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Homework Help: Integration problem am I correct?

  1. May 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Integrate: [tex]\int[/tex]2x*ex2dx


    2. Relevant equations
    [tex]\int[/tex]eudu = eu


    3. The attempt at a solution
    I tried integration by substitution:
    Let u=x2
    Then du=2x*dx

    So then it is:
    [tex]\int[/tex]eu*du

    Which would give me:
    eu+c

    So subbing x2 back in for u:
    ex2+c

    I looked this type of function up online and got a lot of complicated results involving square root of pi or a log function and things referring to an error formula integral or something? :confused: ... so I was just double checking to see if this seemingly simple solution works. And maybe someone could explain what I was seeing online? Thanks
     
    Last edited: May 11, 2010
  2. jcsd
  3. May 11, 2010 #2
    Well first you forgot the dx in the original expression.

    Your work and answer looks right to me. I wouldn't look online for ways to do things or answers because you usually will either find incorrect things or stuff that isn't at your level yet.
     
  4. May 11, 2010 #3

    LCKurtz

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    And, badgermanb, remember you can always differentiate your answer to check it is a correct anti-derivative.
     
  5. May 11, 2010 #4
    Thanks, yeah I remember that now. Worked it out back and forth now. Thanks for the help! It's been a while since I did calc out like this so I keep forgetting my little tips and tricks.
     
  6. May 11, 2010 #5

    Char. Limit

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    Also, badgerman, what you solved (the integral of 2x*e^(x^2)) and what you looked up online (the integral of e^(-x^2)) are different functions and that's why what you looked up was strange. I'm guessing it was something like this?

    [tex]\integral e^{-x^2} dx = \frac{\sqrt{\pi}}{2} erf(x) + C[/tex]

    That's the integral of a different, albeit similar function. Just to clear up what you DID see.
     
  7. May 11, 2010 #6
    Yeah I noticed the negative sign when I was googling it. Amazing how a single negative can change it so much. I think also a lot of the confusion arose from finding a definite as opposed to indefinite integral, in terms of an exact numerical solution. Either way, it's all coming back to me now... once again, thanks for the help :)
     
  8. May 11, 2010 #7

    Cyosis

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    No it is not the negative sign that makes the difference, but the x in front of e^x^2. As you can see by differentiation e^x^2 we obtain 2xe^x^2. There is however no 'standard' function that when differentiated gives e^x^2.
     
  9. May 11, 2010 #8
    That makes sense. A lot, actually. lol
     
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