How to evaluate this integral?

Discussion in 'Math' started by heathhosty, Aug 9, 2009.

  1. heathhosty

    Thread Starter New Member

    Aug 8, 2009
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  2. Ratch

    New Member

    Mar 20, 2007
    1,068
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  3. steveb

    Senior Member

    Jul 3, 2008
    2,433
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    Notice that 3x^2 is the derivative of x^3. That should give you a clue to the easy way to solve this one.
     
  4. heathhosty

    Thread Starter New Member

    Aug 8, 2009
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    So the integral is just exp(x^3)?
     
  5. heathhosty

    Thread Starter New Member

    Aug 8, 2009
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  6. steveb

    Senior Member

    Jul 3, 2008
    2,433
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    Not quite. Watch out for the negative signs. You said it right above - "substitution". In particular, use u=-x^3 which implies  du=-3x^2 dx.

     \int 3x^2e^{-x^3}dx=-\int e^u du=-e^u=-e^{-x^3}
     
  7. Mark44

    Well-Known Member

    Nov 26, 2007
    626
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    Much better as it is a simpler technique to apply than integration by parts.
     
  8. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    Plus a constant...
     \int 3x^2e^{-x^3}dx=-\int e^u du=-e^u + C =-e^{-x^3} + C
     
  9. Ratch

    New Member

    Mar 20, 2007
    1,068
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    heathhosty,

    Yes, indeed. I did not look at the problem close enough, and gave you bad advice. Substitution is the way to go on this one.

    Ratch
     
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