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ParagraphSv 4517.Which is the radius of convergence of the series (Yx"18 Determine whether the improper integral diverges Or converges Evaluate theintegral if ...

Question

ParagraphSv 4517.Which is the radius of convergence of the series (Yx"18 Determine whether the improper integral diverges Or converges Evaluate theintegral if it convergesdx19.Let R= {(xJ):xsysrtlad 1sys3} Find [J,(x'+yJaxdy20 Evaluate the double integral | dydx V 1 +

Paragraph Sv 45 17.Which is the radius of convergence of the series (Yx" 18 Determine whether the improper integral diverges Or converges Evaluate the integral if it converges dx 19.Let R= {(xJ):xsysrtlad 1sys3} Find [J,(x'+yJaxdy 20 Evaluate the double integral | dydx V 1 +



Answers

In Exercises $5-40$ , determine whether the improper integral converges
and, if so, evaluate it.
$$
\int_{1}^{\infty} \frac{d x}{x^{20 / 19}}
$$

In this problem, we're gonna take a look this improper, integral and determine whether or not a convergence or diverges. So the first step that I would take is to convert this into type of function that begins the power rule for so we'll let X b negative 19 over 20 on D X. And so what we're gonna do with this is take, um, limits as let's let our go to infinity. And we have the integral that have one toe are of X to the power of negative 1 19 year, a negative 19 over 20. And so we'll take the anti derivative of this. We'll have our limiting value still, and we add one thio, uh, negative 19 over 20 which gives us X to the one over 20th power. And then we'll divide by one over 20 which is the same thing as multiplication by 20. Um and then we will evaluate this from one to our and so we get the limit as our goes to infinity. Uh, 20 times are too the one over 20th power minus 20 tens one. And of course, the function of art to the power of one over 20 as our goes to infinity would in fact approach infinity. So this particular, um, improper inter girl is divergent

In this problem, We're going to calculate this improper, integral. The improper limit is gonna be zero in this case because you can't divide by zero. So we'll change this into the limit as our goes to zero from the right side. And I always like to change my in a grand if I can to be just a power rule, so we'll let this be X to the negative. Ah, 20 over 19 d X. Yeah, so are anti. Let's see, let's do the limit first with limited as our goes to zero from the right and then our anti derivative will be X to the negative won over 19 and we'll divide that, uh as well. So that same thing is multiplication by a native 19 and we re evaluate from our 2 to 5. Okay, so we had limit as our goes to zero right side of negative 19 times five to the negative won over 19 minus negatives. We've plus 19 times are to the native won over 19. And the issue that we have here is not really with, um, with this piece this pieces numeric, That's okay. This is the problem. You are to the naval one over 19 is really one over R to the positive one over 19. And if we were to apply our limited value as we approach here from the right side, we're gonna get and undefined, um, value there. And so this this particular improper Inderal is divergent.

All right. So here we have another improper into grow because we'll have d x over over zero for this lower bound. So we're gonna rewrite it as the limit as our goes to zero from the right. Uh, aren't 25 of d X over X to the 19 over 20 which equals the limit from our ghost zero Positive. Ah ah, not another equal sign of So this is X to the minus 19 over 20. So we're gonna have ext the one over 20 times 20. So the derivative of that would be this evaluated from five to our and this can be re written is the limit, uh, as our ghost zero from the right of 20 X to 20 times 20 times five to the 1/20 minus 20 times are to 1/20. And as our heads to zero from the right this term is just gonna get smaller and smaller are to the 1/20 is going ahead to zero. And so 20 yards the 1/20 is going ahead to zero, leaving us with just this term. So this is gonna equal 20 times five to the one over 20. And that's what this inter go, this improper, integral comfort just to

To evaluate this integral here. We're gonna use the limit as a variable approaches infinity instead of leaving the upper limit as infinity. Because we can't direct substitute that. So we'll go ahead and rewrite that as the limit from zero to be. Let me go off to infinity then, uh, to evaluate the integral itself. We're going to use the use substitution if we make you X over to and then do you derivative of X over two is just one half dx. Then isolate the X, multiply the to by both sides. We'll get to do you is equal to DX. So that's what we'll go ahead and replace here for this DX too, do you? Then we could even go ahead and push that to to the front. And then we're really integrating Sign of you, Do you? So the integral what do we have to take the derivative of? To get signed, basically, is the ideas that would have to be negative to, but sorry, Negative too. Code side of you. And we'll go ahead and just plug. Are you right back in X over to leave the limit? The approaches infinity out front and this is going to be from zero Whoops. I wanted to do that red zero to be. And so, as we go ahead and evaluate this now we're going to get negative, too. Co sign, be over to and then minus, uh, negative, too. So it'll turn into a plus to cosign of zero. However, we still have this limit as the approaches infinity out front of here is well, so if we go ahead and apply that now, then this is going to change to cosign of infinity. And because that is undefined, it's just gonna oscillator. Then this whole limit is going to diverge because it's not going to collapse onto one specific value.


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