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Thread: integration #2

  1. #1
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    integration #2

    Hi:
    Would somebody check this for me please.



    Also, is this simplified correctly?

    thanks
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  2. #2
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    Quote Originally Posted by stealthmaths View Post
    Hi:
    Would somebody check this for me please.



    Also, is this simplified correctly?

    thanks
    1- You should post this in calculus sub-forum.
    2- You have a sign mistake in the last line.
    3- You should not have that "x" in the second term in the last line.
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  3. #3
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    But if this is in calculus then it is not in pre-university forum. And I am pre-university. Please advise? And how would I now move this thread to calculus?

    How's this?


    And is this simplified correctly?
    thanks
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  4. #4
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    Quote Originally Posted by stealthmaths View Post
    But if this is in calculus then it is not in pre-university forum. And I am pre-university. Please advise?
    Read the "sticky" post at the top of the Pre-Calculus forum. In this American-based web site, calculus is considered university-level, regardless of whether you are actually at university or not.
    Last edited by Opalg; Mar 3rd 2010 at 11:38 AM. Reason: for clarity
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  5. #5
    DBA
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    \int u dv = uv- \int v du

    \int 9xe^{-4x}= 9\int xe^{-4x}

    u=x, du=dx and dv=e^{-4x} , v=(-1/4)e^{-4x}

    9\int xe^{-4x}= 9[-\frac{1}{4}e^{-4x}x-\int -\frac{1}{4}e^{-4x}dx]

    = 9[-\frac{1}{4}e^{-4x}x +\frac{1}{4}\int e^{-4x}dx]
    = 9[-\frac{1}{4}e^{-4x}x +\frac{1}{4}(e^{-4x}(-\frac{1}{4})]+C
    = 9[-\frac{1}{4}e^{-4x}x -\frac{1}{8}(e^{-4x})]+C
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  6. #6
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    I'm sorry DBA. What is it your trying to say. The answer I have above is wrong?
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