evaluate-the-integrals-that-converge-int_-infty0-fracd-x2-x-13

Question

Evaluate the integrals that converge.
$$
\int_{-\infty}^{0} \frac{d x}{(2 x-1)^{3}}
$$

Nick Johnson · Accepted Answer

Okay, so here we have. Well, the integral going from native infinity, Um, zero off DX over two X minus one cube. So we just go where to go ahead and replace native infinity with what would be with a or that I&#x27;m using. Be here. So I&#x27;m taking the limit now. As be, um, tens to Well, it tends to native infinity. Right off the inner girl going from B zero now off the X over the quantity two X minus one pupil. Okay, so you go ahead and we&#x27;re gonna evaluate. Well, the indefinite integral DX over two X minus one cube. We&#x27;re gonna use a u substitution, and I&#x27;m gonna let you be equal to two AKs minus one. They were You was equal to X minus one. Um, that implies that two D X is equal to do you, which implies that the X is equal to one half. Do you? Okay, now we have, or the integral of DX over two X minus one. Cube is equal to one half times the integral off. Just, um, won over you, Cube, is you to the minus three. Do you? Okay, This is equal to one half times while you to the minus two divided my by minus two, which is going to be equal to this becomes a minus one over for you squared. And, of course, while classy. Right, Um, which is equal to then back substituting, um, for you. What is you, Will you was equal to two X minus one. This is equal to negative 1/4 times you, But you is two x minus one. Um, squared. Okay. And then, of course, wealth, of course. Plus C right for the indefinite, Integral. Okay, there&#x27;s no way a man for us because, um, we&#x27;re taking while having a limit here to take the limit. Now, remember as well as be, um I was gonna go to infinity. Oh, well, of, um, negative 1/4 times two x minus one square. And now we&#x27;re evaluating from B to zero. So, um, this is gonna be equal to well, again. We have. Of course, the limit as be tends to infinity. Oh, well, we get a negative 1/4 times zero minus one square, which is becomes a minus 1/4 and then Mary plus 1/4 times to B minus. one birth. So we have plus 1/4 times to be minus one quality squared. Okay, Now we go ahead and we take the limit. This is equal to well, negative. 1/4. There&#x27;s no lipstick there. Then we have while plus the limit as bi coastal infinity of 1/4 times B minus one squared. But as, um as b goes to infinity, Right. The denominator here isn&#x27;t Goldman. Infinity times. Infinity minus one. Infinity squared. Infinity times for infinity. We have one over infinity, which is just zero. This is native 1/4 plus zero, which is equal to negative 1/4. So the integral here converges and it converges to Negative. Juan fourth. All right, take care. Are you

Nick Johnson · Answer

Okay, So, um, wealthy. The given in a girl can be were written right for the integral from three to infinity of two. Over X squared, minus one. The X can be rewritten as the integral while, um, as the limit right is be, approaches infinity of the Negro now from three to be off. Two over X squared minus one, the axe. And then we&#x27;re looking at the in a grand here, two over X squared, minus one. We can go ahead and use partial fractions. Right. So too over X squared minus one. Well, that is equal to weaken. Factor that nominated at this is to over X minus one times X plus one. Right. And then set that equal to well, a over the first factor of X minus one plus B over the second factor of acts plus one. Okay. And then we go ahead. We find any and be on, but that a is equal toe one. And then that means that a plus music with zero stiffer B is equal to negative one. So we get that two over X squared, minus one. Right. That&#x27;s equal to one over X minus one minus one over X plus one. So therefore, um, we can have our integral become. We have the limit as be approaches infinity now off, while of the integral from three to be okay within our inner grand then becomes, well, one over X minus one minus one over ax plus one DX. Okay, so now we can integrate both of these. Um, right. This is a natural log here of X minus one, and basically you substitution. Right and natural log here of X plus one. So what we get this is equal to the limit as B approaches infinity. Uh huh. Off while the natural log of the absolute value of X minus one over X plus one. And then we are evaluating from three to be, Um Okay, so this is equal to again the limit as be approaches infinity of the natural lot off while B minus one B plus one minus the natural log off the absolute value of three minus 1/3 plus one. So this becomes the limit, as be approaches infinity. Uh huh. Um, of well, off the natural log of the absolute value of one. Um, we can have one minus one Overbey over one plus one. Overbey minus the natural log of just one half. All right, that&#x27;s all inside our limit. Um, so this is equal to what would be approaches infinity, right one over B 40 as well as one when it will be 40 So 11 minus 01 plus you. So this is just equal to the natural. Log off. One natural log, one minus the natural log, um, one minus the natural log of Chu. Right. Because one Overbey again approaches zero, as be approaches infinity. So therefore the value of the improper integral, um, does exist because what is this? The natural log of one? That&#x27;s again, This is law based E. So either what power is 10 right, So naturally have one is zero. This is zero minus. While the natural log 10 we have minus a negative natural luck of tooth. This is just zero plus. And that&#x27;s what to which is just a natural one of two. This is just equal to the natural log I&#x27;m off to. So therefore right, the value of the improper integral does exist that we have a finite number here. Um and the integral converges and the value of the integral at the integral from three to infinity of two over X squared minus one. The X is equal to the natural log of to how you take care.

Nick Johnson · Answer

Okay, So are given in a girl. Here is the Negro from negative in spending twin affinity of X over X squared plus three d x. So since we haven&#x27;t more native infinity and infinity as both of our endpoints here, we&#x27;re gonna break this up and say, Well, this is equal to on the integral where we have we&#x27;re going from negative infinity up 20 of acts over X squared, plus tree the X and then plus the integral while going from than zero Chew infinity. Oh, ax squared X great Blustery square the X plus in a woman&#x27;s Theo Trinh 30 of acts over X squared plus three squared the X. Okay, so we go ahead. We actually evaluate these in a girl&#x27;s, um we would get here for the first one is this is equal to negative 16 and Chris and a girl, we get 1638 native 16 plus one sex, which of course, is equal to zero. So therefore, um, are given improper integral, um, converges to 20 So the answer is just beautiful to go. Actually, maybe we&#x27;ll put these functions in. Maybe does most of something in graph form and you&#x27;re gonna see that area under the curve. Um, on to the to the left of the of the Y axis is going to be negative. Then be under the X axis and to the the area under the curve. Um, to the right of the y axis is going to be above the X axis and both of those areas, we&#x27;re gonna be equal. So you they basically cancel out in the total area is, well, zero.

Nick Johnson · Answer

one over X minus 1 to 2 thirds DX. Okay, so here we see you be having infinite discontinuity at one. So we go ahead and we take the limit as well be approaches one off. Well, the inter grow now, going from zero to be of, well, one over X minus 1 to 2 thirds. That&#x27;s X minus one to the negative to dirt&#x27;s. Okay, um, we go ahead and your value this integral. So this is going to be equal to the limit as be approaches one off while we get three times X minus one to the one third. Um, And then we are evaluating from well, from zero to Okay, so this is gonna be equal to he&#x27;s limit. Its be approaches one over when we have three times B minus one to the one third, minus three times zero minus one to the one third. This is the limit And be a purchase. One off, three times e minus one to the one third, um, and then well minus a native three becomes plus tree. Okay. Just about the limit. Here. Always be approaches one, while one minus one is zero. So it&#x27;s one third is your right Three times zero is zero. This is even to zero plus three. Where you go to three limit here is equal to three years and therefore the limit is equal to three. If I number so. Therefore the given integral is convergent and convergence to three right digger.

Nick Johnson · Answer

Okay, so we have an improper and around we go ahead and take the limit, as be tends to negative infinity off Now the integral going from zero to be of, well, each of the X dx over three minus two times e to the X. So we go ahead. And when your value weights, um, this integral first and or to do so, I use a u substitution. So I think if we let you here be equal to three minus two e to the X. Okay, that implies that negative too. Me, um, to the X. The X is equal to do you, which implies that, um, e connects the axe e to the x t x is equal to negative one half d You okay, so now are integral here. The integral of each of the X DX over three minus two. Equally X just becomes, or the negative one half come out front of the integral. So we have negative one half have the integral of just d you over you Okay, well, this is not too bad, artist. Um, soft, right. This is just equal to negative one half time, integral of eu over you it&#x27;s just one over you. Right? Which is the natural log of use. We have native one half times the natural log of the absolute value of you. And, of course, plus a constant. That that&#x27;s the cast is not really gonna matter for us here. So we can go ahead and back. Substitute that you waas equal to three minus two e Could be X. So this is equal to negative one half times the natural log of the absolute value of you. But you is three minus two. Those three minus two e to the axe. Okay, um, so we have, um Well, we&#x27;re taking the limit, as be tends to negative infinity, um, member of well of the integral from B zero off E to the X, the X over three minus two e to the X. So this is equal to the limits as he tends to negative infinity. Oh. Ah, Well, negative one half times the natural laws of the absolute value of three minus two e to the X. And we&#x27;re evaluating from B to zero. Okay, So plenty zero first cracked off what you get when you plenty and be and take a limit here. Um, this is going to give us This is equal to the limit, as be tends to negative infinity. Oh, uh, of well, negatives. One half natural log. Um, off. Just walk three minus two, e 20 which is just three minus two, which is just one right. This is the natural Log off one. Then we get it. Plus one half times the natural log off the absolute value of three minus two. He chew the B okay, and then go ahead and actually take the limit now. So this is going to give us what it was. One half come out front, and then we still have which there&#x27;s a limit. So if this is one half times limit as B tens to negative infinity, um, of the natural, log off the absolute value of three minus to B to the B. Um, So now Oh, taking a limit. Now we just get this is equal to one half times the natural log of well of just three minus zero, which just broke three. Right. This is equal to, um while the next lot of three times one half or Ellen of three over to so the given improper Integral is convergent. We have a finite answer here and the in agro converges to the natural log of 3/2. All right, take care.

Nick Johnson · Answer

we have the integral going from zero to infinity of well, one over X squared DX. So both of the endpoints here, right? We have a problem if we have while the lower end point is zero why it gives us an improper integral because we have in the nominator and also the upper end point was infinite, improper, integral on both of our endpoints. So therefore we can take well is the limit Um, twice here, right, This is equal to the limit as well. Um, let&#x27;s say be approaches zero of the integral going from be up to, let&#x27;s say, one of this option D x over x squared and then plus the limit as well be approaches infinity Oops, infinity of then while going from one, um to of one to be what I will be once I can do something Let&#x27;s see, we just maybe maybe k so the arm limit as people to zero of the integral going from beetle one of d x or X squared, plus the limit. Thats a kay approaches that infinity of then going from where we stopped one to you invented of the same function here D x over X squared. Okay, so then Well, evaluating this mineral we get this is equal to the limit as be approaches zero off. Well, negative. Um, won over. Axe evaluated from b toe one plus the limit as K approaches approaches infinity of well, negative one over. X waiting from 12 Okay, okay, so we get this is equal to I&#x27;m the limit as he approaches zero where we get we get negative one. Just negative 1/1 plus one over b and then plus the limit as que approaches infinity off while negative one. Okay. Plus one. Over. Okay, Well, native of one is just negative one, right. Plus one over. Be taking the limit here. We just get well. This is equal to the native. One can come outside. The limit is going to be equal to the limit as beef approaches. Zero off just well. One Overbey. And then we have wild minus one. Right. And then here was K approaches infinity. Negative one over infinity. That&#x27;s close to zero plus one. So this is just the limit as being put zero, what will be minus one and then this limit, just like I was the one we have plus one. Well, one over. Be right. It goes to infinity. So infinity minus one plus one. That&#x27;s just mean minus one plus one is zero. If you want to think about this doesn&#x27;t matter, Right? This is just equal to the limit as the approaches zero off, one over be well, that&#x27;s equal to infinity. Right? So therefore the given integral integral is guy Urgent, diverted. Get a We don&#x27;t get a finite answer for a limit. Begin infinity! So therefore the improper integral we have is divergent.

Problem

Evaluate the integrals that converge. $$ \int_{-\…

Problem 9 Medium Difficulty

Evaluate the integrals that converge.
$$
\int_{-\infty}^{0} \frac{d x}{(2 x-1)^{3}}
$$

Answer

$-\frac{1}{4}$

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$-\frac{1}{4}$

Calculus Early Transcendentals

Grace H.

Anna Marie V.

Kristen K.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Howard Anton, Irl Bivens, Stephen DavisCalculus Early Transcendentals

Grace H.

Anna Marie V.

Kristen K.

Joseph L.

Howard Anton, Irl Bivens, Stephen Davis

Calculus Early Transcendentals