use-a-graphing-utility-where-helpful-to-find-the-area-of-the-region-enclosed-by-the-curves-xy3-y-x0

Question

Use a graphing utility, where helpful, to find the area of the region enclosed by the curves.
$$
x=y^{3}-y, x=0
$$

Gregory Higby · Accepted Answer

eso won and graft this equation already be careful. Sonata functions You don&#x27;t wanna say that. And then X equals zero is just the y axis. You get a visual there, and what&#x27;s important is you want to see that the bounds that where they cross are from negative one, they do cross at zero. But I&#x27;m not gonna do that with my integral show. You a way around that, um, and positive one. And the reason why I&#x27;m going to ignore that is I just copy this equation real quick if I were to create a new equation, but right as the absolute value of that function, you can actually type into your calculator so the areas will not cancel each other out. Eso then all you&#x27;d have to do is find your integral key, which mines under the functions miscellaneous and egg roll from negative one to one of the absolute value of that function. And then you after I d y, um, and you get the exact answer of only five. Just so you know, there&#x27;s other ways of doing this. If I hide that function and go back to this, um, some of you might be saying there. What if you go from negative 120 You can absolutely no 1 to 0. But think about. Okay, well, we want the area between the Kurds and areas always positive. So then double that. That&#x27;s why those areas you do not want to cancel out family was thinking that, um ah, and you can actually do the integral of this. It&#x27;s 1/4 y to the fourth minus one half life squared and plug in one and negative one and see that you&#x27;ll get possible in half. I feel like I&#x27;ve over explained this problem to this point, so hopefully you&#x27;re happy with what I&#x27;ve said so far.

Gregory Higby · Answer

since we&#x27;re allowed to use a graphing calculator, I&#x27;m just gonna do this. Um, so one option that you could do it is go from the general from negative 2 to 0 because that&#x27;s where the grass intersecting you do. You know, excuse minus four X, um, and find the integral that way, and then I would actually could do is because this is an odd function. You could just double that answer because, uh, we want you show you a different way of doing this is you could just do the absolute value. And if you do, the absolute value has what showing up in green. It just takes everything that was a negative value. Makes it above the X axis, Makes it positive. So anything that&#x27;s below flips up and to be a positive um, So basically what I&#x27;m boiling it down for you is you can I just hit the integral from negative to to to of perhaps a value function and nothing will cancel out and you&#x27;ll get the exact answer of eight. Um, the other option is telling you to dio is you could just do the role. Or was that the general from negative 2 to 0 of this function. Um, looks and he&#x27;s You do have to tell the calculator. And what I would actually use this right to in front to double that area. Eso I gave you a couple options? Uh, definitely would encourage you to work this out by hand. I think we also did this problem in an earlier example. Um, that I would encourage you to look at that in the previous videos. Well, if you want to work it out by hand,

Gregory Higby · Answer

since you&#x27;re allowed to use a graphing calculator, I figured I&#x27;m not as well. I just use a graphing calculator to get the area. Um, So what you want is the area that IHS um and then a Sorry Shero Talib&#x27;s No Michael zero. But that&#x27;s obviously the X axis. So, um, you only want the area between these two curves. So what you could do is the integral from 0 to 1 of that upper function and subjecting zeros the gaps and then the integral from 1 to 3, um, and then the gate, Everything because it zero minus that function or what I&#x27;m about to show you it, you could just type in the inner rule. You find that by hitting the function tool from 0 to 3 of the absolute value of this function, I&#x27;ve been BX Now may I should clarify what the absolute value does is it takes anything that&#x27;s below the X axis and flips it above. So you don&#x27;t have to worry about which ones parts of her wish one&#x27;s negative. Nothing will cancel each other out. Uh, now the downside to my calculators, it only gives me the desk, will answer. But I have the luxury of already having a fractional answer of 37. 12 and you can see that it matches. So, uh, depending on if you teacher requires a fraction or the decimal. Um, there you have both options.

Gregory Higby · Answer

eso since you can use a graphing calculator, I&#x27;m just gonna go ahead and do that, Um, where you can see my red curve and my blue curve, which match up to these and you can tell there&#x27;s a cubic and quadratic and they&#x27;re just sideways, So don&#x27;t let that freak you up. They intersect at three different places. All you really have to do is figure out ah, the furthest ones away from each other where they intersect. So that is why equals zero must get the white coordinates and y equals four. Um, and what that tells you is the bound. So if you pull up your integral, which is under functions under miscellaneous from 0 to 4. And the reason why we can skip this middle part is I&#x27;m gonna absolute value this attraction of the functions and what that does is makes both areas positive. So I don&#x27;t have to. I&#x27;m just saving, um, a lot of time. Now, what you do want to make sure you do is make sure you put parentheses in the second one because you have to make sure that you distribute that minus two each and then tell your culture to do in terms of why, um and this is just a decimal approximation. That looks like 56 though, um, so I got a denominator of six of the 66 plus 5 71 6 Uh, so an exact answer would be some people in six for this. But this calculator just gives you the decibel approximation. There&#x27;s a lot of thought that gave me this answer. Um, if any of your sitting there and saying, Well, why don&#x27;t you have to put the first one in parentheses? Um, you don&#x27;t because there&#x27;s nothing to distribute in front. But if it helps, you can type it in there, and it will not change your answer. Um, so, yeah, look at your outside bounds. It&#x27;s all you need for one in a girl for area between these two curves.

Gregory Higby · Answer

Uh, okay, so I really graft. I used a graphing calculator to ndgraf both functions. And what&#x27;s important here is ah, where the x values were. These graphs intersect. Um, so maybe I should, uh I could show you a swell is if you create another equation. Um, so if you notice from 0 to 1 the red curve, the X cubed function is greater. Um, and then from 1 to 3, the quadratic is greater. It&#x27;s above the red function. Um, but May I could point out for you is if you do the absolute value of these things and they also put in parentheses. Happy that and then subtract off with parentheses. This function, what you&#x27;ll see is my new graft takes everything and puts it above the X axis. And, um, we hide that again. Um, you can actually see that the area in between these when you add them together stays positive. Um, so this might be a good visual for you to understand, as you know, to the, um, in a role which is under my function tool, Miscellaneous. Um, from where the grass would have intersected if I still had them up on the screen, but actually makes sense that they intersect because at zero in three, because that&#x27;s if they intersect their equal. If you subtract them from each other, if they&#x27;re equal and you subtract, you get zero for that white corner. So does actually make sense. That 0 to 3. Are you going? You reverse the video to do that, and then I could just copy that over. Um, and you can see the answer of 3.83 Um, but I believe as a fraction is 37 calls. So depending if your teacher likes decimals or fractions. Um, that&#x27;s what I would do. Teoh. Get the answer. Use a calculator.

Gregory Higby · Answer

eso. Since you&#x27;re allowed to use a graphing calculator, I would just go ahead and use a graphing calculator. You can see the exit of the X squared function is the blue curve in the absolute values of V shaped green curve eso as faras finding the area between the curves. You just go to your integral, um, between zero. You can tell that they cross at zero, and this is actually should be exactly if you set them equal to each other of 56 Um, 56 If you don&#x27;t believe me, I&#x27;ll just give you 56.833 Um, and maybe it&#x27;s not exactly 56 million. I should just write out 560.833 ankle getting answer. That&#x27;s close up anyway. Um, and so then the upper function is the absolute value, But I&#x27;m just going to rank two X. You can leave it as the absolute value. This country let you do that. But that&#x27;s the positive version, so we don&#x27;t need to do the absolute value minus this function, which I&#x27;m just going to copy and paste. It&#x27;s actually helpful to put in parentheses. I don&#x27;t know if you actually need parentheses and then helicopter to do D X. Now this just looks like mumbo jumbo decimals. But looking at it, if you actually did you substitution for the Annan drift of these plugged it in, you would see that the exact answer is natural. August 2, um, minus one half. Um oh, it&#x27;s a little bit off. So may a better answer was using 56 year, as I mentioned earlier. Quick, and see if I get the No, no one still off just a little bit. Um, but the important thing is finding these values and understanding how to get there. Um, so far, you out just around my answer to be 0.193 instead of the exactly into that the book says.

Problem

Use a graphing utility, where helpful, to find th…

Problem 23 Easy Difficulty

Use a graphing utility, where helpful, to find the area of the region enclosed by the curves.
$$
x=y^{3}-y, x=0
$$

Answer

$$
\frac{1}{2}
$$

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Calculus Early Transcendentals

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Anna Marie V.

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Integration Review - Intro

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$$\frac{1}{2}$$

Calculus Early Transcendentals

Catherine R.

Anna Marie V.

Caleb E.

Kristen K.

Integration Review - Intro

Definite vs Indefinite

Howard Anton, Irl Bivens, Stephen DavisCalculus Early Transcendentals

Catherine R.

Anna Marie V.

Caleb E.

Kristen K.

$$
\frac{1}{2}
$$

Howard Anton, Irl Bivens, Stephen Davis

Calculus Early Transcendentals