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Either calculate the improper integral or show that the improper integral diverges. y dy: y2 4[Note: The integrand is y/Vy2 4.]...

Question

Either calculate the improper integral or show that the improper integral diverges. y dy: y2 4[Note: The integrand is y/Vy2 4.]

Either calculate the improper integral or show that the improper integral diverges. y dy: y2 4 [Note: The integrand is y/Vy2 4.]



Answers

Evaluate the improper integral or state that it diverges. $$\int_{1}^{\infty} \frac{d x}{x^{4}}$$

Okay. We're going to evaluate improper integral or show that it diverges. No. If we look at our problem, we're focused on either our limit goes to negative infinity or infinity for improper in a role. Or we have to approach vertical ascent open. So consider how the bottom would be equal to 0, -2 times eight would be zero. And if eight equals X squared then x has to equal plus or minus the square root of eight. So that definitely the positive one falls. Actually the negative one falls between zero and negative four. So I also notice our inner Groll is is not the norm. Normally we go from a smaller number to a bigger number. We are going kind of integrating an area backwards. Okay so let's go ahead and show Kind of how you find that zero. So you subtract 16 from both sides. Then we'll divide both sides by -2. And then we take the square root. So here we see the plus or minus. So we are going to rewrite our inter girls. And what we're gonna do is we're going to consider that be will go towards the negative square root of eight from the negative side. So that allows us to integrate from zero to that negative square root of eight. Now a separate inner girl we're going to approach the square root of eight from the other side from the positive side and then continue on until we get to -4. Wait A Second. Let's check this. We have to approach we have to approach from the positive side here and then we're going to be approaching from the negative side here again because they reverse them kind of confused me there. Okay let's go on. Okay so we're gonna re we're going to look at our actual integral. So um let's go ahead and said are you equal to R. 16 minus two X. Squared. And remember when you're trying to integrate usually want to clean up that denominator. So that makes our D. U. Equal to a negative four X. Dx. So we see our X. And R. D. X. So what we're gonna do is we're gonna divide both sides by negative four. So that we end up with that negative four D. You being substituted for the X. Dx. Okay so let's go ahead and write out that integral here to the side. We have that negative 1/4, we have the D. You on top and then we have the you in the denominator. What is that case? This is their special case and it's our case of alan so as we um place that back into our limit problem we'll have our negative 1/4 out front and then we'll do the Ln of the absolute value of our 16 minus two X squared, bringing back our exes that we substituted for you. And then of course we have to consider our other limit. Now remember if any piece um goes to infinity, it means that the whole thing goes to infinity. So if I think about putting that negative square root of eight in square do the piece. I have the Ellen of zero inside my limiter as I take my limit here. Well the Ellen of zero. Remember at zero you have a vertical ascent ope. So that does go to infinity. So we know that everything else goes to infinity. And so this diverges.

We're going to evaluate our improper integral or show that it diverges. So notice that we are integrating from 0-4. Well this function does have a denominator and we do need to make sure that it doesn't equal zero anywhere between zero and four. So let's go ahead and set that equal to zero. So to solve we will have to subtract two from both sides and then divide by a negative three. So it looks like at X equals two thirds. That is an important place. And we do have a vertical acid tope there. So we will be looking at um approaching that 2/3 from both sides. But first we're going to separate it into the two inter girls and then we can write it with our limits. So as we start to write it with our limits, I'm thinking that we should actually kind of figure out what the integral of this is first. So let's go ahead and think about this as a Um -1 3rd power. So if we were going to integrate our two minus three X. All to that negative one third power. We can see that we can use power rule but we also have to use U. Substitution. So in your use substitution. Notice that you have a negative three in front of your ex. So if this is your you that inside function your D. You would be a negative three D. X. You're gonna have to replace your dx with a negative one third D. U. So you get that negative one third. That comes out. And then we also have the situation where we go up a power and then we multiply by the reciprocal. So you can see that we're actually gonna have a negative one half out in front when those threes cancel out. So that's gonna save us a little space and time because we did that integral to the side. But now that we have that anti derivative, we can place it into our formulas and notice B. Is approaching um two thirds from the negative side where A. Is approaching it from the positive side. So we're approaching our vertical ascent opened, not going through it. Okay, so now we're ready to place some values in, so we can go ahead and we can place our B value into our statement here. So we have that to minus three, B to the two thirds. Now, when we put zero in we can see that it's just gonna be a two to the two thirds. Now on the other side, um we have to place a four in and when we place four and you can see that that's going to be a to minus 12, So that's going to be a negative 10 inside. Of course it gets raised to the square and then to the third power and then we have our A term. Okay, so let's think about what our when we put our B. Value and are a value in what's going to happen. So with that two thirds multiplied by the three, we'll get a two. And so we have a tu minus two. So that zeros out. In fact they both zero out in both situations but zeros are fine. It's 1/0. Is that we have to be concerned about. So we do have a zero and then we have that positive one half times two to the two thirds power. And now notice our next term has a negative 10 but it gets squared and then to the third power. So that's the same as 10 squared to the third power. So we can actually drop that negative because it's not useful because of the squaring. So right. My final answer, I am going to um factor out the one half And then I'll write to to the 2/3 power -10 to the 2/3 power. So our improper integral actually converges.

We're going to evaluate the improper integral or show that it's divergent. So notice that we're integrating from four to infinity. So we're actually going to write this as a limit. We're going to do our limit as b goes to infinity. And then we'll integrate from four to be No. Um let's look at how to actually integrate R D X over pi minus X. All to the two thirds power. So we really need to use substitution. Be a lot nicer if that denominator was just a simple you but through you substitution we see that are derivative is just a negative dx. So we do need to bring extra negative in and then we're going to bring that you to the two thirds power and make it a U. To the negative two thirds power. So now that we can go up a power so we go to one third and then we multiply by the reciprocal so we'll have a negative three out front. Okay? So now that we've done that in terms of you were going to go ahead and replace are you with that pie minus X and place it back into our limit problem. So now we have the limit as B goes to infinity of negative three times pi minus X. All raised to the one third power we're going to first place are being and then we're going to subtract placing the foreign Mhm. Something. Mhm. Okay so now we need to consider what happens as B goes to infinity will notice that we have that pi minus B. So as you place numbers in there they are going to be negative but they are going to get bigger and bigger and bigger. So what happens is um that piece will go to infinity. It's actually like a negative three times negative infinity. But that will go towards infinity. And our second piece, even though it does go to an actual value. If one part of it goes to infinity, the whole piece goes to infinity. So this improper 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 at 80 and then we'll take the limit of the new integral. As the variable replacement approaches infinity or negative infinity. So this will become the integral from teacher three of x cube over x rays to the fourth therapist one dx. And then you will take full limit as T approaches negative infinity. Now our next step will be to apply substitution. We want to let U equal to x rays to the fourth power plus one And you will be four x cubed DX. This gives us 1/4 DU equal to x cubed dx. And then from here we will change our bounds from X to you. So if X is equal to T we have you equal to T to the 4th power plus one. And if X is equal to three we have you equal to three to the fourth power. That's 81 plus one equals 82. So from here we have limit as the approaches negative infinity of the integral from T to the 4th para plus one 2 82 of 1 4th do you over you. And this is equivalent to the limit as T approaches negative infinity of 1/4. L N absolute value view evaluated from T to the fourth era +12 82. Now. So when U is equal to 82 this will be limit as he approaches negative infinity of 1/4. L N absolute value of 82 minus 1/4. L N absolute value of T. To the fourth power plus one. And if you plug in infinity this will be 1/4 l. Of 82 minus 1/4. L. N off negative infinity raced to the fourth power becomes positive infinity and you add one that's still infinity. If you take the absolute value that's still infinity. And if we take Ellen of infinity this is infinity. So if we subtract this from 1/4 Ellen of 82 this will give us a value that is negative infinity. And because this is infinite, we say that the improper integral diverges.


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