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I1-3 using 7.8 Improper Integrals techniques and #4 using 8.1 Arc Length techniquesf9 ed] 01 J1 d 2 2 x +r-659 3 Cxe dv=7(6 +r)9 Deterrnine the length of 189 < X...

Question

I1-3 using 7.8 Improper Integrals techniques and #4 using 8.1 Arc Length techniquesf9 ed] 01 J1 d 2 2 x +r-659 3 Cxe dv=7(6 +r)9 Deterrnine the length of 189 < X < 875

I1-3 using 7.8 Improper Integrals techniques and #4 using 8.1 Arc Length techniques f9 ed] 0 1 J1 d 2 2 x +r-6 59 3 C xe d v=7(6 +r)9 Deterrnine the length of 189 < X < 875



Answers

(a) Sketch the curve $ y^3 = x^2 $.
(b) Use Formulas 3 and 4 to set up two integrals for the arc
length from $ (0, 0) $ to $ (1, 1) $.Observe that one of these is an improper integral and evaluate both of them.
(c) Find the length of the arc of this curve from $ (-1, 1) $ to $ (8, 4) $.

Okay, let's say I wanted to find the integral of these two values here. Well what I can do is it for finding the definite integral could find the integral of each of these terms separately. So from -3-0 of two And then also from negative 3-0 of squared of nine minutes, X squared. Ok, so first if I'm looking at the function too, if I were to graph this right, this would just be a straight line And I just care about the values from negative 3-0. So the area under the curve here, even though this is using negative X values, I'm still above the X axis. So three to the area of a rectangle three times two is just six. So this is just six plus. What is the geometric figure here? Well, this is the same as why squared plus X squared equals nine or equals three squared. So this is the equation of a circle With a radius of three and this is just from um 0 to -3. So I just care about this portion here. Um And when I only care about this portion here, I'm only looking at quarter of the circle. So what that is? The area is high squared Times nine or 9 pi All divided by four. So 9 pi over four plus six will be the area of this integral. Just by using geometry and calculus. Next let's say I'm finding the integral here. Okay, again I can take the integral of each term from negative to two of just the constant one. I could pull up a constant negative three negative three times the integral from negative to two of the absolute value of X. Okay, So again let's just look at each other girl separately. The function one just has one above it and I'm going from negative to to to. So this will just be a rectangle with dimensions for in height of 14 times one. This area here is just four minus three times what's the area under here? Well note that I go from negative to to to so I'll have to different triangles. They're both above the X axis. So area of this triangle two times two divided by two is two, saying here two plus two is four. Okay so the area both of those added together since they were above the X axis is for, So this is just for -12. This area will end up being -8.

In part A. We were asked to sketch the curb y cubed equals X squared. Well, to do this, you want to use a graphing calculator, or what I'm going to do is just grab it by hand roughly so this won't be as accurate. So we have our X and Y axes, and we have the When X is equal to zero. Why is equal to 02 of a point at the origin? We see that when next is equal to plus or minus one, why is equal a positive one with points at one one and negative 11? And the way this goes is is that this is like a square root function in that it's increasing slowly, but now it's only increasing for X greater than zero when it's decreasing for X less than zero. So the graph looks something like this. We have a cusp at zero Part B were asked to use formulas three and four, set up to integral for the arc length from 00 to 11 and then to evaluate them while using these formulas, we have that you can first rearrange why is equal to X to the two thirds. And so this implies that the Y DX is equal to two thirds X to the negative one third. And so we have that one. Plus D Y. D X squared is equal to one plus 4/9 x to the negative two thirds. And so it follows that according to this formula, L. The length is the integral from X equals 0 to 1 of the square root of one plus 4/9 x to the negative two thirds the X now notice that when X is equal to zero, that X to the negative two thirds is not defined. And so, in fact, this is an improper integral. It's the kind where D function is not bounded near a finite value. Now, another way to compute this integral is by rearranging and finding the X. Well, this could be plus or minus why they're the rehabs. But because we're just looking at the section of the Ark from 00 to 11 we're just looking at positive. Why to the three halves and so it follows that de y dx or sorry dx dy y it's going to be three halves. Why to the one half and so one plus dx Dy y squared is equal to one plus 9/4. Why? And therefore the arc length is the integral. From what we see, why ranges according to our graph here from zero 21 of the square root of one plus 9/4? Why de y notice that this is a proper, integral and in fact can be evaluated easily. So, evaluating the second integral, we see that taking anti derivatives This is equal to skipping a few steps here. 4/9 times two thirds times one plus 9/4 Why and this is to the three halves evaluated from zero toe one and so, substituting in values we get. This is eight 27th times one plus 9/4 to the three halves, which gives us 13 Ruth 13/8 minus and one of the three halves, which is simply one and multiplying out the factor we get. This is 13 route 13 minus eight all over 27. And so this is the exact value. So we evaluate the second integral. The first integral can also be evaluated, although not as easily. So we have that this integral l is the same as the limit as T approaches zero from the right of the integral from t 21 of the square root of one plus 4/9 next to the negative two thirds DX and notice that multiplying top and bottom inside. Bye x to the two thirds we get the limits as T approaches zero from the right of the integral from T 21 of and then in the numerator we have mhm nine x to the two thirds plus four square rooted and then over the square root of X to the two thirds which, because t is greater than zero. This is going to be positive since the positive square root which is simply next to the one third DX. And so now we see we could make a use substitution here. In fact, this is actually over nine x to the two thirds should be three X to the one third, and for you, substitution will let you be nine next to the two thirds inside the top. And we have therefore that do you is going to be two thirds of nine, which is six x to the negative one third DX and so it follows of this integral is equal to the limit as T approaches zero from the right of the integral from t toe one of the square root of U plus four. And then we have that this is going to be over three X to the one third and we have times DX is the same as 1/6 times X to the one third. Do you and so the X to the one thirds cancel out. And so, in fact, actually made a mistake here. Um, this should be U of t, which is 92 3rd and this is you have one, which is nine. So this is the same as the integral from natural limit as t purchase here from the right of the integral from 92 the two thirds 29 of the square root of U plus four over 18 do you? And now this is no longer an improper integral. So this is the same as the integral from 0 to 9 of the squared of U plus four over 18 do you and then taking anti derivatives, we get 1 18 times two thirds u plus four to the three halves evaluated from 0 to 9 and plugging in I get mhm can sell the two with that 91 27 times. Nine plus four is 13. So 13 Route 13 minus four to the three halves, which is the same as 44 which is four times two or eight. So once again we get 13 Ruth 13 minus eight over 27. Yeah, finally, in part C, whereas to find the length of the ark of this curve from negative 11 to eat four well, we have that. This is going to be integral from zero toe one first of all because it's symmetric about the Y axis of square root of one plus 9/4. Why de y plus integral from 0 to 4 Because we're looking up to the 0.84 of square root of one plus 9/4. Why de y? So you can see how this has really broken into two different sections. We have the left hand cusp here, and then we have the right hand here. So those are the two values now by part B, we have that this first integral is simply 13 Route 13 minus eight over 27 and the second integral. Well, if we take anti derivatives, we get eight 27th times, one plus 9/4. Why to the three halves evaluated from 0 to 4 and so substituting in, we get 13. Route 13 minus 8/27 plus and then eight 27th times and then 10. Route 10 minus one and adding things. Together we get 13. Route 13 plus 80 Route 10. Then we have negative eight, minus eight. It's negative. 16 all over 27.

And this question we're asked to draw a curve, find two intervals for arc length and then find the arc length from another point to another point. Let's see how to do this question. I plotted the graph for you already and we need to find two integral is for the arc length. Let's first remember what our arc length formulas are. We know that the arc length is equal to one plus Y. Prime of X squared in the square ridge. We also know that it's equal to one plus ex prime of Y squared. So we know both of these are true. The issue is we need to have yeah. Oh yeah, we need to have why? In terms of X alone or X. In terms of why alone? So let's start with the first integral. I'll just call this one, number one And I'll call this one. # two. Okay, For Integral # one. If we have wise, if we have y cubed equaling X squared, we're going to have to take the cube on both sides. So we're going to have Y equals X to the two thirds And differentiating both sides. We got that y prime equals just 2/3 Times X to the 1/3. Now you have to set up our integral are integral will be going from 0 to 1 and we're going to have the square root of one plus dy dx squared. So we have two thirds times X to the Times X to the -1 3rd. The whole thing squared. And what that does is that will simplify to Will have one plus. Yeah, We'll have four over nine nine X to the 2/3 And we can get a common denominator for that by having an integral from 0 to 1 of the square root. We'll have the square of nine X to the 2/3 plus four. All of us divided by three X to the one third integrated with respect to X. Now, as you can see this one is improper because I have to if I put in zero exactly, I'm going to be getting a division by zero error and we can't have that. Yeah. So what we need to do is we need to convert this into a limit question. This is going to become well we'll do that after substitutions. Let's make a substitution first. Our substitution will be U. Equals nine X to the two thirds plus four. And our D. You will be equal to the constant goes away. We'll have nine times two thirds. Which is going to be six times X to the minus one third dx. Mhm. This will work out nicely. So our limited integration will be changing from if I put in zero, I get four. If I put in one You will become just 9-plus 4 or 13. Therefore are integral becomes The integral from 4 to 13. We'll have the square root of you Over three X. to the 1/3. And D X. And dx is going to be do you over six X. To the negative one third, but the negative one third can flip up so we have X to the one third Times six. Do you? Yeah. Right. Mhm. Yeah. Uh huh. As you can see here, however, since our limit is not defined, Well since our equation was not defined at zero We have to use a limit at four. So that's our new limit. So our question just becomes the limit As a goes to four. Mhm. Oh okay. From the right hand side. Just the integral of descriptive you over 18, do you? And that's going to be from 8 to 13. So this is going to be Mhm. The integral of you is just U. To the three halves times two thirds Times 18 times 1/18. This is evaluated from a. to 13. And let's simplify this, this becomes one, this becomes nine. So we'll have this is equal to 1/27 times 13 To the 3/2 minus the limit As a goes to four plus of A. to the 3/2. Fortunately this limit exists and we're going to get that we have that this is equal to simplifying this. This is going to be equal to 13 13 minus eight. Mhm. All of this over 27. So this works okay now for B. Now for our second integral recall that we had white cubic willing X squared we want to get X as a function of why? So we're going to take the square of both sides. In other words we're going to have that Y to the 3/2 equals x. And also for calling our formula, we need to differentiate Acts with respect to why? So ex prime with respect to Y is equal to 3/2 times Y. To the one half. Obviously there are no integration. There are no domain issues here. So our length is equal to the integral from 0-1. We're going to have one plus three over to why to the 1/2 All of this squared integrated with respect to why. Mhm. And we can simplify this so that this becomes the integral from 0 to 1. We're going to have one plus nine y divided by four dy Which one we simplified that. Getting a common denominator will result in the square of four plus nine y over over the square root of four. Which is just to now this is much easier to integrate because all we need to do is we just need to bring the one half outside. Yeah and then we're going to integrate four plus nine Y. And this will be raised to three house power and this will have to be multiplied by 2/3. And don't forget we also have to divide by the derivative which is 1/9. You can verify this with the substitution, canceling these out. We're going to get that we have to evaluate this from 0 to 1 but don't don't be negligent and cross out the zero because we will not get any zeros When we do so we're going to have that. This is one of the 27 times four plus four plus nine To the 3/2ves minus four plus zero to the three house. And when we do that we're going to Evaluate this to the very same answer. This is 1313 -8 divided by 27. Okay. Yeah. Now for part C for part C you want the arc length? We want we're going to want the arc length from negative 11 284 This is actually very straightforward. Mhm. Okay. We can consider this As two separate integral. We're going to evaluate it first from negative 10 From negative 1 to 0 and then from 0 to 4 on the white side we will deal. We're going to use the y bounds so it's going to be an integral from negative 1 to 0 plus. An integral from one from 0 to 4. Yeah. And because of symmetry, The integral from -120 is equal to the integral from 0 to 1 which we found earlier. All right, So what we have to do is we need to evaluate this. So we have 1313 -8 divided by 27 plus the integral from 0-4. And we can again use the same set up as we did last time, as we did in the second integral. We have just the square root of 44 plus nine Y, divided by two. Uh huh. And we used and we and we just use the simplifications from part B. Yeah. And evaluating this again we can use the same anti derivative. It's going to be won over 27 times four times four plus nine Y To the 3/2ves Evaluated from 0 to 4. And when we do this this is going to be one of the 27 times four plus 36. Which is going to be 40 To the 3/2, -4 to the 3/2. Think okay. Happy. Yeah. Okay. Yeah. Mhm. And when we simplify this we're going to get that this is all equal to This is all going to be equal to 13, 13 Plus 10, route plus 80, route 10 minus 16. All of this will be divided by 27. That's how you do this question.

When the phone function why is equal to eight over? Ax squared from the interval from one of four sort of start off was winter, my primary directive eight over X squared, which gets *** 16 over X cubed. And so then we know that our plants is an integral from 1 to 4 of the scurries of one plus 90 16 over X cubed square T X. Get us into girl from 1 to 4 squirts over one plus 256 over excess six D x in certain. Over. Put on your calculator. We will get approximately 8.71 as our answer.


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