evaluate-the-integrals-that-converge-int_0infty-frac1x2-d-x

Question

Evaluate the integrals that converge.
$$
\int_{0}^{+\infty} \frac{1}{x^{2}} d x
$$

Nick Johnson · Accepted 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.

Nick Johnson · Answer

the improper integral here. Nickel going negative one to infinity of X over one plus x squared DX. So we go ahead and we take the limit. Um, as well as be approaches infinity of the inner grow. Now from negative one to be lips, um, to pee of, well, X over one plus x squared dx. Okay, so we&#x27;re just gonna go. This is equal to the limit, as be approaches infinity. Um uh, well of one half times the integral from negative one to be of two X over one plus x square the x. Okay. Um, so now, well, this is equal to the limit as be approaching infinity of well, we get. So again, we have the limit here. Right? Okay, are going limit limit has be approaches. Infinity of well of one half times is integral becomes the natural log of one plus x squared. And then we are evaluating from negative one to be so this is equal Chew, um, one half times the limit, as be approaches infinity. Um, off. Well, the natural log of one plus b squared minus one half times the natural log, um of to write, which is which is equal to infinity. So we see here that delimited here approaches infinity. Therefore, the given improper integral is die a virgin. So we are divergent and we cannot compute it because, well, we are governed.

Nick Johnson · Answer

So you have here the integral going from zero oz of infinity off axe times E to the negative X squared DX. Okay, so evaluate this Improper. And after all, we go ahead and dish especially just change our infinity. Right to be there wasn&#x27;t enough for me to be so and let it be tends to infinity. So we have the limit as B goes to infinity of well, the inner grow now from zero to be of x times eat the negative X squared e x. Okay, so this is going to be equal to the limit as he goes to infinity. Um, Of what? While the integral from zero to be, uh, we got a general here. Well, um, were you substitution? Right. Um, this is gonna be equal to one half times while times chew ax e to the negative X squared DX. Okay, um which is just equal Teoh while the one half come out front of our integral. So this is gonna be equal to one half time. The limits as B goes to infinity, um, of the inner grow from zero to be of two acts e to the negative X squared dx Okay, so this integral right? Well, just evaluates chew. Um, negative e to the negative. X squared right by substitution is to cancel out here. This is equal to one half times the limits as B goes to infinity off negative e to the negative X square. And then we&#x27;re evaluating from zero to be So we have one half times the limit be goes to infinity of one of negative e to be square. Right. So right here. I guess this is gonna be equal to one half times the limit as B goes to infinity. Oh, get negative E to the negative b squared. And then they get a plus a plus, right? Because Well, um, yeah, What&#x27;s attracting off minus negative is gonna be a plus, just e to the zero. So plus e to the zero, which is just equal to well, one half times. This is just one minus zero. This is equal to one half times one minus zero, or just one half have one, which is equal to one half. So therefore, the given improper in integral is convergence, and it converges to the value off one. All right, thinker

Nick Johnson · Answer

have the integral going from zero to infinity of E to the negative two x d x so well, we haven&#x27;t improper in a girl here of me every have about when it&#x27;s being infinity. So, um, we just take the limit here so that we take limit as well as be approaches infinity of the integral while going from zero to be o e to the negative two acts the X Okay, well, what is the integral off eats the negative treks. Well, that&#x27;s even native directs the but of my native to this is equal to the limit as be approaches infinity O e actually negative two x divided by negative too. And then we are evaluating this. I&#x27;m from zero to well, to be okay, So what is? It was good to the limit as be approaches infinity is equal to deliver it as the approaches infinity off. Well, e to the negative to be, um, opening up to minus each of zero over. Native to that&#x27;s gonna be e to the negative to be divided by negative too. Plus, while e to the zero over native to, um is just a one over negative minus right minus a negative. One half becomes a plus. One half. Okay, um, and then well, putting in I&#x27;m taking the limit here. This is just equal to one half. So, um, evaluate and we get the answer of one half. Um, So since this limit is finite, right, the Inter the integral convergence and its value is one I take her.

Nick Johnson · Answer

from one to infinity of D X over X times the square root of X squared minus one. Well, our lower limit of one right is gonna make the denominator is zero. That&#x27;s bad. And also we have infinity as the upper limit. She didn&#x27;t break this up and say, Well, we can take to limits here because again one right makes the nominative zero. So let&#x27;s go ahead. Let&#x27;s take the limit as, let&#x27;s say be approaches one. Oh well, the integral going from B to somewhere, right? Because going to infinity. But we have to stop somewhere. In the meantime, take another limit because we have the upper limit needs to take a luminous wall because you have an affinity as the upper limit. Start to say we have the integral going from being to AIDS that I wanted ability or from 1 to 2, um, of this function here, the X over x times, the square root of X squared minus one and then plus the limit plus the limit, plus the limit. As let&#x27;s say, K approaches infinity of then the inner girl going from we stopped to to infinity of this function DX over X times the square root of X squared minus one. Okay, Well, um, taking the integral here one over X times square of excrement is one. This is just inverse c can&#x27;t of X. So this is equal to the limit. As be approaches one of in verse E cans of x. Uhm evaluating now from B 22 plus the limit as que goes to infinity off in Versi Can&#x27;t of X evaluating now from two to a Okay, so where we get here? Well, this is equal to again. Little limit as be approaches. One of what? Well, um, plugging in to the skin in verse. Seek it of too. So hey, this is equal to invent universe. He can&#x27;t of two minus in verse. E can&#x27;t of beef the limit there. And then what we have plus live it as k goes to infinity of inverse, he can&#x27;t of k minus in versi Can&#x27;t of two. Okay, well, both what we get me if you take the limit here. What? We get the limit as be approaches one of oh, well, hinders, um seeking to That&#x27;s just mean And so the kinds and right we just This just becomes the limit as be approaches one of negative inverse seek and of B which is just zero. So this first term, just close to zero, let me have zero plus Well, as K goes to infinity, this is gonna go too high over to we just have zero plus pi over two which of course, is just pile up too, right? This is just equal to pi over two. And therefore the given into girl is convergent We take a limit getting find an answer So are in ago is convergent and converges to pi over two

Nick Johnson · Answer

here. We have a group going from two to infinity. Just take the limit right as he goes to infinity off the integral from two to be off one over x Times square root of Ellen X DX. Okay, so if we evaluate a square root, um, if you evaluate the limit, the integral of one over X times good of a necks. Um, that is going to be well to, um, we do a u substitution. Right. Let, um, and we get this is two times the square root of X plus C. Right. Okay, so, um, now we&#x27;re taking the limit. What we have here is this is equal to the limit, as be tends to infinity of well of to times the square root of Alan X. Okay. And now we are evaluating from two to be So this is equal to again. We have the limit as be heads to infinity. Now, we evaluate this. This is going to be, um, to while two times the square root of Ellen B minus, two times the square root of, um, Ellen, too. So, um, we can do is give it a two outside of the limit. So this is actually gonna be equal to two times the limit as B goes to infinity off the square root of Ellen B minus, two times the square root of elan of To. Okay, so now taking the limit here and letting be Go to infinity. Well, this is just equal to um right. Ellen of infinity is infinity squared. Independence Square root of infinity is infinity. This is just equal to two while two times infinity minus two times the square root of L and two. OK, well, two times infinity is definitely infinity. Right? Then minus while some finite value here is gonna be still infinity. So this is gonna be so That limit here is equal to infinity. Right? And sister, get infinity Here. Um, the given, improper integral is divergence. We can evaluate it because the integral diverges. So we have this integral. It is die virgins. All right,

Problem

Evaluate the integrals that converge. $$ \int_{1}…

Problem 29 Medium Difficulty

Evaluate the integrals that converge.
$$
\int_{0}^{+\infty} \frac{1}{x^{2}} d x
$$

Answer

Divergent

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