Can someone help me determine if an improper integral converges or diverges?

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Can someone help me determine if an improper integral converges or diverges?

Mathematics
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Hey
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ok I will provide the example :D just give me a sec

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Other answers:

\[\int\limits_{1}^{\infty}(x ^{2}-6x+1)/(x ^{2}+4)\]
I seriously dont understand the whole concept :P
No, it doesn't.
can you explain ur reasoning?
Sure. 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.
= -x^-1
Evaluated at the endpoints gives us?
-1/a +1
Perfect. Now apply the limit. Does it converge, and if so, to what?
it converges it equals 1
right?
Jemurray?
Sorry for the delay. Yes, that's correct. Now do the same for the integral \[\int_{1}^\infty \frac{1}{x} dx \]
Well = lnIxI
Okay, keep going :)
lnIaI-lnI1I =lnIaI
And if you apply the limit then we see it diverges
Indeed it does. You seem to have it down pretty well. :)
well this is easy but I am getting confused with this one
I also dont know when a function is slowly hitting zero as it approaches infinity
LIke how wld you know if it is approaching slowly or quickly?
You 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?
N must be greater than 1. But i know that because my book tells me so
what wld happen if n would be equal to 1?
That was the second integral you calculated :)
oh lol I see
You 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.
oh i see yes thats clear
but lets return to the original problem
How would i know like which part of the function dominates?
What is the limit of the integrand as x approaches infinity?
of which function?
Your original problem.
well there are so many x's in this problem. Some will be negative and others will be positive
Try dividing top and bottom by x^2 and then taking the limit, ignoring terms that go to zero.
sorry what do u mean?
Well now i realized that when x approaches infinity x^2 dominates the numerator
Dividing yields \[ \frac{1 - 6/x + 1/x^2}{1 + 4/x^2} \]
the others are insignificant in comparison
Yes, that's correct. So what's the limit?
x^2/x^2=1 so it equals =x and so it diverges as x approaches infinty
what i meant to say was that the integral of 1=x
Well, 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.
So how wld you solve it?
I wanna learn because i dont feel like i am doing it correctly. I am missing the basics in this area
It's not a question of solving it or not. It doesn't converge, that's the end of the story.
LOL hehe ok got it
so for example \[\int\limits_{50}^{\infty}dx/x ^{3}\]
This we can see automatically converges
since as x becomes larger than 1/x^{3} is getting closer to 0
No, because that would imply that 1/x converges too. Just remember what we said, it will converge if n > 1.
well ok so if i had to explain a reasoning i wld say that /[1/x^{3}/] approaches 0 quickly as x approaches infinity?
Quickly is a bad term. Instead, say that the antiderivative approaches zero, not the integrand itself.
ohhhhh i seee fine
Thanks for your help
I am just trying to logically see this in my brain
So that I will be able to solve on my own
ok gn. If I have any more questions I will be back :D
I hope u dont mind
Sure, 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 :)
ok Great thanks
I finished my hmwrk So i am heading to bed :D
I dont need anymore help :D

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