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Interval of the definite integral

  1. Jul 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Let F(x) = [tex]\int[/tex][tex]^{x}_____________{0}[/tex] x*e^(t^2) dt for [tex]x\in[0,1].[/tex] Find F''(x) for [tex]x\in(0,1).[/tex]

    My only problem is the x, because the interval of the definite integral goes from 0 to x, and x is in the integral, even though the integral is with respect to dt. So I'd just like to know what happens in this case? Is x a constant (as if the interval went from a to b, rather than 0 to x)? Or something different?
     
    Last edited: Jul 20, 2008
  2. jcsd
  3. Jul 20, 2008 #2

    tiny-tim

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    Hi Nan1teZ! :smile:

    You can rewrite the integral as F(x) = x*[tex]\int^{x}_{0}[/tex] e^(t^2) dt.

    Then just use the product rule. :smile:
     
  4. Jul 20, 2008 #3
    Re: Integral

    Okay so i get F''(x) = (2*e^(x^2))(1+x) + x

    Is that right?
     
  5. Jul 20, 2008 #4

    tiny-tim

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    I don't think so …

    can you show your detailed working? :smile:
     
  6. Jul 20, 2008 #5
    Re: Integral

    oops just a stupid mistake!!..

    okay I got F''(x) = (2*e^(x^2))(1+x^2).

    If that's wrong I'm gonna show the detailed working..
     
  7. Jul 20, 2008 #6

    tiny-tim

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    :biggrin: Woohoo! :biggrin:
     
  8. Jul 20, 2008 #7
    Re: Integral

    Thanks! :D
     
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