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# Use a computer algebra system to find the exact area enclosed by the curves $y = x^5 - 6x^3 + 4x$ and $y = x$.

## $12 \sqrt{6}-9$

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Applications of Integration

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##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

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's. We'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're not going to use the approximations that a graphing calculator give us for the points of intersection. So we're going to have to determine those by ourselves. So let'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'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'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'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'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're going to take the positive square. So the first will be three minus the square root of six over there. And then we're also interested in excess three, plus the square root of six. So these are the points of intersection that we're interested in. Now we can We are ready to determine the area. So let'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'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'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'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'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's area. One area too is equal to the integral from we'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's just going to be the opposite of what we use in area one. So we'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're we just have to do into Gran'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'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'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'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.

University of Toronto

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Applications of Integration

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