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Calculating the integral of x^n e^(2x) from 0 to 1

  1. Aug 25, 2006 #1
    Hey, I got a couple of problems on integration that I can't seem to figure out.

    1. Suppose [tex]I_n = \int_{0}^{1} x^n e^{2x} dx[/tex] Evaluate [tex]I_n[/tex] in terms of [tex]I_{n-1}[/tex] for any natural number n


    2. Suppose [tex]I_n = \int_{1}^{e} \left[\ln x\right]^n dx[/tex] Evaluate [tex]I_n[/tex] in terms of [tex]I_{n-1}[/tex] for any natural number n

    Not sure what to do with these. Do I need to integrate [tex]I_{n-1}[/tex]? What do I put into [tex]I_n[/tex] after I integrate [tex]I_{n-1}[/tex]??
     
  2. jcsd
  3. Aug 25, 2006 #2

    TD

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    With

    [tex]I_n = \int_{0}^{1} x^n e^{2x} dx[/tex]

    then

    [tex]I_{n-1} = \int_{0}^{1} x^{n-1} e^{2x} dx[/tex]

    Use integration by parts to get a relation between I_n and I_n-1.
     
  4. Aug 25, 2006 #3
    Yup, I did that and I got [tex]I_n = \frac{1}{2}x^ne^{2x} - \frac{n}{2}I_{n-1}[/tex]

    For the second question, I let u = ln x, but what would I let dv/dx equal to?
     
  5. Aug 25, 2006 #4

    TD

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    It's a bit unclear to me what you mean with dv/dx, but do you mean identifying the factors for integration by parts?

    Remember that one can integrate ln(x)dx by looking at it as ln(x).1dx and taking u = ln(x) and v = 1, dv = dx.
     
  6. Aug 25, 2006 #5
    Yup! [tex]I_n = x[\ln x]^n - nI_{n-1}[/tex]
     
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