find-the-areas-bounded-by-the-indicated-curves-y6-3-x-x0-y0-y3-3

Question

Find the areas bounded by the indicated curves.
$$y=6-3 x, x=0, y=0, y=3$$

Tyler Moulton · Accepted Answer

we want to find the area bounded by the curves? Why equal six minus three? X X equals zero, Y equals zero and Y equals three. This question is challenging our understanding of applications of integration in particular how to find the area between curves. So let&#x27;s not appropriate definitions. The area between two curves A. Is given the vertical components as equal integral to be. Why do you mind if I won the X identically for horizontal components? It&#x27;s a equal to go see the D extra minus X 24 to identify the appropriate bounds and curbs. To integrate. We&#x27;re going to sketch this out first, then solve. So sketching this out. We have our area shaded in yellow as the relative region. Why equals 03 and X equals zero are marked. We also have X equal to minus one third. Y marked. To solve what we&#x27;re going to do is rewrite, X equals two minus one. Rewrite are like 63 X. As I&#x27;ve already having left exactly two minutes one third y and integrate from 0 to 3 from 03 This interval has ended around two y minus y, squared over six, from 0 to 3, or plugging in 4.5.

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

Robert Daugherty · Answer

section seven. Not one problem. 15 9 So they asked us to find the area that&#x27;s bounded between two curves for the X values from 0 to 4. So what I wanted to do first is just take a look at both of these curves. So the first function we had was what you see here, X squared over X cubed minus three acts. So that is the function you see, hear and read. And then if you look at the next curve we see right here Ah, one over X cubed minus three x That is the curve that you see in blue. And we&#x27;re integrating from the value of two over the to the value of four. So over that particular X interval. So where you can see is if I knew the area beneath the red curve and then I subtracted the area that was beneath the blue curve, I would have the area between the red and the blue curve. So what it tells me is the area between those two curves is going to be the integral X squared over X cubed minus three x the X minus the integral of one over X cubed minus three x dx. So I just need to find the integration and subtract those and then evaluate all of that from 2 to 4 to get my final answer. So let&#x27;s go work that out. So that was our graphical exploration here and now we&#x27;re just going to go back and do the algebraic and the calculus calculation. So I&#x27;ve just determined here that the area between these two curves is going to be the integral from 2 to 4 of x squared over X cubed minus three X minus one over X squared minus three x dx. So the way I&#x27;m gonna do is since I&#x27;ve gotta work on these inter gross wishes, work on these individually and then bring everything back together. So I&#x27;ve got to come up with to anti derivatives. So how do I find the anti derivative of X squared over X cubed minus three X t X. And then how do I find the anti derivative of one over X squared minus three x the X. So let&#x27;s just take a look at these in the first case, Um, I could, right. This is the integral So X squared, and here I could factor out an X on I&#x27;m left with what? X squared minus three DX. That is helpful because then this turns into the integral of X over X squared minus three DX. And now I can make a substitution. I could say let you equals X squared, Uh, minus three. Then you know that d you is going to be two x the X. So in order to find a two x d x How would need to modify this integral in that way? And now I see this is in the form of 1/2 and what you see here is just one over you, Do you? That&#x27;s pretty straightforward. So this is going to give me 1/2 natural log, absolute value of you, plus a constant of integration. So that&#x27;s just going to be 1/2 natural log of you, which is X squared minus three plus a consummate immigration. So that&#x27;s the anti derivative for the first part of this problem. Now, let&#x27;s go look at the second part of this problem. So how do I deal with this? And they&#x27;re sorry. That little bit of a typo here, this is X cubed, huh? X cubed minus three X. So how do I work on, um however, work on this, Um, What we can do? Let&#x27;s look a little bit of a trick first. But I could factor out this x cubed so I could write. This is the integral one over X cubed. And then I go right this one minus. And this will be three x to the minus two Power DX. So check that when I multiply x Cuban one, I get the X cubed when I multiply x cubed and three. Excellent minus two. I get three x to the first power. So doesn&#x27;t you? Might not think Well, that&#x27;s not really much of a help, but it is a help because now I could rewrite this as the integral of X to the minus third power one minus three x to the minus two DX and you&#x27;re still saying, How is this helping? Let&#x27;s let you equal one minus three x to the minus two. Then do you is going to be when you different shape this you&#x27;re going to get six next to the minus third DX. So if I saw a six here and a 1/6 here, then six x to the minus 30 eggs. That&#x27;s a d u. So this turns into 16 And then what you see here is one over you, Do you? And you know that this is going to be 16 natural log of you, plus a constant. And so this is 16 natural log. What is you? You is one minus three over X squared. So you have one minus three over X squared, plus a constant of integration. So now I have both, um, indefinite, Integral is. And now my gold is to use this formula to figure out what the exact area is. So I know from now that the integral from two 24 of X squared over X cubed minus three X minus one over X cubed minus three x. The X is going to be equal to 1/2 natural log X squared minus three and then plus excuse me, there&#x27;s a minus sign right here with general minus 16 natural log one minus three over X squared. And then all of this has to be evaluated from 2 to 4. So, calculus, part of this problem is over. Now, let&#x27;s just go substitute in and see what we, uh we come up with. So I&#x27;m going to substitute the values for two and four and see what simplification I come up with. So if I substitute a four, I&#x27;m going toe have 1/2 natural log four square to 16. 16 minus three is 13. Okay. And then minus 16 When I substitute of four, I&#x27;m going to have one minus. That&#x27;s 3/16. So that&#x27;s 16 16 minus 3 16 So this is the natural log of 13 over 16. How much substitute A to into this equation, you get 1/2 natural log. Two squared is four for minus three is one I like that one because that&#x27;s going to simplify. And then I have minus 16 if I substituted to in his equations So natural Lago one minus 3/4. So that&#x27;s 4/4 minus 3/4 which is 1/4. So what I have here is 1/2 natural log of 13 minus 16 Natural log of 13/16 and this term is simply zero. And then I&#x27;m gonna be left with plus 16 natural log. 1/4. You can work with these natural logs and try to simplify and use properties. But when you evaluate this, you kids should get an answer about 1.8, um, as the area between these two curves. So quite a complicated problem here. But where it started out Waas I looked at my graph and that told me that it was the The difference between these two definite Integral is would give me the area. So I proposed that. And then I realized, OK, I&#x27;ve got to find two anti derivatives. I had to use some algebraic manipulation. Some use substitution to get the answers. I came up with the lager rhythms as manager derivatives and then substitute in to get my final answer for the definite integral.

Dillon Huddleston · Answer

s so in this case, we can integrate with the state. Why so X equal to two I squared is a problem with access upper proble with access been paused about X axis eso integrating between ask equal to the square excluded Got equal to zero wishes wth e Why access and why Call the three s o. We can integrate just two I squared zero is too. I swear has entered your motive 2/3. Why cute? This could be interpreted from the intersection of X zero with exes to I square that why zero and why&#x27;s three is a buyer in question? So we&#x27;re really right between zero to me. Of course, this zero value did y zero. So me on you consider the upper limit three So three tube is 27 divided by three is on. I am multiplied by 2 18 and we can see that this value of 18

Bobby Barnes · Answer

we want to find the area enclosed between the breach and X is equal to two by squared excessive zero and why is equal to three? So the area were interested in will be located right here. The part that I&#x27;m shading in lack right now. So there&#x27;s two ways that we could go about this. We could eat there, get everything in terms of why, I mean, in terms of X and then solve. Or we could do this in terms of X goes, we could think about it in the same way in terms of the upper minus the lower bound. But the upper is going to be the first to the right, and the lower is gonna be the first to the left. So if we start on the left, they keep on going. We first enter our region at X equals the rooms. That&#x27;s gonna be our lower function. And we keep on going out until we pass through, and then we pass through this red so we can actually do this room. So since that point, there&#x27;s just gonna be zero, and we know our upper bound in this case is three. For why so it&#x27;s going to be, too. Why squared? Minus Cyril, which we&#x27;re just leave us with our function so we could go ahead and integrate this now to find all the area in that region we have shaded. So that&#x27;s gonna be 2/3. Why cute from power rule 0 to 3. Then we can plug in three. So it&#x27;s gonna be 27 divided by three, which is nine times to 18. Logan zero, we get zero. So that area, the region should be 18.

Problem

Find the areas bounded by the indicated curves. $…

Problem 15 Easy Difficulty

Find the areas bounded by the indicated curves.
$$y=6-3 x, x=0, y=0, y=3$$

Answer

$$A=\frac{9}{2}$$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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Joseph L.

Integration Review - Intro

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$$A=\frac{9}{2}$$

Basic Technical Mathematics with Calculus

Grace H.

Kayleah T.

Kristen K.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

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

Grace H.

Kayleah T.

Kristen K.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus