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(4 pls) Find the antiderivative that satisfies the given condition f()er-2r"+I; F(0) = [(4 prs) Suppose the interval [1, 3] partitioned into subintervals What ...

Question

(4 pls) Find the antiderivative that satisfies the given condition f()er-2r"+I; F(0) = [(4 prs) Suppose the interval [1, 3] partitioned into subintervals What is the subinterval length Ax? List the Efid points XuXX X% and x4. Which points are used for the lefi, right and midpoint Riemann sums?(4 prs) Use the figures calculate the left and right Riemann Sums for fon the given interval and for the given value of n. (s"

(4 pls) Find the antiderivative that satisfies the given condition f()er-2r"+I; F(0) = [ (4 prs) Suppose the interval [1, 3] partitioned into subintervals What is the subinterval length Ax? List the Efid points XuXX X% and x4. Which points are used for the lefi, right and midpoint Riemann sums? (4 prs) Use the figures calculate the left and right Riemann Sums for fon the given interval and for the given value of n. (s"



Answers

Suppose the interval [1,3] is partitioned into $n=4$ subintervals. What is the subinterval length $\Delta x$ ? List the grid points $x_{0}, x_{1}, x_{2}$ $x_{3},$ and $x_{4} .$ Which points are used for the left, right, and midpoint Riemann sums?

For this problem we are asked to approximate the integral from 0 to 1 of x minus one half squared by the midpoint rule with n equals Or excuse me for this problem is actually by the trapezoidal rule. Um then to find the exact value by integration. So to begin we want to find our delta X. So that's just going to be 1/4, having our delta X. We want to find our list of points will be evaluating our function at. So those should be just 01 quarter, one, half, 3/4 and one. Now all we need to do to evaluate with the trapezoidal rule with n equals four. Uh We should have delta X over to outfront times our function evaluated at the initial point plus two times the function evaluated at all. The mid points plus two times or not. Two times. Excuse me. Plus just the function evaluated at our end point. So when we uh when we execute that, expanding everything out uh This is including doing the multiplication by delta X. We should have in our some 5/1 28 Plus 1/1 28 Plus 1/1 28 Plus 5/1 28 Adding all of those up. The final result should come out to about 0.09375. Now finding the exact integral From 0 to 1 of x minus one half squared the X all we need to do, we can make the substitution that U equals x minus one half, which means that will change the boundaries of our integration. Zero is going to become negative one half. One is going to become one half. We're integrating you d you which first of all we can recognize that actually this is just going to be the same thing as two times Two times integral from 0-1 half of you square to do not you do you. So we should get that. That's going to be two times U cubed over three, evaluated from 0 to 1 half, which is going to be two times 1/2, cubed, divided by three and plugging that in evaluating it out as a decimal form number, the result should be 0.8333

So. Hillary solve the integration from 0 to 2. Of the spirit of X. Cubed Plus one. Yes. Using the representative room Where N. is given as four. So and did you close it for Now? He travels the world on the radio. As the integration from a. to b. x. d. x approximates to the XR two of fx zero plus 12 X. Wonders do with extremist Perfect street all the way to effects and which is the last run for the first and the last time everything is multiplied by two. Um And it's from there that the X. Is the most they were in which in this case the the upper hand but it is too. And a learning blue which is zero over end. That you're not for Dedicate your door 4" half. Oh now you have delta X. You can find your X coordinates. Excellent points. So extras from zero going to have. So you just need to add um half every time. Sure to have one three by two. Finally to so these are for exponents towards the upper hand. Love you. So it stops there using this you can find your by coordinates such as the Earth ex uh apartment using By plugging in the experience into the equation which is the root of x cubed plus one. So I'm gonna skip a step and find the tone that you need to plug in into. The Jeff is a little formula. So test of X. Not as living in the home. That is yeah as f as usual. which gives your route of one which is one now two or two and two Type of X. one as to enter from half. Now when you put it in you get your answer as to your door to three. Yeah for two and next term is to end half of extra. There's three and 2. We just write that okay, three into s. Fault one. This gives you There were probably three. One. Sorry about this guys, I meant to say to to winter FFX three cycle computer that set to into that's one. Yes. When you put this into the formula you get and to into F of X four, this still tend to Lot of three x 2. You got that from now you get root of 70 upon to and finally, First and the last time you don't need a month, a month that I do. So it will say X5, which is the last one, which is F. Of X and and S of two. I have to when you love to enter the formula To keep us on Telus. Fabio gives you three. Now you have the drums which need to plug in after the. Uh huh. They're forming up so that's what we are going to do. There's approximates to Delta X over two and that's the excess half. So half forward to give you one more far into that. You have the values of one class, three years or two over to yes. Photo flash, 70 or two last three and this many plantation into the calculator. You get your answer as 3.23 and that can support is given in the text book. So you guys understood that and thanks for watching.

Okay, This question wants us to evaluate this Integral using the reminds some definition with the right partition. So writing this is a remind some. The area under the curve is just the limit, as we have infinitely many rectangles of the some from K equals one to end of our function sampled at the specific values we want times the interval will, or rather, direct tingle with where Delta X is the integral with divided by n or in our case, to minus zero over and or to divide of. And so now we can write orange a girl as the limit is an approach infinity of we can bring the two over end in front of the sun because it does not depend on cable. And then we still have our except case star. But in our case, we know that we're dealing with a right remind zone. And in the right re months, some we're sampling the right ends of every rectangle. But we don't have to worry about the starting value because it zero So in our case, Ext RK is just que Delta X so we can replace that and are some now. So now we have the limit has an approaches Infinity of two over and times that some from K equals one to infinity of four times K Delta x que and I'll pull that four out in front. Does that does not depend on? Okay, so we're just left with with some K cube times. Delta X cute. But we know Delta X. That's just two over end. So now we can pull that out front too. So we have eight over and times two over and Cube times The sum from K equals one toe end of que que And this is again our third of the special sums that we discovered. So doing this, we have the limit as an approaches infinity of I'll simplify it like terms here. So we have eight times eight So 64 over end of the fourth times this cake cube some which is and squared times and close one squared, divided by four. And now let's just expand this out to find the limit. And we could be done so 64 to bottom before That's just 16. So we have 16 times. Well, we just have an end squared times and end squared. So we have ended the fourth, plus some Lord, some lower order stuff that's not going affect our limit because the denominator has an end of the fourth inning. So we just take the ratio of 16 to 1, so 16 is our definite integral value.

We want to approximate the given integral numerically using the trapezoidal and Simpsons rules. The integral is 0-1 square root, one plus X cubed dx with n equals four. So two numerically integrate, we have to make the following notes to access b minus A. The bounds are integral minutes or one at Jackson square root one plus X cube Rx ir X equals zero X 22.25 All the way up to X three export has given. Now we have all the information to plug into the episode of the Simpsons rules. So first episode of rule is the integral is approximately delta X over two times. Fx not plus two. Fx one plus two. F X two plus two. Fx three plus fx four or 1.11699 The Simpson's rule is the articles approximately delta x over three times fx not plus four. Fx one plus two. F x two plus four X three plus fx four. This gives a solution 1.11144


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