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Let f be defined on (0, 1] by f(x) (rsin,4 ) 2r sin,- } cos Show that the improper Riemann integral of f converges on (0, 1], but that the improper integral of |fl ...

Question

Let f be defined on (0, 1] by f(x) (rsin,4 ) 2r sin,- } cos Show that the improper Riemann integral of f converges on (0, 1], but that the improper integral of |fl diverges 0 (0, 1]:

Let f be defined on (0, 1] by f(x) (rsin,4 ) 2r sin,- } cos Show that the improper Riemann integral of f converges on (0, 1], but that the improper integral of |fl diverges 0 (0, 1]:



Answers

$$
\begin{array}{l}{\text { Comparing lntegrals Let } f \text { be continuous on the interval }} \\ {[a, \infty) . \text { Show that if the improper integral } \int_{a}^{\infty}|f(x)| d x} \\ {\text { converges, then the improper integral } \int_{a}^{\infty} f(x) d x \text { also }} \\ {\text { converges. }}\end{array}
$$

We are going to evaluate the improper integral or show that it diverges. So we are integrating from zero to pi and notice our denominator we have cosine X minus one. Well when is co sign equal to one? The coastline of zero is actually equal to one. So instead of integrating from zero to pi we're actually going to approach zero from the positive side. So we're going to write that as the limit as a approaches zero from the positive side of the integral of a two pi of R D X over Cosine X minus one. Okay so we have a little bit of a problem here because what we have is that um co sign X minus one is in our denominator. But if we set that as are you we would need signs somewhere. So this is where you have to go back to some of your trig identities and change the form of your trig function. Um and see if you can find a different way to or different form it should be in. So you can integrate it. So a great way to change cosign X minus one and something else is to multiply it by cosine X plus one. Now if you do that on the bottom you also have to do that on the top. So notice that that denominator is made up of um the coastline X minus one and the coastline X. Plus one. So that's equal to cosine squared minus one. Now you can use your um staggering identities. Remember sine squared plus cosine squared equals one. So that means that co sine squared X minus one actually equals a negative sine squared X. And so now we finally clean this up to having a little bit nicer denominator. And we also noticed we have both sign and co sign in our problem now um I'm thinking if my you would Synnex my d you would be co Synnex. But what about that? Plus one? Well we're going to have to separate this into two separate inter girls in order to integrate it. So one of the other girls will be Cosign X. Over sine squared X. Where I can make my use Synnex and I'll put that as a square so that we can set up that. And then the other one is Co Seacon square Tax which we it is somebody's derivative. So it's easy to take the derivative. So now that we know that are you as our Synnex we really have a U. To the negative two. So as we write this we think about going up a power reciprocal so that you is going to be now to the negative one. And then and then we'd actually multiplied by a negative one. So make that positive. And then we have minus. I almost did the derivative of just a regular co sigint. Um The integral of Co sigint and did L. N. And then I stopped myself said no. The integral of cosine squared is actually co tangent and it's actually a negative co tangent. So we're also going to have to make that one positive. Okay so now we're going to be putting in pie and then we're gonna be subtracting putting in zero now one over sine of pi. So the sine of pi though is zero. So before I go on and do all the other pieces, I've already gotten to something that is an issue. This guy is 1/0 which means infinity. And if one piece goes to infinity and probably looks like the co tangent would too. But at this point we just know that this improper integral diverges.

We will be evaluating the improper integral or showing that it diverges. So as we look from zero to pi over two, we can see that if we place pi over two into our denominator we get the value of one. But when we put zero into our denominator We get 1 -1. So we do have a vertical ascent tobe at zero. So we will be taking the limit as a approaches zero from the positive side of the integral from a to pi over two of our function. So now we need to consider it. How are we going to actually integrate this? Well integration is very helpful with our use substitution and making your denominator um cleaner and nicer is usually the way to go. So if U equals one minus cosine X. D. U. Is sin X. Dx. So our whole numerator gets replaced with D. U. And then our denominator becomes the you. So this is our case of Ellen. So we have Ellen of the absolute value of you plus C. We'll go ahead and put it back into our limit statement but now we'll do Ln of our one minus cosine X replacing that you And we're gonna be going from pi over 220. So first putting the pi over two in, we have the Ellen of one minus zero. So Ellen of one is zero. And then we have the Ellen of one minus a co sign a bank. So now when we put the co sign um of a put equal zero in we do get a one minus one which is zero but it's inside of L. N. Well as we approach zero in L. N we're going towards negative infinity. So because of that piece being divergent, the whole integral is divergent.

To evaluate this type of integral. The first thing we have to do is to replace the infinite bound with a variable. That's a T. And then we will take the limit of the new integral. As the variable replacement approaches infinity or negative infinity. So in here we have this integral from to over pie to tea sign of one over X over X squared. And then D. X. And then you will take the limit as T approaches infinity. And then from here we have to do um U substitution. We want to let U equal to one over X. That's the same as x rays. The negative one. So if we take the differential of this we have D U equal to negative x rays to negative two D X. That's the same as negative D U equal to one over X squared dx. And then you will change our bounds from X to you. So if X is equal to two over pie then you will be 1/2 over pie. That's a reciprocal equal to 5/2. And if x is equal to T Then we have you equal to one over T. So from here we have this limit asked he approaches infinity As the integral from Pi over two two. One of routine of sign of you times negative do you? Which is equal to the limit as T approaches infinity of co sign of you. This will be evaluated from five or 2 2, 1 over T. So if you is one of her T. This is equivalent to the limit as T approaches infinity of co sign of one over T-, go sign up by over two. This is equivalent to co sign of one over infinity- Cosette five or 2000. And because one over infinity goes to zero, then the value of co sign of one over infinity approaches one, So we have 1 0 equal to one. And because this is why night we say that the improper integral converges.

We know that if and it's the number of rectangles in the width of the wrecked tank would be Delta Acts is one minus here, over on which is one over an There for the area of the first rectangle would simply be, and therefore we know we have the sum from aye equals one. Don't forget the limit. We know this is greater than the limit. As UN approaches, infinity on equals infinity as you can see the first term and some go to infinity. So it's not Integra ble on a certain interval if it goes towards positive or negative infinity. Therefore there's no limit, so it's not in terrible on this interval 01


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