find-the-areas-bounded-by-the-indicated-curves-yx-4-sqrtx-y0-3

Question

Find the areas bounded by the indicated curves.
$$y=x-4 \sqrt{x}, y=0$$

Tyler Moulton · Accepted Answer

we want to find the area bounded by the curves, Y equals x minus four. Route X and Y equals zero. This question is challenging our understanding of applications of integration and particularly testing our ability to find the area between two curves which is for vertical components equal to B Y U minus Y one X. And the horizontal components I equal interval C two d x two minus x one dy. If we sketch these curves out we can identify the area that&#x27;s founded and then we can identify the bounds and relevant functions. Didn&#x27;t do so. If we sketch this out we have four minus X. Going down to the origin and up between Y equals zero. In this curve we see the small mountain shape underneath the X axis which has intercept 16. Thus if we integrate from 0 to 16 4 minutes forward. Ex excellent. Forward X. We can solve that. We have a equals equals 0 to 16 X minus forward X equals expert over 28 30 X three half to 16 or area 128 33 where taking the magnitude or absolute value so that the area is not negative.

Dillon Huddleston · Answer

eso this case, we can integrate respect. Why now? To find the bounds of integration. Uh, we can solve for why? Actually, in fact, we we will be with respect. That&#x27;s not why, um so to find the bounds of integration eyes fax wi told both equations for why and said results people one another. So that gives for X squared minus for or four minus four. X squared is equal to, uh, X to the four minutes one. So we have a cortical equation, then zero equals, except for some plus four x squared, um, minus five. So, uh, that then is, um x weird. Then x squared. That&#x27;s five times x squared plus on which yields again, except for minus for X squared minus five is expected and shall we solve for again? Since that&#x27;s equal zero, we have the roots ah x squared equal to x ray. Get expertise for the five and extorted with minus one. I have the X squared equals minus one. Gives roots positive minus I. That is imaginary roots, not meal roots, which you couldn&#x27;t ignore. S we have only to consider the square roots. Well, the values X is putting minus Route five. And those, um, those are br pounds of integration Exes minus five. Positive route. Uh, two verified the values of Bryton posit Mayes. Uh, room five million on X word is five. We have four times five is 20 four minus twenties, minus 16 for why Stands on quick correction that are cortical equation is X to the war plus four x squared, Um, minus five, not minus four. It&#x27;s it&#x27;s weird s o then we factor prize. That is X squared. Plus five Mother Bye bye x squared by this one which gives roots husband through five I, Patrick ignore and husbands, what in there are substituting those roots and just to check, we have purposes plus minus one that gives from first equation four times amplifies force y zero. And, um, putting in plus minus one to the floor is one minus lives. One also gives y zero. And just like the other two routes on that fugitive plus minus ah, square root five I square Now we have minus five. So lettuce 20 plus four is 24 is lie and then we have minus five. Squared is 25 someone is 25 is minus 24th mind swelling to get wise. 24 s o the bounds of interrogation then are between exes minus one and positive on because we ignore the plus finance screwed five i since they&#x27;re not real. So, um, next thing is to determine which curve is the upper and pushes the lower, um So the curve why is excited for minus well is an upward pointing Kartik, whereas the function why is four minus for squared is a downward pointing. Quadratic this with a quad drive because our upper curve in the quarter over Kurds. So we then want to integrate the difference of those which is four minus for X squared, minus extra before um, monies a minus. My insolence, sir. Plus one, it&#x27;s altogether than me and five minus four. It&#x27;s weird minus X to the fore and integrating done five eggs 65 over five a month is 4/3 excuse. And again it&#x27;s we buy from, uh, X is mine. It&#x27;s Wonder Woman. Um, now, since this region of integration Rex between minus one on one, it&#x27;s messing about why access then the anti derivatives has only on powers of X, that it&#x27;s odd we conclude that its value at minus one is the negative of that value at one. And so we can actually just evaluate, uh, by this, the cemetery property who can just evaluate the integral from zero and multiply the result by two. And of course. But you did it. Zero a, uh, anti driven and vanishes. Just evaluating one. Were Chanel starting the coefficients of the powers backs. That&#x27;s five minus of such 25 months when it&#x27;s 24 over five minus four over three. So that&#x27;s, uh, 75 over 15. It is 20 over 15 so it&#x27;s 55 over 50. Wish is 11 over three, actually, since we had 24 over five, that&#x27;s 72 over 50 minus 20 over 15 which is 52 over 50 and we multiply that by two, peeled 104. And given that five divides 105 and three divides 102 we conclude that this fractions fully reduced form and also know that you for but by 15

Bobby Barnes · Answer

we want to find the area enclosed by these two curves that they provide us. So the area were interested in to this area in green here. So now the first thing we&#x27;ll need to decide is Do we want to integrate this with respect acts or with respect to why so for me, I would think to find that area there. It would be easier to do it with respect to X because notice that if we look at this going from bottom to top, so we enter the region at only one point the blue line and then we exit through on Lee at the red line. But if we were to do this from left to right than notice going from left, we enter the region through the red and then through the red again over here. Then down here we enter the region through the blue and then through the blue again, so we would have to set this up into two different intervals. But in this case, we can only use one, so it&#x27;s probably better idea to go about that way. So that&#x27;s going to tell us, though, that the blue line will be our lower and the red is going to be our upper. Well, we need to get this in terms of X. So let&#x27;s go ahead and do that. So if we were to subtract the four x squared over, we&#x27;d have Why is it too or X square got subtracted. So four minus for X squared. And then for the blue, we would want to add the Y over subtracted one. So we get why is equal to X to the fourth minus one. Now we still need to figure out Well, what are our points of intersection right here. And to find those, we would want to set these two equations equal to each other so it have, or minus or X square equal to X to the fourth minus one. And if you were to move everything over to one side factor, everything this should give us that X is equal to plus or minus one. So let&#x27;s go ahead and well, everything in now. So this is gonna be negative one toe, one for our bounds. And we said our upper function is that red lines of four minus or X square, and they were going to subtract off the blue so negative X to the fourth plus one dx and let&#x27;s go ahead and simplify all this Damn. And you might notice that we&#x27;re going to be left with and even function that we&#x27;re integrating. So this here is even and by even what I mean is if we replace all of the exes with negative X, this will give us the original function. And the nice thing about this is we can rewrite this as the integral from zero to our upper bound, and the rest of the function states the same. And we might want to do this because plug it and zero is normally easier than plugging in any other number. Now we just integrate everything using power rule. So it be two times five x minus 4/3 X cubed minus 1/5 x to the fifth. We evaluate from 0 to 1, so plug it in one. We should get five minus 4/3 minus 1/5 and then we subtract off when we plug in zero, which just gives us Syria. So this last term just goes away. And if we were to add all of this up, I believe it should give us 10 4/15 or the area of that region

Tyler Gaona · Answer

find the area of the regions and closed by this curve like X squared of a squared minus X squared. And why equal zero? That&#x27;s just the X axis. First, let&#x27;s just try to sketch a picture of what&#x27;s going on. So first for this this curve, let&#x27;s just try to see I mean, for road values of X. Does this even make sense? So we have to have that X squared is less than a squared for this argument to the square root function to be positive. So basically, we need the absolute value of X to be less than a for this to make any sense. So let&#x27;s see what happens. And, of course, if you plug in excess, eh? Or exes minus say here, we&#x27;re going to get zero so years, eh? We&#x27;re giving. That is positive. Here&#x27;s minus a. Also noticed a few plug in x zero here, you get, you get Why&#x27;s he called a zero. So if we sketch this over here, we&#x27;re gonna have I don&#x27;t know, something like that. And then notice a few plugin Negative X. You just factor that out and you get negative of why. So this is actually just an odd function. So it&#x27;s going to look something like this. So if we want to find the area between these two guys, Well, what we want to do is just integrate. Well, maybe I won&#x27;t do that in red. You just want to integrate for minus a a dysfunction minus the function Y equals zero. So this is just what we&#x27;re doing, Okay? And since this is an odd function, it turns out that this is the same. It&#x27;s just twice if we just look att going from zero to a All right, So now how do we evaluate this cinder girl? Well, this seems like a good candidate for substitution, because here I haven&#x27;t X squared in some constant under the radical. And I even have a two x dx hanging out. So if I make the substitution, you is a squared minus X squared. Andy was minus two x d x. I can replace two x t x by minus. Do you? I can replace square root of a squared minus x squared by just squared of you. And then let&#x27;s change the limits. So if I plug in x zero, I get it a squared for you if I plug in excess, eh? I get zero for you so I can flip around the limits if by absorbing the minus sign to you. So the anti derivative is just you to the so I raised the power by one gives me three halves. Divide by that power I want evaluate this between a squared and zero. So if I plug in a squared, that&#x27;ll cancel with this 1/2 power. And so I have to a cubed over three giving me my area.

Bobby Barnes · Answer

we want to find the areas enclosed between the region. Why is it was zero. And why is equal to X times the square root of base with my sex word where a is strictly larger than zero. So the region that we&#x27;re interested in is the one I&#x27;m going to shaded red over here. And so normally we would just use the area between the two curves by the upper Meisler. But in this case, we have to separate regions that were working on. And it may help for us to think about this in another way. So the balance for that we&#x27;re just gonna be negative. Eight A. And we could think about it as instead of breaking up into two regions as absolute value of our first function, minus our second. So in this case, it doesn&#x27;t matter how we write this, but the reason why I&#x27;m goingto want to write it like this is it will help simplify everything a little bit. So minus zero isn&#x27;t going to really do anything. So we could get rid of that. And now the reason why I did this is about to be a little bit Maura parent now Oh, I don&#x27;t know what happened with that. We&#x27;re discreet right now. The absolute value of X times screwed of a squared my sex where this is an even function on a symmetric interval. So remember, even means f of negative X is equal to f of X. So if you were to plug in negative X for all the exes, you&#x27;ll see that this is true. So we can rewrite this as two times are integral from zero to a Remember, we can only do this since it&#x27;s a symmetric interval. And then it would be X times a square minus square root of X squared DX. So have this here now. And the rumors and why I wanted to do this is on this interval here it was negative, but on the interval, restricted to now it&#x27;s just all positive. So we don&#x27;t need the absolute value anymore. And we can just get rid of this and then go about using a U sub to integrate this. So I think it would be a good idea to let you equal to a squared minus X squared. And then we get the differential. D&#x27;you is equal to negative two x d x. Now the only thing we&#x27;re missing is that negative. So why don&#x27;t we just multiply each side of our differential by negative one? And so then that&#x27;s going to give us that two x d x. Is he going to negative to you? And that, too we&#x27;re getting is outside there. So all that becomes, do you and then we&#x27;ll have the square root of you, do you? And then negative. So there are negative on the outside There. Now let&#x27;s figure out what our new bounds are gonna be. Well, when X is equal to zero, then you is going to be a square, are lower bound, is a square. And when X is equal to a well, we&#x27;ll get you is even too. So a squared minus a squared, which is serious or upper bound is zero. And another nice thing is, when we have a negative, we can use that to flip our boundaries here so we can rewrite this as negative zero to a square of the square root. And I&#x27;m gonna rewrite the square root of you as you to the 1/2. Do you Now let&#x27;s go ahead and move this up here so we can integrate that now. And actually, I got rid of that negative to put those bounds now, integrating that will be with power. Also BU to the three halves times the reciprocal of it&#x27;s new power. So 2/3 and we evaluate from zero to a squared. So plug it in a square that should give us 2/3 a cube and I&#x27;m plugging in zero. We just get zero. So this here would be our area.

Dillon Huddleston · Answer

eso here we can with respect to why I know to find any bounds of integration in solve for X of huge curve. And I said the results of the equal tonight yields four minus four wives squared equals, um one minus one of the four which we can be expressed as where did the four minus four y squared? Um, plus me or uh, no. Why did Before, minus four y squared plus de eso We can then factor that as Why squared monies 30 I squared minus one. I&#x27;ll give this again, That&#x27;s where. Minus, why didn&#x27;t four months for y squared plus me? And so we conclude then that the roots of that equation satisfy y squared, uh, three. My squared equals one. That is why is plus minus one and plus minus Route three. Then we could verify that by Princeton&#x27;s first putting in plus minus one gives X Plus four is four. Rex is your impersonation, and X plus one is one again. Exit on the second equation. And for possible sweetie express four times three, uh is for Rex is mine and say in the second phase, X plus nine is one. So organics is minus eight. Um, so O r uh, the bounds of integration For why? Or the, uh, the pretend the potential points One who through immigration are my little shrew three minus 11 Route three. And we also have the constraint that X is not negative. Um, s So first we can consider that wish curve belies lies below the other, which, uh, Polish values of why so we have one minus y to the four is a downward pointing quadratic. Uh, it intersex X equals zero a plus, minus one. And we have four minus four by squared that were pointing quadratic, which also just like their X equals zero. Uh, what is plus one? It&#x27;s one s. So we actually own any int integrated Therefore, between his minus wise minus one wasp Positively. Uh, because further values of why by accident And we again, we already checked for pas minds through three values of X for both curves. Those being there are other points of protection. Reminded say that yes, we have only one integral to evaluate between my eyes minds 11 for instance, why zero the quadratic has value for on the court has why once all the quarters are over function. So we have into grand four months for Why Square? That&#x27;s right before So why did the four minus four y squared plus three, which we can integrate, wanted five, but about outside on us for Kurds. Want cute? Plus, do you want? Unless we want to about right between minus 11 again, since for the roots minus my room three and impact all values. That&#x27;s why I created this one. Of course. Find bodies, Baxter Negative. And we&#x27;re bound for ex non need. Um, so we get anyone by between months on one. Now, since that interval symmetric about the X axis and the function has only with a little meal without orders because only odd Howard&#x27;s of why Ah, such a function is odd. And given this in this mansion interval about the ex access on weaken just integrate from zero on multiple result by two well, now evaluated it. Zero. Yeah, too derivative. Zero. So much is about a rate of one, which means something. The coefficients That&#x27;s 1/5 but it&#x27;s 4/3 which is three minus 20. So minus 17 over 50 ah, plus three. That&#x27;s 45 minus 17 is 28 divided by 15 and, uh, which we about to get 56. No. Five divides 55 3 divides. 57. We conclude that this faction is, uh, really reduced. It also that this value 56 to vote by 15.

Problem

Find the areas bounded by the indicated curves. $…

Problem 17 Easy Difficulty

Find the areas bounded by the indicated curves.
$$y=x-4 \sqrt{x}, y=0$$

Answer

$$A=\frac{128}{3}$$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Grace H.

Catherine R.

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Integration Review - Intro

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$$A=\frac{128}{3}$$

Basic Technical Mathematics with Calculus

Grace H.

Catherine R.

Heather Z.

Caleb E.

Integration Review - Intro

Definite vs Indefinite

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

Grace H.

Catherine R.

Heather Z.

Caleb E.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus