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AmTran_Bus

  • one year ago

Help with integration problem (by parts)

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  1. AmTran_Bus
    • one year ago
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    |dw:1433866464450:dw|

  2. AmTran_Bus
    • one year ago
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    I know|dw:1433866481311:dw|

  3. AmTran_Bus
    • one year ago
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    So it is uv-integral vdu, correct?

  4. johnweldon1993
    • one year ago
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    That is the correct formula yes \[\large uv - \int vdu\]

  5. AmTran_Bus
    • one year ago
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    |dw:1433866588759:dw|

  6. AmTran_Bus
    • one year ago
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    |dw:1433866619694:dw|

  7. AmTran_Bus
    • one year ago
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    Am I good so far @johnweldon1993

  8. johnweldon1993
    • one year ago
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    Looks good to me just simplify a bit from here

  9. AmTran_Bus
    • one year ago
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    My book says v=x. Why and how??

  10. anonymous
    • one year ago
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    You should have \(dv=dx\), not \(dv=x\). Somehow you separated \(\ln x\) into \(\ln \) and \(x\), which doesn't make sense.

  11. AmTran_Bus
    • one year ago
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    Hum. Ok, that makes much sense.

  12. johnweldon1993
    • one year ago
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    Oh yeah I was wondering about that...if this is ln(x) or ln*x

  13. AmTran_Bus
    • one year ago
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    so|dw:1433866830247:dw|

  14. anonymous
    • one year ago
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    Yes, carry on :)

  15. AmTran_Bus
    • one year ago
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    |dw:1433866867187:dw|

  16. AmTran_Bus
    • one year ago
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    So the two x's on the right can cancel, right? xlnx-integral dx?

  17. AmTran_Bus
    • one year ago
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    so xlnx-x+c?

  18. johnweldon1993
    • one year ago
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    Looks good...I would factor out the 'x' quick to make it look better I guess? \[\large x(ln(x) - 1) + c\]

  19. AmTran_Bus
    • one year ago
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    Ok. Thanks. I think I got this.

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