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precal Group Title

Improper Integral

  • one year ago
  • one year ago

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  1. precal Group Title
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    |dw:1350067985984:dw|

    • one year ago
  2. precal Group Title
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    I have the solution, looking for help to understand the process

    • one year ago
  3. experimentX Group Title
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    lol .. "Improper Integral" is not quite the term you describe it.

    • one year ago
  4. ipm1988 Group Title
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    break x^4 in (x^2)^2 then use substitution

    • one year ago
  5. precal Group Title
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    |dw:1350068067914:dw|

    • one year ago
  6. ipm1988 Group Title
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    yes

    • one year ago
  7. ipm1988 Group Title
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    are you familiar with last three special integrals formula

    • one year ago
  8. precal Group Title
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    three special integral formula? not sure which one you are referring to

    • one year ago
  9. ipm1988 Group Title
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    did you use double substitution

    • one year ago
  10. experimentX Group Title
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    let x^2 = u sub ... this would end up into inverse typerbolic function.

    • one year ago
  11. precal Group Title
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    |dw:1350068248916:dw|these two?

    • one year ago
  12. experimentX Group Title
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    well ... you can try that. but I'll give you a shortcut.

    • one year ago
  13. precal Group Title
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    ok I will love to see the shortcut

    • one year ago
  14. experimentX Group Title
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    look at the the inverse hyperbolic of sine http://en.wikipedia.org/wiki/Inverse_hyperbolic_function

    • one year ago
  15. ipm1988 Group Title
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    @experimentX me to please

    • one year ago
  16. experimentX Group Title
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    |dw:1350068353134:dw|

    • one year ago
  17. ipm1988 Group Title
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    @experimentX nice one

    • one year ago
  18. precal Group Title
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    I see it is the first one

    • one year ago
  19. experimentX Group Title
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    |dw:1350068400466:dw| probably you would get inverse hyperbolic for this types ... but generally formula is given as |dw:1350068503882:dw| ... lol all these forms are inverse hyperbolic function. people usually don't use these.

    • one year ago
  20. precal Group Title
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    |dw:1350068751083:dw| solution manual has this as a second step

    • one year ago
  21. experimentX Group Title
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    hmm ... it really was improper intgral!!

    • one year ago
  22. precal Group Title
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    Is this because of the square root function's domain being continuous on [a, infinity)?

    • one year ago
  23. myininaya Group Title
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    |dw:1350068740846:dw| Start with a right triangle Label the sides such that the inside of that radical you have above appears somewhere. |dw:1350068780160:dw| So we will use substitution \[x^2=\tan( \theta) \] |dw:1350068798338:dw| Now we need to take derivative of both sides of our substitution \[2x dx=\sec^2( \theta) d \theta \]

    • one year ago
  24. experimentX Group Title
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    this doesn't converge.

    • one year ago
  25. precal Group Title
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    ok I am following you so far

    • one year ago
  26. precal Group Title
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    yes in the end, I know this diverges. I know the final answer, just trying to learn the process or understand the process (is a better description)

    • one year ago
  27. myininaya Group Title
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    \[2x dx=\sec^2(\theta) d \theta = > x dx =\frac{1}{2} \sec^2(\theta) d \theta \] Now you go to your little expression and make your "suby's" happen.

    • one year ago
  28. experimentX Group Title
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    |dw:1350068915623:dw|

    • one year ago
  29. precal Group Title
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    |dw:1350069081160:dw|

    • one year ago
  30. precal Group Title
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    I guess I am not sure in the set up why [a, infinity) was used vs (-infinity, b] or (-infinity, infinity) is it because of the domain of the square root function

    • one year ago
  31. myininaya Group Title
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    @precal I'm not sure what you are asking

    • one year ago
  32. precal Group Title
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    In the definition of Improper integrals over infinite intervals I have 3 definitions

    • one year ago
  33. myininaya Group Title
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    It is improper integral because of that infinity thing.

    • one year ago
  34. experimentX Group Title
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    probably with definition of improper integral ... when you have limit goes to infinity ... you have improper integral of first kind.

    • one year ago
  35. precal Group Title
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    |dw:1350069479305:dw| provided the limit exists

    • one year ago
  36. precal Group Title
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    I should mention that if f is continuous on [a,infinity), then for the above first definition

    • one year ago
  37. precal Group Title
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    2nd definition If f is continuous on (-infinty, b], then|dw:1350069615971:dw| provided the limit exists.

    • one year ago
  38. experimentX Group Title
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    |dw:1350069732672:dw| this is called improper integral of First Kind.

    • one year ago
  39. precal Group Title
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    3rd definition If f is continuous on (-infinity, infinity), then |dw:1350069715530:dw| probided both limits exist, where c is any real number

    • one year ago
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    |dw:1350069792525:dw|

    • one year ago
  41. precal Group Title
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    |dw:1350069861584:dw|

    • one year ago
  42. experimentX Group Title
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    as long as you have infinity as limits ... just relax .. they all are of same type of improper integral. probably you are confused with improper integral of second kind.

    • one year ago
  43. precal Group Title
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    |dw:1350070134924:dw|maybe it is because it should have stated the following limits, I think I see a typo from the source

    • one year ago
  44. experimentX Group Title
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    the second kind appears when you have singularity in domain.|dw:1350070237179:dw|

    • one year ago
  45. precal Group Title
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    |dw:1350070363244:dw|

    • one year ago
  46. precal Group Title
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    now I see the connection, it has to be establish - no wonder you made the statement earlier about it not being improper

    • one year ago
  47. experimentX Group Title
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    |dw:1350070434570:dw|

    • one year ago
  48. precal Group Title
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    ok I thinkg I got it, it was the set up that threw me off. I think I understand the solution better now. Thanks

    • one year ago
  49. experimentX Group Title
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    try watching this http://www.youtube.com/watch?v=KhwQKE_tld0

    • one year ago
  50. precal Group Title
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    ok thanks, I can use all the help

    • one year ago
  51. precal Group Title
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    Thanks, I enjoyed that lecture.

    • one year ago
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