find-the-areas-bounded-by-the-indicated-curves-using-a-vertical-elements-and-b-horizontal-element-12

Question

Find the areas bounded by the indicated curves, using (a) vertical elements and (b) horizontal elements.
$$y=4 x, y=x^{3}$$

Lucas Finney · Accepted Answer

for this problem we are asked to find the areas bounded by the curves, Y equals four X and Y equals X cube. So our first step here is to find their points of intersection. So I will be when four X equals X cubed or when four x minus x cubed equals zero. Or can factor out the X. We&#x27;ll have then that that will be when X times four minus X cubed equals zero. So that then means that we need to have either X equals zero is probably our left hand point of intersection. And or we need x cubed to equal four. So we need X two equal forward to the power of 1/3 Cube Cubed Root of four. So to begin we need to have Yeah or actually let me just do a quick sketch here. We know we have four X which all just draws a straight line. We have x cubed. So we know that we&#x27;ll have one point of intersection at 00 and the other point of intersection will be at X equals four to the power of 1/3. I&#x27;ll write that down here Actually We had four to the power of 1/3 Or cubed root of four. And the corresponding why value would just be why equals for so for part A we&#x27;re asked to find the area between the curves using vertical elements. So we know that we&#x27;re integrating from zero up to 4th power of 1/3. Our upper function there is our straight line. So we&#x27;ll have four x minus X cubed dx integral of four. X would be four X squared over two. So two X squared integral of X cubed will be X power 4/4. We&#x27;re integrating from zero up to 4th power 1/3. Yeah, so we&#x27;ll have two times four to the power of 1/3 squared or so. That should be two times four to the power of to over three minus four. To the power of 4/3 over four. So that would just be four to the power of 1/3 which has a value Decimal value of about 3.45. Then we are asked to go through the same process but using horizontal elements instead. Well we were given that Y equals four X. So we need to express this as X equals Y over four and we have Y equals X cube. So that will become X equals Y to the power point over three. And we can see that using horizontal elements. Our upper function, so to speak is going to be that wide power of 1/3 and will be integrating for why between zero and four. So we&#x27;ll have wide span of 1/3 -Y over four. Dy now integral of white power of 1/3 will be wide span of 4/3 divided by 4/3. So that&#x27;s times 3/4 Then integral of why over four will become y squared over eight. 1/4 times y squared over two. So you have minus Y squared over eight, evaluating that from zero up to four again will give us a final result equivalent to 2 to the power of or two times 4 to the power of to over three -4 to the power of 1/3, Or approximately 3.45.

Tyler Gaona · Answer

So it&#x27;s probably easiest to write thes to curves as functions of why, in terms of X, so I&#x27;ll just write thes as Why equals X cubed and here wise equal to three X squared minus four. So to find the area of the region between them, the first thing we should do is figure out where these curves intersect. So for that I have to set them equal, which, if, after moving everything to one side, is equivalent to writing, writing this. Okay, so now let&#x27;s see, how can we solve this equation for X? It&#x27;s not just a quadratic, so we can&#x27;t just plug into the quadratic formula, but let&#x27;s see what we can do. So maybe one thing we can try here if we don&#x27;t have, say, access to a computer, which Khun just solve this for us has made. Let&#x27;s see if we can maybe just find a solution just by looking at it, we might try some easy numbers to plug in. So if you plug in, say, just one, for instance, you get one minus three plus four. Okay, so that&#x27;s one minus three is negative two plus fours, too, so it&#x27;s not zero so that&#x27;s not a solution. But say, if we had a minus one minus three plus for that would be a solution. And you can check if you plug in X is negative one. That&#x27;s what we get. So few plug in negative one. Here you get minus one minus three plus 40 So tell this negative one is a solution. Meaning we can factor like this. So x minus minus one. All right, Hoss. Explosive one. And then we&#x27;re gonna have some quadratic term. Okay, so I know that I have to have an x cubed. So that&#x27;s why no, they&#x27;re coefficient. Here is one. Because when I multiply these, the only way I&#x27;m going to get next cubed is from thes two terms. And then I have some linear term. And then I know also that to get a four here, so I want thes to multiply thes to to multiply out and give me execute minus three X plus four. To get this for is the constant. While I have one times something so that something has to be four. Okay. And so let&#x27;s figure out what this is actually supposed to be, so we can just multiply these out and then compare the coefficients of there What we get when we multiply it with the polynomial we started with. So if I multiply these out, I get next Cube. And now let&#x27;s see what we have on an X squared I have on a X squared plus a one x squared. So write that like this. And what about the linear terms? Well, I have a plus four x one person X and then I just have my plus four for Constant. Okay, so this is what we get if we multiply these two out, and now let&#x27;s compare the coefficients well, appear on the linear term. The A is there is no linear term. So this four plus a should be equal to zero, which just tells you that a should be minus for okay and we can check out, You know, if we look at the quadratic term, a minus four works there also. Okay, So that means then we factor this properly as explosive one x squared minus for X plus for equal zero. OK, And so now we have this quadratic and it looks like we should be on a factor this is exposed. Swon. This looks to me like it&#x27;s going to be X minus two squared. Okay, so we have the intersection. Points are when X is equal to minus one. And then we have a multiple route when X is equal to two. But still it&#x27;s intersecting at minus one, and X equals two. So now that we know where the curves intersect, we can if we can set up an integral to evaluate the area of the region. So it&#x27;s C a first. Let&#x27;s just try to sketch a picture. What&#x27;s going on? So here is negative one. Here&#x27;s to Okay, so then I have X cubed. No, which looks something like that. And then I have three x squared, minus four sets of parabolas, some sort. Okay, so this is what we have. So this proble So it&#x27;s a problem of this three is stretching it a little bit and minus four is taking it down. And since we know execute, I mean, you can just plug in X equals zero here and see that zero is greater than minus four. That tells you the green function is on the bottom. The red function is on the top. And so then the area is going to be the integral from minus 1 to 2 of the red function that I&#x27;ve drawn here X cubed, minus a green function. Okay. And I&#x27;ll leave the evaluation of the center grow up to you. Since we&#x27;ve there are plenty of examples in the previous videos evaluating in a girls like this, the important point is just being able to get here and setting it up. So we found the points of intersection figured out which graph which function is above the other. And then we use that to set up our integral that gives us the area of the region between the curves.

Bobby Barnes · Answer

we want to find the area enclosed by these two curves that they provide. So that area that were interested in would be this area right here. And in this case, it probably be better too integrate with respect to X. Since solving for X is gonna be pretty straightforward as well as if we were to start from the very bottom down here and then passed through the region, the blue line is always gonna be on the bottom, and then the red line is always going to be on top as opposed to If we go from left to right at some point, the right ID is on the bottom, blue on top, and then we just have the blue all together on the bottom, so becomes kind of confusing on how to set the intervals. So we&#x27;re gonna let this be our lower and the red be our upper. So we need to solve for y in each case for the red, it&#x27;s just gonna be why is equal to x cute and then the blue we should get why is equal to three X spared minus four now to figure out what our points of intersection here. So it should be like these points right here. We would set our two functions equal. So we have execute is equal to three X squared, minus four. And so you move everything over to one side factor, do all that. And in the end, though, we should end up getting that X is equal to negative one and two. So we know what our bounds of integration are gonna be. So let&#x27;s go ahead and set it up now. So it should be negative one, too. Of so he said, our upper function is execute and our lower function was that quadratic. So three X squared plus four d X. Now we can use powerful to integrate. So the 1/4 ex to the fourth minus X cubed and then plus four X we&#x27;re gonna evaluate this from negative one toe, too. So plug it in Tuesday to to the fourth of 16. Divide by four gives us four and minus eight plus c. Then we subtract when we plug in negative one, which you give us 1/4 must one minus a war. And so well, first these eights cancel. Oh, and then what do we get with? It looks like it gives us 27 over four or that area that we shaded in

Tyler Gaona · Answer

so to find the area of the region between these two curves, it looks like I would prefer to write thes as functions of X in terms of why rather than the usual why in terms of X, because I don&#x27;t want to take square roots on the why. So I can write thes curves as X is equal to three minus y squared and this one would be X is equal to minus y squared over four. Okay. And so to find the area of the region between them, the first thing we should do is figure out where these two curves intersect, so we can set them equal. And so this is giving us three is equal to Let&#x27;s see, I have minus y squared over four plus four y squared over for gives me three. Why squared over four? So then why Squared is equal to four gives me wise Plus and minus two are the points of intersection. Okay, so then the question is well, which function is above in which function is below. We could just maybe test the point in between since there are no other intersections. So maybe if we just plug in wise equals zero. We see here when y is equal to zero, X is equal to three. And here X is just equal to zero. So this one will be the upper function, and this will be lower function. So now the area then set up by this animal where we&#x27;re going to be integrating. Why between minus two and two and we&#x27;ll take our upper function minus or lower function. So that&#x27;s three minus y squared. Plus, why square before you should be giving us this, Okay. And then we can just find the anti derivative by reversing the power rules. So you raise the power by one and divide by. That s so Yes. So let&#x27;s see here we should have. Why Cuba and then divide by three. Just let me just Only that is why cubed over four. Okay. And then we&#x27;ll plug in between wise too and wise minus two. And so this gives us six. Mine is too cute. This eighth Air Force Two. And when we plug in negative two, we have a minus six here and then a plus two. So at the end of the day, it&#x27;s just 12 minus four, which is equal to eight

Bobby Barnes · Answer

We want to find the area enclosed by these two curbs here. So that&#x27;s gonna be this area and red. And so you might notice. Looking at these equations, it&#x27;s easier to solve in terms of X as opposed to a why. And also even looking at this region here, if we were to start on the left and then go towards the rate, we have only one lower function and then only one upper function. But if we were to go from bottom to top, well, we&#x27;d have to break this up in tow to separate intervals as well, a cz solving for why we&#x27;d have, like, a plus or minus for each due to the square roots. So it&#x27;s probably a better idea to Seoul. Everything&#x27;s in terms of why and then integrate with respect to why. So let&#x27;s go ahead and figure out what are two points of intersection right here are now. We can do that by setting them equal. Let&#x27;s solve for X first we&#x27;d get X is equal to negative 1/4 why squared and then for the green equation over here, we should get X is equal to three minus y squared. So we set those two equal. So you get negative 1/4. Why Squared is equal to three minus y squared and you can go through and solve how you normally solve quadratic. But this should give us that. Why is he Goto plus or minus two? So we go ahead and plug everything in now. So our bounds of integration gonna be negative. 2 to 2 here upper function is that green. So three of minus y squared and then our lower function is the negative 1/4 y squared. Subtract that which wouldn&#x27;t make it actually adding it and then, well, go ahead. A combined aren&#x27;t like terms. Now, anytime we have a symmetric interval, it&#x27;s a good idea to check to see if this is a even or odd function, because that only simplifies the calculations by a lot. And it just so happens what we are integrating is even so if we were to check to see if this is f of negative, why, If we were to replace all the wise with negative why we would just get our function out again and that would turn this into two times integral from zero to our upper bound to and the rest of the integral states to say, And we would want to do this because it&#x27;s easier to plug in zero than it is to plug in negative, too. Right now, we can use powerful to integrates would be two times three. Why minus 1/4 y que because we would divide by three of those cancel. Then we evaluate from 0 to 2 so plugging in to we should have six minus two, and then when we plug in zero, we get zero. So that should simplify down to eight or that area there.

Tyler Gaona · Answer

let&#x27;s write thesis functions of X in terms of why and then we can integrate with respect the why to find the areas. So it&#x27;s fine where these two curves intersect. So I don&#x27;t think this is a simple was just maybe solving a quadratic equation and factoring. But I think we can just maybe observe where these two points intersect. So if you just look, if you plug in, why equals one, you know, raising one to the fourth power here it&#x27;s raised to the squared. We still get one. And if if you plug in negative one also you get one. And we had wood. In either case, we would have one on the left is equal to minus one on the right. So those so why equals plus or minus one satisfies both of these equations, and you can check by graphing both of these. The&#x27;s give you the points of intersection. Okay, so now to find the area, you want to set up the integral from minus one toe one, we should figure out which of these is the function, our top function, which is our bottom function. So imagine if you just plug in, say, Why equals two here? We&#x27;re going to get something. Very. Ah, sorry. We&#x27;re not going that for if we plug in Y equals zero here we get X equals zero. And here we get X equals two. So too is greater than zero means that this one is our top function. So we have two minus y to the fourth and always attract. Why did the 2/3 D Why? So we can find the anti derivative as to why, minus y to the fifth over, five minus here we would have wide of 5/3. Dividing by that power gives us 3/5 this evaluated between one and minus one. So we have to minus of fifth minus 3/5 minus a negative, too. And then here we would have plus 1/5 plus 3/5. Because these air odd powers. So we really just get for minus 2/5 minus 6/5 which is 20 minus eight over five, just 12 5th

Problem

Certain physical quantities are often represented…

Problem 42 Easy Difficulty

Find the areas bounded by the indicated curves, using (a) vertical elements and (b) horizontal elements.
$$y=4 x, y=x^{3}$$

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