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

1. anonymous

Hey

2. anonymous

3. anonymous

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

4. anonymous

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

5. anonymous

I seriously dont understand the whole concept :P

6. anonymous

No, it doesn't.

7. anonymous

can you explain ur reasoning?

8. 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.

9. anonymous

= -x^-1

10. anonymous

Evaluated at the endpoints gives us?

11. anonymous

-1/a +1

12. anonymous

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

13. anonymous

it converges it equals 1

14. anonymous

right?

15. anonymous

Jemurray?

16. anonymous

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

17. anonymous

Well = lnIxI

18. anonymous

Okay, keep going :)

19. anonymous

lnIaI-lnI1I =lnIaI

20. anonymous

And if you apply the limit then we see it diverges

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. 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?

26. anonymous

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

27. anonymous

what wld happen if n would be equal to 1?

28. anonymous

That was the second integral you calculated :)

29. anonymous

oh lol I see

30. 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.

31. anonymous

oh i see yes thats clear

32. anonymous

33. anonymous

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

34. anonymous

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

35. anonymous

of which function?

36. anonymous

37. anonymous

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

38. anonymous

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

39. anonymous

sorry what do u mean?

40. anonymous

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

41. anonymous

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

42. anonymous

the others are insignificant in comparison

43. anonymous

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

44. anonymous

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

45. anonymous

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

46. 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.

47. anonymous

So how wld you solve it?

48. anonymous

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

49. anonymous

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

50. anonymous

LOL hehe ok got it

51. anonymous

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

52. anonymous

This we can see automatically converges

53. anonymous

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

54. anonymous

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

55. 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?

56. anonymous

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

57. anonymous

ohhhhh i seee fine

58. anonymous

59. anonymous

I am just trying to logically see this in my brain

60. anonymous

So that I will be able to solve on my own

61. anonymous

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

62. anonymous

I hope u dont mind

63. 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 :)

64. anonymous

ok Great thanks

65. anonymous

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

66. anonymous

I dont need anymore help :D