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

- anonymous

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

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- anonymous

Hey

- anonymous

What's your question?

- anonymous

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

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## More answers

- anonymous

\[\int\limits_{1}^{\infty}(x ^{2}-6x+1)/(x ^{2}+4)\]

- anonymous

I seriously dont understand the whole concept :P

- anonymous

No, it doesn't.

- anonymous

can you explain ur reasoning?

- 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

= -x^-1

- anonymous

Evaluated at the endpoints gives us?

- anonymous

-1/a +1

- anonymous

Perfect. Now apply the limit. Does it converge, and if so, to what?

- anonymous

it converges it equals 1

- anonymous

right?

- anonymous

Jemurray?

- anonymous

Sorry for the delay. Yes, that's correct. Now do the same for the integral
\[\int_{1}^\infty \frac{1}{x} dx \]

- anonymous

Well = lnIxI

- anonymous

Okay, keep going :)

- anonymous

lnIaI-lnI1I =lnIaI

- anonymous

And if you apply the limit then we see it diverges

- anonymous

Indeed it does. You seem to have it down pretty well. :)

- anonymous

well this is easy but I am getting confused with this one

- anonymous

I also dont know when a function is slowly hitting zero as it approaches infinity

- anonymous

LIke how wld you know if it is approaching slowly or quickly?

- 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

N must be greater than 1. But i know that because my book tells me so

- anonymous

what wld happen if n would be equal to 1?

- anonymous

That was the second integral you calculated :)

- anonymous

oh lol I see

- 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

oh i see yes thats clear

- anonymous

but lets return to the original problem

- anonymous

How would i know like which part of the function dominates?

- anonymous

What is the limit of the integrand as x approaches infinity?

- anonymous

of which function?

- anonymous

Your original problem.

- anonymous

well there are so many x's in this problem. Some will be negative and others will be positive

- anonymous

Try dividing top and bottom by x^2 and then taking the limit, ignoring terms that go to zero.

- anonymous

sorry what do u mean?

- anonymous

Well now i realized that when x approaches infinity x^2 dominates the numerator

- anonymous

Dividing yields
\[ \frac{1 - 6/x + 1/x^2}{1 + 4/x^2} \]

- anonymous

the others are insignificant in comparison

- anonymous

Yes, that's correct. So what's the limit?

- anonymous

x^2/x^2=1
so it equals =x and so it diverges as x approaches infinty

- anonymous

what i meant to say was that the integral of 1=x

- 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

So how wld you solve it?

- anonymous

I wanna learn because i dont feel like i am doing it correctly. I am missing the basics in this area

- anonymous

It's not a question of solving it or not. It doesn't converge, that's the end of the story.

- anonymous

LOL hehe ok got it

- anonymous

so for example \[\int\limits_{50}^{\infty}dx/x ^{3}\]

- anonymous

This we can see automatically converges

- anonymous

since as x becomes larger than 1/x^{3} is getting closer to 0

- anonymous

No, because that would imply that 1/x converges too. Just remember what we said, it will converge if n > 1.

- 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

Quickly is a bad term. Instead, say that the antiderivative approaches zero, not the integrand itself.

- anonymous

ohhhhh i seee fine

- anonymous

Thanks for your help

- anonymous

I am just trying to logically see this in my brain

- anonymous

So that I will be able to solve on my own

- anonymous

ok gn. If I have any more questions I will be back :D

- anonymous

I hope u dont mind

- 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

ok Great thanks

- anonymous

I finished my hmwrk So i am heading to bed :D

- anonymous

I dont need anymore help :D

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