrate, we can solve sketching this out. We have the area in yellow as our founded area. We see that we&#x27;re going to have to integrate on two different areas. First we&#x27;re going to integrate from negative 2 to 12 minus x minus x squared to minus six Y. One. Or rather Y two X squared is Y two Y one. Then from negative 1 to 1. We integrate two minus x minus Y equals one. So this kid&#x27;s area two x minus X squared over two minus X, equal to negative one plus x minus X squared over two from negative 1 to 1 or 19/6.

Amit Srivastava · Answer

in this part we can draw. The graph is something like this. So ah, burning wounded part will be like this. And it is here x equal toe to park because if illiquid, why won&#x27;t acquired to try to all temple gate cans Also x equal to two whereby one is take personal first cup and why do it a second cup? So finally area will be Why one minus jailer due to DX No, it is did little to x minus two by x x squared minus two x plus two DX difficult to X minus two or DX excess choir minus two x Bless Toby X So it is If we consider that square managed to express to as dirty for two x minus two dx will be ditty later on. We&#x27;ll change the limit. X equal to zero equal to do X equal to two for six minus four. Answer to seen. So if the limit our scene Ifill port deal a report Execute Kojiro thought it Final answer. Maybe possibility, because limiter here always seem so. The final lots of baby occurred to Chico

Dillon Huddleston · Answer

eso here, The two Kurds, Um will the of a curve will be the constant sexualized too. And the lower curve will be the problem. Expert man, it&#x27;s too. Find the exc ordinance on any range of integration. We just set the two pervs equal. So we have to equal, I explore, advised to or explored his four ex is possible eyes to so integrated from X equal money is 2 to 2 the function to minus extra mai&#x27;s too or, uh, four minus X squared. So the anti derivative for that is for ex Miss X cubed but by three. In this way we get enough bob evaluate between exes minds two and two an impact. We can see that the value of the inter derivative them honest you, given that the derivative that&#x27;s odd is equal to the value of the anti derivative that to, uh, soon we subtract the value of the anti derivative. But mine is to just get another copy of the value of the inter do bit of it too, so we can just take the ladder and multiply it by two. Um, so then we have four tonnes. Two is eight by this 8/3. So 24 3rd say thirds and 16 3rd actually, then multiplied by two. To get 32 3rd you can see that this value 32 3rd

Bobby Barnes · Answer

So they want us to find the area enclosed between the regions of why is equal to X squared minus two, and why is able to? So I went ahead and already sketched a graph of this cause I feel like sketching. It gives me a better idea of how to approach these. So the area were interested in Is this area right here Now? Area between two functions is going to be the region or the area of the upper function, minus the lower function. So that&#x27;s the red line is our upper function each case, or in this case, and our green line is our lower function, so we can go ahead and write this essay. We&#x27;ll figure out our balance in a moment, but so it&#x27;s two minus X squared plus two DX, and that would simplify down to for minus X squared DX. Now let&#x27;s figure out what our bounds are. No, If you were to actually grafted still bit better, you probably already know what these two are. But in the case of you&#x27;re not having that, all we need to do is set those two functions equal to each other. So we&#x27;re gonna do to is equal to X squared minus to add the two over we get to is equal to x squared. Apply the square root property on each side. I&#x27;m sorry. This should be four here not to, but we apply the square root property which says X is equal to plus or minus two. So our balance are gonna be negative to two. And one thing you might notice Let me write that up Periods will. That four minus X squared is an even function. So why even? I mean, if we do f of negative X, that should give us f of X. And because of that, we can go ahead and write this as two times since we&#x27;re on a symmetric interval, 0 to 2 of four minus X squared DX. And we just want to do that because playing and zero is easier than Blufgan and negative too. So we integrate this now, though, so four gets integrated for X X squared is gonna become X cube 1/3 and we evaluate from 0 to 2. So plugging in to so going through all the algebra should give us 32/3, and then when we plug in zero, that should just give us zero. So get rid of the zero and our solution

Dillon Huddleston · Answer

eso for this question. We can integrate with respect. Why? And then I can see that the lower of the two curves will be acceptable. I swear, Blind song. Because that&#x27;s a upward facing proble I wish has value minus one when? Why zero? You want your eggs axis whereas the other function? Absolutely. Why? It has a positive skirt mind so I swear. Ah, it&#x27;s always not negative. Uh, for the more Yeah, really expect, uh, we only consider the right hand function between why equal to minus 11? Because only in ay Jay&#x27;s is one minus y squared positive. They&#x27;re well on negative and therefore does thes square root Will my soul I squared ah, equal a real number The wise at the complex values will only be considering, uh most ways of why between minus 11 and, uh, what we could know. Furthermore, is that our function the square root of woman is I square is zero. Why equal the plus minus one? And the problem y squared minus one is also zero i weikel plus minus one eso those air two points of intersection of the two points of intersection of the two curves another woman. I want to square. Square root, of course, is a circle on the X Y plane. It would be the right half of the This is the part of the circle to the right of the y axis, um, of the inner circle and of course, have to value. Why increases from 0 to 1? Why increases from zero to plus minus one? And so the function, that&#x27;s why screw Pullman Square root woman. It&#x27;s Y squared zero plus minus. One is zero Uh, the proble minus 10 s. We have just two points of intersection between these two curves. Um, no. So the underground will be a two body. Why square with minus last squared? Ah, Les Waas, where? Bonus one. And we can split this into the two house for a husband minus y. And what we can see further more is that this function both functions are even with us back to why s O then also is the difference of the two functions. And since we&#x27;re integrating from one equals minus on the positive one, uh, by symmetry since the interval, symmetric about the exact says As to the underground, we can actually just take integral from 0 to 1 multiplied by two for that purpose her into grand and zap survey. Why was your husband? Why? So why square root woman? It&#x27;s life squared. Plus one Les Waas. Where? On the tree derivative of that will be we&#x27;ll have. Why? Minus y cubed, divided by three. Ah, then will have. Ah! My national Monets lot squared minus 2/3 one minus one square 23 It is a checked appreciating the last term with respect. Why, if 2/3 lots of blood by three hearts as one the square root of all mine, it&#x27;s life squared. And then we want the bottles in my mind as to why that means we have, uh to I s So, in fact, we actually Lisa tried 1/3 of this difference. Um, so again, we have three halves multiply by third. It&#x27;s 1/2. Well, mostly I squared Swear. Ooh! And then we model bye bye, minus two. Why? So the two cancel the have the minus cancels this attraction and we&#x27;re left with, uh why? Time to square one minus y. I swear I was expected. Okay? And so this has we have to evaluate between zero most by the result by two eso evaluated at one with minus five squared zero sum. You only consider the 1st 2 terms, which is one minus other words. We have 2/3. Never that we subtract AA minus 1/3. So we had 1/3 army of one because of my sincere squares under three ounces one, we have one from the back, the term every multiply that by two to get to let me see that this value too

Bobby Barnes · Answer

We want to find the area enclosed by these two curbs here. So that&#x27;s going to be all of this area right here Now, since everything is in terms of X, what we should probably do is start on the very left of our region have been go to the right and once we enter this region, that&#x27;s going to be our lower bounce of the green and then keep on going through till we pass through. And so the blue is going to be our upper, and we would then need to figure out what are these two points of intersection. And to do that, we would sent these equations equal to each other like this, and the first thing we&#x27;d want to do is square everything. And once you go through all the steps, we should get wise, you know, to just plus or minus one so you may end up with some extraneous solution since we&#x27;re square and get. But if you were to go and check them in the end, only these two would work. So our new bounds are going to be negative 1 to 1 and our upper function that absolute value of why square root of one minus y squared and then we subtract are lower function. So being minus Y squared plus one deed, Why now? First noticed that this is going to be even function and by even I mean, we were to replace all the wise with negative wise, we should end up with the exact same equation. The nice thing about even functions is that we can write as two times the interval or the integral of zero to our upper bowel, which is this case is one. And the reason why I want to do that is really to deal with the absolute value that we have there. Because before we had negative values that were plug it in so we couldn&#x27;t get rid of it. And we would have to split this up until, like, two really big into roles if we still had the negative parts that were plugging it. But since now we&#x27;re on just interval 01 all of our outputs for the absolute value. Why? Well, it would be the same thing is if we just dropped the absolute value. So now, at this point, what I&#x27;m going to do is distribute this integral across due to the linearity of it. And the reason why I want to do that is because this first part we&#x27;re gonna have to use a use up to integrate. And the second part we could just integrate directly. So it should be two times intro from 0 to 1 of why sward of one minus y squared, do you? Why? And then plus two times in a row of +01 negative y squared plus one. Do I now this here. Like I was saying, we need to do a use of and we&#x27;re gonna let you eagle toe one minus y square. So that gives us d. Yu is eager to negative too. Why square do you know why? And so, if we were to move this too inside here and then multiply by negative one twice this ends up becoming he used was going to replace that we have negative, integral of so plug in and zero into here. So when why is equal to zero? We should get you. Is he goto? One start lower bound should be one. And then what? Why is he the one you should be equal to zeros or upper bound and zero. And that&#x27;s gonna be the square root of why we&#x27;re squared of you D u. And this second interview over here. Well, we can just use power rule to integrate everything, so it&#x27;s gonna be negative 1/3. Why cubed? Plus why? Evaluated from zero to what now? Integrating this second interval, we should get well. The square root of you is really you to the 1/2. So we&#x27;re going to use power rule for that as well. So it should be you to the three halfs power now and then we divide by its new power. So we&#x27;d get 2/3 up from how we evaluate this from 1 to 0 aren&#x27;t now plugging. And one for this. We should get negative 1/3 plus one and imploded and zero. We should just get zero and over here, plug in and zero. We should get zero put the plug in and won, which are not with plus two thirds because we&#x27;re subtracting it and one minus 1/3 is 2/3 that I should give us plus 4/3 and then adding those together would give 6/3 or two or the area of that region

Problem

Find the areas bounded by the indicated curves. $…

Problem 20 Easy Difficulty

Find the areas bounded by the indicated curves.
$$y=x^{2}, y=2-x, y=1$$

Answer

$$A=\frac{37}{6}$$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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Integration Review - Intro

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$$A=\frac{37}{6}$$

Basic Technical Mathematics with Calculus

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Integration Review - Intro

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

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Kristen K.

Michael J.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus