use-a-cas-to-find-the-exact-area-enclosed-by-the-curves-yx5-2-x3-3-x-and-yx3

Question

Use a CAS to find the exact area enclosed by the curves
$y=x^{5}-2 x^{3}-3 x$ and $y=x^{3}$.

Gregory Higby · Accepted Answer

uh, So I went ahead and ah, typed in the graph. The equations of these graphs that doing is a little bit more, um, agency that X cubed is in red. The extra 50 IAEA is in blue, and what&#x27;s important is finding the points of intersection that are outside of on the furthest. Now they do intersect at X equals zero. But you can actually ignore that in this problem. Make sure I don&#x27;t think the Intersect again. Doesn&#x27;t look like it. Okay, so I really can&#x27;t negative X equals name 1.947 in positive 1.9477 So starting new equation hit functions miscellaneous right here. Negative 1.947 to possible in 0.9 course of it now because I&#x27;m doing the problem this way. You want to hit the absolute value again that&#x27;s under functions, Um, or no, it&#x27;s not. It&#x27;s right here. But I actually used a calculator shortcut. And what you can do is it doesn&#x27;t matter which function you take. Um, actually, let me take this Alice to the top. Because if you do the top one, there&#x27;s one more stuff you have to dio minus in Corinth theses because you want to make sure you subtract of everyone in these pieces. Um, I&#x27;m hitting command. See, by the way, I&#x27;m copying and pasting that right here. And you also just say the X now by doing the absolute value what it does, it is. It always takes, um, the positive version of these answers. So if you look, these two areas would cancel each other out. If I drop the absolute value and just delete those absolute values your quick and notice you get an answer. Uh, ups you get an answer zero because they cancel each other out. So that&#x27;s why you need tohave the absolute value in there so that they don&#x27;t cancel each other out. So about 14.770 if you around a three decimal places, um, and other graphing calculators would give you the perfect answer. This one does not like a fractional answer. This one just gives you a decimal. I don&#x27;t know how to fix that.

Kenneth Kobos · Answer

We want to find the area of the region and closed. But these two functions the 1st 1 in blue y equals X to the five minus six X to the three plus four X and the second function in green y equals X. So use a graphing calculator. We can get a quick idea of what the region looks like and we find actually that there are foreign closed regions with five points of intersection. So the points of intersection are here, here, here, the origin one here and one here. So we actually want to find the area of all foreign closed regions. However, what we noticed is that, um are two functions. Here are odd functions. Ah, which means that this red region here has the same area as this a green region here. And this region here has the same area as this large region over here. So actually we only need to determine the area of two of the region&#x27;s. We&#x27;re going to be looking at the red regions and then once we determined that area or we have to do is double our answer. Okay, So we were going to do area equals area one plus area two. Okay, we do want to figure the exact area. So we&#x27;re not going to use the approximations that a graphing calculator give us for the points of intersection. So we&#x27;re going to have to determine those by ourselves. So let&#x27;s do that. We want to find out what points the two functions cross. So the first function was X to the five minus six x to the three plus four X, and the second function was just X. So what we&#x27;re going to do is solve this equation by bringing everything to one side. Zero equals x to the five minus six x to the three plus three x we confected rodent X we&#x27;re left with next to the four minus six X squared plus three. Okay, so right away we see that one solution coming from here is that X equals zero. So that&#x27;s what we expected. We saw that the grafts cross that the origin, but now we want to solve this quadratic. We wantto determine at what X values is this. Uh, is this quartet function equal to zero, but notice that if he rates a equals x squared, this shows that this shows us that we actually have a squared minus six. A plus three equals zero. So this is actually just a quadratic in disguise, so we can solve this. So to solve this, we can use the quadratic formula, which, after using the quadratic formula, will give us that a is equal to three plus or minus the square root of six. But remember, a is X squared. So this tells us that X is going to be the square root of three plus or minus route six. But since we took the square root, we have to put a plus or minus here. So we find actually that we have all five solutions explicitly. The 1st 1 was X equals zero. And this expression over here gives us all of the other four solutions by taking the old possible combinations of plus and minus. So, looking back over here, we&#x27;re interested in the red regions. So what we actually want is this intersection, which is the origin we want that intersection point and this intersection point. So, looking at our solutions that we have here, we are interested in X equals zero X equals. So we want positive values so we&#x27;re going to take the positive square. So the first will be three minus the square root of six over there. And then we&#x27;re also interested in excess three, plus the square root of six. So these are the points of intersection that we&#x27;re interested in. Now we can We are ready to determine the area. So let&#x27;s take a look at area one. Area one is equal to the integral from zero to our second point, which is this crew of three minus read six three minus route six. And it is the top function, minus the bottom function. So which regional we looking at? We&#x27;re looking at this region here, this small red one. So we see here that the top function is blue and the bottom function is green. So we&#x27;re going to subtract that as usual. Ah, it is the top function, which was the Quintus X to the five minus six X three plus three X. And then we&#x27;re going to subtract the bottom function, which is the green one, which is just y equals X. Okay, So what we have here is I&#x27;ve already subtracted the X since the Queen Tick actually had a plus four x over here, but then plus four x minus. X gives us three X. So that&#x27;s area. One area too is equal to the integral from we&#x27;re going to integrate from our remaining points. Ah, that we saw for. So the bottom point is three minus two screwed six and the top point is screwed of three. Plus the square root of six. Okay. And notice that now the green function is on top and the blue function is on the bottom. So when we do the top minus the bottom, it&#x27;s just going to be the opposite of what we use in area one. So we&#x27;ll just copy that with a minus sign extra the five minus six text to the three plus three x dx. Okay, so each of these we can solve algebraic Lee because we&#x27;re we just have to do into Gran&#x27;s of polynomial Sze. Well, we just have to do anti derivatives of Paul no meals which we know how to do. And then we&#x27;re just plugging in these radicals. Uh, but we are allowed to use an algebraic calculator. I believe so. When we actually determine the area area is equal to area one plus area too and we simplify everything. What we&#x27;re going to get is the exact area of 12 times described of six minus nine as the exact area of the enclosed region. Notice that it&#x27;s exact because we determine these points of intersections Algebraic lee. So we know that those are the exact values and, um, we just plugged them into our polynomial Sze Ah are the anti derivatives of the polynomial, so we actually get an exact area. There was no approximation here whatsoever.

Sky Li · Answer

All right. Now sketch the region enclosed by those two curves. So this is our XY plane. And the first function is why he goes to X Cube. So you should look like this. This is our by equals two x Q And why he goes to X. Okay, it&#x27;s a street line. This is why it goes x in this those two eco region since everything is symmetric, uh, with respect or region. So those two are exactly the same. Um, they have the same area, and now we need to find this shaded region area. Um, so they will be, um yeah, we only need to calculate one side. So wait time and then times two. So it will be two times. Um, this shaded region. Yeah. Um, and, uh, I want to calculate this region by taking integral with respect to X. Since everything is represented bags All right. Right. The next step, we need to find the boundary. Correct. Clearly, X goes from zero to this point this point x two all right and x two should satisfy the those two functions, which is X square equals two x, so x two should satisfy this equation. Yeah. Since extra is not zero, we can cancel one X Then we got X squared equals one, which gives us X two eco swat. Since X two is part, so are integral. That goes from zero to one. Okay, um, so then the thing inside the integral is the upper curve, minus the lower curve. So our upper curve here is a straight line X minus glower, curve xq. Yeah. Then let&#x27;s find anti duality of that. That will be up half black Square, minus fourth. Thanks for you had a boundary one and zero. So Well, X equals one we plug in. This will give us one half minus or quarter, which is a quarter times two is about half right minus, we know, actually called zero. This term, zero times two is also zero. So it&#x27;s like one half minus zero. Our answer will be just 7.5

Amrita Bhasin · Answer

okay. We know we&#x27;re only looking at the positive side because we know in this context, we can only have positive area. Okay, so now we over area is from 01 x cubed Jax. And I remember how we integrate increase The exploited by one&#x27;s own. Just four instead of three were dividing by the new explain it, which is four now we&#x27;re plugging in. This is officer just zero, Which means we have 1/4 is our solution.

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.

Kenneth Kobos · Answer

we want to do a quick sketch of these two equations, and then we want to determine the area of Dean closed Region s. So let&#x27;s take a look at our 1st 1 So we know that as X goes to infinity and also X goes to negative infinity, we know that the symbiotic behavior is that it goes to zero. So for this one, we know that we&#x27;re going to zero both on the right and the left. Okay? And we also know that it&#x27;s going to cross the origin. What else do we know? We know that the denominator one plus X squared is always greater than the numerator, uh, for all acts. So we&#x27;re never going to pass the values between minus one and one. So this is what the the equation looks like. Now, let&#x27;s take a look at the 2nd 1 Um, for the 2nd 1 we also see the same behavior is X goes to plus and minus infinity. We see that it&#x27;s going to zero because excuses in the denominator, it&#x27;s a good in the next squared. Ah, but we also have this one vertical aspirin tote. Since the denominator can be zero at X equals negative one since negative one cubed is minus one. So we should draw in that ass in tow. This is that X equals minus one. Let&#x27;s draw this one in blue. So to the left, we see the X is going to minus infinity and it goes to zero. But on the right, we also see that this function must touch the origin. And it exhibits this kind of behavior. It kind of goes up, and then it goes back down. Eso we do see that there&#x27;s this enclosed region right here, and that&#x27;s the area that we want to determine. So we need to figure out our limits of integration. We no one will be the origin, and we need to figure out the other. So to solve for these, we equate that you equations. So let&#x27;s do that. So our 1st 1 was X over one plus X squared. The 2nd 1 was X squared over one plus x cubed. So we&#x27;re just going to multiply the denominators. So here we have X times one plus execute equals X squared times one plus x squared Good. Okay, X plus X to the fore equals X squared plus X to the fore. Okay, so we see that these x to the force cancel. We bring everything to one side. X squared minus X equals X times X minus one. So we see that our values, our X equals zero, which we expected and X equals one. So we see that this point over here is X equals one, and this one is X equals zero. So we&#x27;re just going to integrate from X equals zero X equals one area equals into rose 0 to 1 of our top function minus or bottom function. Our top function here was the green one. So this was our 1st 1 And then the bottom function is our blue one. That&#x27;s our second function. So we&#x27;re just going to subtract those. The 1st 1 was X over one plus X squared. The 2nd 1 was X squared over one plus execute DX. Okay, let&#x27;s break this up into 200 girls and salvage separately. X over one plus X squared DX minus into grows 0 to 1 of one plus X cube under X squared D x. For this 1st 1 we&#x27;re gonna do a U. Substitution of U equals one plus x squared so that d u equals two x d x And over here we&#x27;re going to do also u substitution Yukos one plus X cubed and d&#x27;you equals three x squared DX. So we see that we don&#x27;t have a three up here, So we&#x27;re going to multiply by three and then divide by three. And so now we can just carry out these into girls. Oh, uh, also, over here, we don&#x27;t have a to X in the top, so we&#x27;re gonna multiply by two and divide by two. And now we can carry out these into girls. Okay, Equals here we had 1/2 integral. We want to switch to the U variable now. So, looking over here, when x zero you is one and when X is one us too. So here we&#x27;re integrating from 1 to 2. Now we have this two x d. X Over here, that&#x27;s a d u. And we have a you in the bottom. So this is just d&#x27;you over you. We&#x27;re adding there was a 1 30 year into girl. Oh, it&#x27;s a subtraction, not adding, uh, minus. Here we have our three x squared DX. That&#x27;s our d&#x27;you and then we have a u N the denominator. So this is d&#x27;you in the numerator. You in the denominator. Oh, and we need our limits of integration. So, looking over here, we see that when x zero you is one and when x is one used to So again, we&#x27;re going from 1 to 2. Okay, so actually receive that these two into girls are the same. So oh, um, what we can do is just combine them into one piece. So we have 1/2 minus 1/3 which works out to be will have his 36 and whether it is to six. So we have 1/6 integral from 1 to 2 of do you over you. Okay, so let&#x27;s just calculate this one over six. The anti derivative of one over you is the natural longer them or lawn of you. And we&#x27;re going from 1 to 2. So we have won over six multiplied by lan of two minus long off. One loan of one is just zero. And so we have. Our final solution equals one over six, one of two. That&#x27;s the area of the enclosed region.

Problem

Find a horizontal line $y=k$ that divides the are…

Problem 34 Medium Difficulty

Use a CAS to find the exact area enclosed by the curves
$y=x^{5}-2 x^{3}-3 x$ and $y=x^{3}$.

Answer

$$
14.770
$$

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Calculus Early Transcendentals

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Integration Review - Intro

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$$14.770$$

Calculus Early Transcendentals

Anna Marie V.

Heather Z.

Caleb E.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Howard Anton, Irl Bivens, Stephen DavisCalculus Early Transcendentals

Anna Marie V.

Heather Z.

Caleb E.

Joseph L.

$$
14.770
$$

Howard Anton, Irl Bivens, Stephen Davis

Calculus Early Transcendentals