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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    Hey

  2. anonymous
    • 5 years ago
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    What's your question?

  3. anonymous
    • 5 years ago
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    ok I will provide the example :D just give me a sec

  4. anonymous
    • 5 years ago
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    \[\int\limits_{1}^{\infty}(x ^{2}-6x+1)/(x ^{2}+4)\]

  5. anonymous
    • 5 years ago
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    I seriously dont understand the whole concept :P

  6. anonymous
    • 5 years ago
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    No, it doesn't.

  7. anonymous
    • 5 years ago
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    can you explain ur reasoning?

  8. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    = -x^-1

  10. anonymous
    • 5 years ago
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    Evaluated at the endpoints gives us?

  11. anonymous
    • 5 years ago
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    -1/a +1

  12. anonymous
    • 5 years ago
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    Perfect. Now apply the limit. Does it converge, and if so, to what?

  13. anonymous
    • 5 years ago
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    it converges it equals 1

  14. anonymous
    • 5 years ago
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    right?

  15. anonymous
    • 5 years ago
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    Jemurray?

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

  17. anonymous
    • 5 years ago
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    Well = lnIxI

  18. anonymous
    • 5 years ago
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    Okay, keep going :)

  19. anonymous
    • 5 years ago
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    lnIaI-lnI1I =lnIaI

  20. anonymous
    • 5 years ago
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    And if you apply the limit then we see it diverges

  21. anonymous
    • 5 years ago
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    Indeed it does. You seem to have it down pretty well. :)

  22. anonymous
    • 5 years ago
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    well this is easy but I am getting confused with this one

  23. anonymous
    • 5 years ago
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    I also dont know when a function is slowly hitting zero as it approaches infinity

  24. anonymous
    • 5 years ago
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    LIke how wld you know if it is approaching slowly or quickly?

  25. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    N must be greater than 1. But i know that because my book tells me so

  27. anonymous
    • 5 years ago
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    what wld happen if n would be equal to 1?

  28. anonymous
    • 5 years ago
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    That was the second integral you calculated :)

  29. anonymous
    • 5 years ago
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    oh lol I see

  30. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    oh i see yes thats clear

  32. anonymous
    • 5 years ago
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    but lets return to the original problem

  33. anonymous
    • 5 years ago
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    How would i know like which part of the function dominates?

  34. anonymous
    • 5 years ago
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    What is the limit of the integrand as x approaches infinity?

  35. anonymous
    • 5 years ago
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    of which function?

  36. anonymous
    • 5 years ago
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    Your original problem.

  37. anonymous
    • 5 years ago
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    well there are so many x's in this problem. Some will be negative and others will be positive

  38. anonymous
    • 5 years ago
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    Try dividing top and bottom by x^2 and then taking the limit, ignoring terms that go to zero.

  39. anonymous
    • 5 years ago
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    sorry what do u mean?

  40. anonymous
    • 5 years ago
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    Well now i realized that when x approaches infinity x^2 dominates the numerator

  41. anonymous
    • 5 years ago
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    Dividing yields \[ \frac{1 - 6/x + 1/x^2}{1 + 4/x^2} \]

  42. anonymous
    • 5 years ago
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    the others are insignificant in comparison

  43. anonymous
    • 5 years ago
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    Yes, that's correct. So what's the limit?

  44. anonymous
    • 5 years ago
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    x^2/x^2=1 so it equals =x and so it diverges as x approaches infinty

  45. anonymous
    • 5 years ago
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    what i meant to say was that the integral of 1=x

  46. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    So how wld you solve it?

  48. anonymous
    • 5 years ago
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    I wanna learn because i dont feel like i am doing it correctly. I am missing the basics in this area

  49. anonymous
    • 5 years ago
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    It's not a question of solving it or not. It doesn't converge, that's the end of the story.

  50. anonymous
    • 5 years ago
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    LOL hehe ok got it

  51. anonymous
    • 5 years ago
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    so for example \[\int\limits_{50}^{\infty}dx/x ^{3}\]

  52. anonymous
    • 5 years ago
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    This we can see automatically converges

  53. anonymous
    • 5 years ago
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    since as x becomes larger than 1/x^{3} is getting closer to 0

  54. anonymous
    • 5 years ago
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    No, because that would imply that 1/x converges too. Just remember what we said, it will converge if n > 1.

  55. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Quickly is a bad term. Instead, say that the antiderivative approaches zero, not the integrand itself.

  57. anonymous
    • 5 years ago
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    ohhhhh i seee fine

  58. anonymous
    • 5 years ago
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    Thanks for your help

  59. anonymous
    • 5 years ago
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    I am just trying to logically see this in my brain

  60. anonymous
    • 5 years ago
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    So that I will be able to solve on my own

  61. anonymous
    • 5 years ago
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    ok gn. If I have any more questions I will be back :D

  62. anonymous
    • 5 years ago
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    I hope u dont mind

  63. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    ok Great thanks

  65. anonymous
    • 5 years ago
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    I finished my hmwrk So i am heading to bed :D

  66. anonymous
    • 5 years ago
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    I dont need anymore help :D

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