evaluate-the-given-definite-integrals-int_275325-fracd-xsqrt36-x1-3

Question

Evaluate the given definite integrals.
$$\int_{2.75}^{3.25} \frac{d x}{\sqrt[3]{6 x+1}}$$

Dwijendra Rao · Accepted Answer

Hello Friends. We have to evaluate to given definite integral that is And they get sent off 2.75 two. Okay. Dx upon Give Rudolph six x plus one. Okay so we can write it in the years and off. Six X plus one. to the power of -1 x three of dx Limit is 2.75. 3.25. So the integration of this will be six express went to the power of minus of one by three plus one upon- of one x 3 plus one multiple with coefficient of X. The limit is 2.75. 3.25. Okay so we can write it six X plus one to the Pavlov two or 3 a bomb 2 14-6 2.75 and two. Um it is a parliament is 3.25. Yeah So this belief one upon four six x plus one to the power off To buy three limited 2.75 2 3.25. So this will be when we put the limit 16 to 3.25 Plus 1 to the power of two x 3- of 16 to 2.75 Plus 1 to the power of two by three. Yeah. 16 to 3.25 plus Fun. Well with 215 to the power two or 3- of 16 to 2.75 plus one to the power of two or three. Mhm. Yeah. Mhm Yeah. Yeah. Okay so we can solve it 71 folks To the power of two x 3. Mhm. 20.5 to the power of to buy three minus of 17.5 hour of two or three. So this will be a cost to 125 multiple with .74985. So this will be cost 2.18 746 So this is the answer. I hope you understood. Thank you.

Patrick Vaughn · Answer

in this problem. We are looking at the interval of six X cubed DX over root X squared plus one from the limits Syria to route three. If we use the use substitution on the Dominator where U equals X squared plus one, then t you will equal to X DX. Our limits are gonna become one Teoh three, Route three squared plus ones. That&#x27;s for the numerator. So too is gonna come out of the sex. Little be left with three and X is going to come out of the X cubes. Obi X squared. We look over here. X squared is gonna equal u minus one you and on the denominator we have review. Now we can distribute this route you to both terms in the numerator. Let&#x27;s stay the same. They have three you. So the 1/2 power minus you to the negative 1/2 hour, Do you? We can now use the definition of the inter role of a polynomial to evaluate this. So that&#x27;s gonna become stony that anywhere. So we get three times 2/3 you to the three halfs minus to over one times you to the 1/2 evaluated from one before. Now we evaluate the limits. This becomes to you. So we get two times route for is to so to to the third that&#x27;s eight minus two times two, which is core and then the lower limit. It&#x27;s since we&#x27;re just plugging at one. We had to minus two. So this chairman, zero and 16 minus four, is 12 so in of that 12.

Amit Srivastava · Answer

I could do integration off DX upon 36 effect a common here it will be a six Xs guy minus Piper six. We have taken six has outside but by six. So final part will be under root X squared minus pipe upon six bullets As we know in Dickerson off and I dont x squared minus a square will be love X plus on Darrelle their square minuses. So one by six look express and adult That&#x27;s the squared minus five past six for the square.

Amy Jiang · Answer

Okay, so, no. So we have hacker box sign interests, and we know that the derivative of that is one over this term here, so we can use yourself. We&#x27;ll let u equal. High gridlock sinjin verse affects. So do you. That&#x27;s equal to one over square root of X squared plus one. Yeah, So I could rewrite every in a girl. Uh um, you do you? And that&#x27;s difficult to you squared over two plus e. But you write that in terms of you. So that&#x27;s singe in verse of X over two. And then this is squared and we&#x27;ll evaluate that. This has over four on 5/12. So we get sine inverse of 3/4 squared over two minus. Sign in verse of 5/12. Word over, Chip.

Patrick Vaughn · Answer

in this problem, we have the integral of one over Exodus six times X to the fifth plus four DX. So the point of this problem is we want to manipulate this in a way that we can start using partial fractions. The reason why we can&#x27;t really use partial fractions now is because of this extra the fifth plus four term. We don&#x27;t really have a good way to factor that. So we&#x27;re gonna want to play with this a little bit. So suppose we let u equal Teoh extra 50 plus for Because it&#x27;s kind of the problem child here. So then d you would equal five extra before uh, D X. Okay, So what would happen if we multiply a top and bottom of our fraction by excellent forth. So that would get us exit the fourth over x to the 10th times X to the fifth plus four DX. So this is kind of what we&#x27;re looking for. So notice that what is X to the fifth equal to so X to the fifth is equal to U minus four. So here we have an extra fifth squared. Um, here we have a you and here we if we put the d X in the top. So here we have a 1 50 You kind of get us what we want so we can rewrite this head roll as 1/5 d&#x27;you over u minus four quantity squared times You all right? So now we&#x27;ve got the problem into a form where the Inter Grant is a polynomial divided by call. No meal so we can use partial fractions. So we&#x27;ll have one. Let&#x27;s not use the 1/5 Let&#x27;s pull the 1/5 outfront just cause it&#x27;s easier doing math with one seemed easier than doing math with. So have you minus four squared you is gonna be equal to, so we&#x27;ll have some constant over U plus another constant over U minus four plus 1/3 constant over U minus for quantity squared so we can use the heavy side method to find A and C. So then a is gonna equal one over negative for square. So it&#x27;s gonna be 1 16 c is going to be one over for So Okay, so now we just need to find be on. We&#x27;ll be there. Um, so let&#x27;s go ahead. Multiply the left and right hand sides by the dominator of the left hand side. So we&#x27;re gonna get one is equal Teoh A No, we don&#x27;t have any day 1/16 times U minus 41 squared, Uh, plus B times you times u minus four uh, plus CC was 1/4 you. So far so good. Okay, so we need to start pulling things that Arby&#x27;s and everything else out. Um, if we just look at that, Constance will have an easier Actually, we just want to look at the U terms or the U squared. I lets go for you squared. So we&#x27;re gonna have being used Squared here will have 1/16 you swear is equal to your zero use words. That means that B is equal to negative 1/16. So we got all the pieces. Let&#x27;s rewrite our generals are Interval is going to be. We have a 1/5 in front. Don&#x27;t forget that. What was a a is 1/16. So we&#x27;re gonna have ah 1/16 1 of the you. What was BB is negative. 1/16 1 over U minus four. And finally see, So we&#x27;re gonna have plus 1/4 1 over u minus for quantity squared, do you? Right. So from here, we need Teoh for this term we&#x27;re gonna have to do a substitution is we&#x27;re gonna be equal to U minus four. And for these two, we could just integrate them directly. So have won the times 1/16 l an absolute value U minus 1/16 Ellen Absolute value U minus four. Uh, but some constant. Um, plus 1/20 5th Groll of one over the square. Davey. So far, so good. Okay, so we have one. We can combine these. So 1/80. Who? That&#x27;s five times 16. Ellen. Absolute value of you over absolute value. U my passport. There&#x27;s a constant completely integrate. This will have minus 1/20. Uh, one over already. Got the constant. Okay, So what was the V was U minus four. So 1/80. Ln absolute value U over U minus four, um, minus 1/21. Over U minus four. Awesome concept. What was you? Let&#x27;s go up here. You was equal to X the fifth house for So let&#x27;s go ahead and plug it in. So I have one over a l. An absolute value extra fifth plus four close absolute value all over. So if we have excellent fifth plus four minds for that&#x27;s the same as extra The fifth, uh, minus 1/21 over X the fifth. Sloppy. Thanks. Here we go. Brought some constant, and that completes their come.

Tom Greenwood · Answer

okay, Evaluating this integral in The first thing we have to do is take the five X over the X squared minus X minus six, and you are partial fraction decomposition of this. So what we need to do first ISS to factor the denominator and realize that that factors into an X minus three times X plus two and those are the denominators off our decomposition. And then those being linear and not repeated, we would have just a over X minus three plus B over the X plus two. Now we multiply all that by our common denominator to get rid of fractions. So we just get five x on the left hand side. Each of these denominators will reduce, so we&#x27;re left with the numerator times. The other denominator. So eight times the X plus two se x plus two a and B times the X minus three his B X minus three b And now we can equate our terms on inside quickening coefficients. The five x five gotta equal the A plus B x so five equals a plus. B is one equation it would have and then are constants to a minus three b well knows, add up to zero because there&#x27;s no constant on the other side. So now pick whatever method you want to do to solve this system of equations. Whichever method you&#x27;re using, you should end up with a equals three and B equals two, and that should be a minus three b non A plus three b typographical air there cause it was a minus three B. All right, so there&#x27;s our correct system and three into early. It&#x27;s the correct solution here for a partial fraction decomposition. So we have the integral from negative wanted to of three over X minus three. Pause to over exports to the X so we can separate thes and individual and girls. So we&#x27;re gonna have the girl from negative wanted to pulls a three out in front only that one over X minus three D X and then plus again pull the two out in front of the integral. We have two times an agro from negative 1 to 2 off one over X plus two d X. Because now these air teach in the form one over you. Do you? So we have three times the natural algorithm of the absolute value of X minus three plus two times in natural algorithm of the absolute value of exports to and that&#x27;s gonna get evaluated from negative 12 So now we just have to plug those values in. So we have plugging in the to three times. The natural log Tu minus three is negative. One absolute value of negative one is one plus two times a natural log of two plus two is four minus. Now we put in the negative one three times a natural of negative one minus three is negative for absolute value of negative force for Waas to times the natural log of negative one plus two is one. Well, the natural longer than one is zero. So that zero and that zero. So I have to natural before minus three times a natural before, which gives me a negative one natural log off for There&#x27;s

Problem

Evaluate the given definite integrals. $$\int_{2}…

Problem 21 Easy Difficulty

Evaluate the given definite integrals.
$$\int_{2.75}^{3.25} \frac{d x}{\sqrt[3]{6 x+1}}$$

Answer

$\approx 0.18746$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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$\approx 0.18746$

Basic Technical Mathematics with Calculus

Grace H.

Heather Z.

Caleb E.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Grace H.

Heather Z.

Caleb E.

Michael J.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus