anonymous
  • anonymous
Can someone help me determine if an improper integral converges or diverges?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Hey
anonymous
  • anonymous
What's your question?
anonymous
  • anonymous
ok I will provide the example :D just give me a sec

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

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