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precal

  • 2 years ago

Improper Integral

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  1. precal
    • 2 years ago
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    |dw:1350067985984:dw|

  2. precal
    • 2 years ago
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    I have the solution, looking for help to understand the process

  3. experimentX
    • 2 years ago
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    lol .. "Improper Integral" is not quite the term you describe it.

  4. ipm1988
    • 2 years ago
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    break x^4 in (x^2)^2 then use substitution

  5. precal
    • 2 years ago
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    |dw:1350068067914:dw|

  6. ipm1988
    • 2 years ago
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    yes

  7. ipm1988
    • 2 years ago
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    are you familiar with last three special integrals formula

  8. precal
    • 2 years ago
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    three special integral formula? not sure which one you are referring to

  9. ipm1988
    • 2 years ago
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    did you use double substitution

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

  11. precal
    • 2 years ago
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    |dw:1350068248916:dw|these two?

  12. experimentX
    • 2 years ago
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    well ... you can try that. but I'll give you a shortcut.

  13. precal
    • 2 years ago
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    ok I will love to see the shortcut

  14. experimentX
    • 2 years ago
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    look at the the inverse hyperbolic of sine http://en.wikipedia.org/wiki/Inverse_hyperbolic_function

  15. ipm1988
    • 2 years ago
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    @experimentX me to please

  16. experimentX
    • 2 years ago
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    |dw:1350068353134:dw|

  17. ipm1988
    • 2 years ago
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    @experimentX nice one

  18. precal
    • 2 years ago
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    I see it is the first one

  19. experimentX
    • 2 years ago
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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.

  20. precal
    • 2 years ago
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    |dw:1350068751083:dw| solution manual has this as a second step

  21. experimentX
    • 2 years ago
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    hmm ... it really was improper intgral!!

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

  23. myininaya
    • 2 years ago
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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 \[2x dx=\sec^2( \theta) d \theta \]

  24. experimentX
    • 2 years ago
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    this doesn't converge.

  25. precal
    • 2 years ago
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    ok I am following you so far

  26. precal
    • 2 years ago
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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)

  27. myininaya
    • 2 years ago
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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.

  28. experimentX
    • 2 years ago
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    |dw:1350068915623:dw|

  29. precal
    • 2 years ago
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    |dw:1350069081160:dw|

  30. precal
    • 2 years ago
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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

  31. myininaya
    • 2 years ago
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    @precal I'm not sure what you are asking

  32. precal
    • 2 years ago
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    In the definition of Improper integrals over infinite intervals I have 3 definitions

  33. myininaya
    • 2 years ago
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    It is improper integral because of that infinity thing.

  34. experimentX
    • 2 years ago
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    probably with definition of improper integral ... when you have limit goes to infinity ... you have improper integral of first kind.

  35. precal
    • 2 years ago
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    |dw:1350069479305:dw| provided the limit exists

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

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

  38. experimentX
    • 2 years ago
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    |dw:1350069732672:dw| this is called improper integral of First Kind.

  39. precal
    • 2 years ago
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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

  40. experimentX
    • 2 years ago
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    |dw:1350069792525:dw|

  41. precal
    • 2 years ago
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    |dw:1350069861584:dw|

  42. experimentX
    • 2 years ago
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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.

  43. precal
    • 2 years ago
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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

  44. experimentX
    • 2 years ago
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    the second kind appears when you have singularity in domain.|dw:1350070237179:dw|

  45. precal
    • 2 years ago
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    |dw:1350070363244:dw|

  46. precal
    • 2 years ago
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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

  47. experimentX
    • 2 years ago
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    |dw:1350070434570:dw|

  48. precal
    • 2 years ago
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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

  49. experimentX
    • 2 years ago
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    try watching this http://www.youtube.com/watch?v=KhwQKE_tld0

  50. precal
    • 2 years ago
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    ok thanks, I can use all the help

  51. precal
    • 2 years ago
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    Thanks, I enjoyed that lecture.

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