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anonymous
 5 years ago
Can someone help me determine if an improper integral converges or diverges?
anonymous
 5 years ago
Can someone help me determine if an improper integral converges or diverges?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What's your question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok I will provide the example :D just give me a sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{\infty}(x ^{2}6x+1)/(x ^{2}+4)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I seriously dont understand the whole concept :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you explain ur reasoning?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure. Let's try the integral \[\int_{1}^\infty \frac{1}{x^2} dx\] We define improper integrals as \[ \lim_{a \rightarrow \infty} \int_{1}^a \frac{1}{x^2} dx \] Can you find this integral? Don't worry about the limit yet.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Evaluated at the endpoints gives us?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Perfect. Now apply the limit. Does it converge, and if so, to what?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it converges it equals 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry for the delay. Yes, that's correct. Now do the same for the integral \[\int_{1}^\infty \frac{1}{x} dx \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And if you apply the limit then we see it diverges

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Indeed it does. You seem to have it down pretty well. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well this is easy but I am getting confused with this one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I also dont know when a function is slowly hitting zero as it approaches infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LIke how wld you know if it is approaching slowly or quickly?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You just figured that out for yourself, though you may not realize it. We can find the "boundary" so to speak, or exactly how "fast" a function must approach zero such that its integral converges. Take \[ \int_{1}^\infty\frac{1}{ x^n} dx \] What must n be so that the integral will converge?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0N must be greater than 1. But i know that because my book tells me so

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what wld happen if n would be equal to 1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That was the second integral you calculated :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You see, if n is greater than one, then your boundary term will be 1/x^{something}, which approaches zero as x approaches infinity. If n is equal to one, we have a logarithm which diverges, and if n is less than one, the antiderivative has some power of x in the numerator, which obviously diverges.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i see yes thats clear

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but lets return to the original problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How would i know like which part of the function dominates?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is the limit of the integrand as x approaches infinity?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your original problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well there are so many x's in this problem. Some will be negative and others will be positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Try dividing top and bottom by x^2 and then taking the limit, ignoring terms that go to zero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry what do u mean?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well now i realized that when x approaches infinity x^2 dominates the numerator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Dividing yields \[ \frac{1  6/x + 1/x^2}{1 + 4/x^2} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the others are insignificant in comparison

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, that's correct. So what's the limit?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2/x^2=1 so it equals =x and so it diverges as x approaches infinty

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what i meant to say was that the integral of 1=x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, I suppose you can think of it that way. It's not necessarily okay to just integrate the limit of the integrand but you can see that as x gets really really large you're effectively finding the area of the graph of y = 1, which is clearly infinite.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So how wld you solve it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I wanna learn because i dont feel like i am doing it correctly. I am missing the basics in this area

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's not a question of solving it or not. It doesn't converge, that's the end of the story.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so for example \[\int\limits_{50}^{\infty}dx/x ^{3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This we can see automatically converges

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since as x becomes larger than 1/x^{3} is getting closer to 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, because that would imply that 1/x converges too. Just remember what we said, it will converge if n > 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well ok so if i had to explain a reasoning i wld say that /[1/x^{3}/] approaches 0 quickly as x approaches infinity?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Quickly is a bad term. Instead, say that the antiderivative approaches zero, not the integrand itself.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am just trying to logically see this in my brain

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So that I will be able to solve on my own

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok gn. If I have any more questions I will be back :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure, no problem , though I may be away tonight... if I can answer I will, if not, you can ask everyone else, or I'll be back tomorrow. Good luck :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I finished my hmwrk So i am heading to bed :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I dont need anymore help :D
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