Question

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

Answer 1

Hey

Answer 2

What's your question?

Answer 3

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

Answer 4

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

Answer 5

I seriously dont understand the whole concept :P

Answer 6

No, it doesn't.

Answer 7

can you explain ur reasoning?

Answer 8

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.

Answer 9

= -x^-1

Answer 10

Evaluated at the endpoints gives us?

Answer 11

-1/a +1

Answer 12

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

Answer 13

it converges it equals 1

Answer 14

right?

Answer 15

Jemurray?

Answer 16

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

Answer 17

Well = lnIxI

Answer 18

Okay, keep going :)

Answer 19

lnIaI-lnI1I =lnIaI

Answer 20

And if you apply the limit then we see it diverges

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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?

Answer 26

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

Answer 27

what wld happen if n would be equal to 1?

Answer 28

That was the second integral you calculated :)

Answer 29

oh lol I see

Answer 30

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.

Answer 31

oh i see yes thats clear

Answer 32

but lets return to the original problem

Answer 33

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

Answer 34

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

Answer 35

of which function?

Answer 36

Your original problem.

Answer 37

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

Answer 38

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

Answer 39

sorry what do u mean?

Answer 40

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

Answer 41

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

Answer 42

the others are insignificant in comparison

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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.

Answer 47

So how wld you solve it?

Answer 48

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

Answer 49

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

Answer 50

LOL hehe ok got it

Answer 51

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

Answer 52

This we can see automatically converges

Answer 53

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

Answer 54

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

Answer 55

well ok so if i had to explain a reasoning i wld say that /[1/x^{3}/] approaches 0 quickly as x approaches infinity?

Answer 56

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

Answer 57

ohhhhh i seee fine

Answer 58

Thanks for your help

Answer 59

I am just trying to logically see this in my brain

Answer 60

So that I will be able to solve on my own

Answer 61

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

Answer 62

I hope u dont mind

Answer 63

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 :)

Answer 64

ok Great thanks

Answer 65

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

Answer 66

I dont need anymore help :D