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Reduction formula question (int by parts)

  1. Apr 9, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [tex]I_{n} = \int^{2}_{0} x^{n}e^{x} dx[/tex] where n is a positive integer. Use integration by parts to show

    2^{n}e^{2} - nI_{n-1}

    By first finding

    [tex]I_{1} = \int ^{2}_{0} xe^{x} dx [/tex]
    find I2 and I3.

    2. Relevant equations

    I'm sure your all aware of the formula for Int by parts. We'll take the [tex]e^{x}[/tex] function as the one to integrate and the [tex]x^{n}[/tex] as the one to differentiate.

    3. The attempt at a solution

    So use integration by parts to find In:

    [tex]x^{n}e^{x} - n\int^{2}_{0}x^{n-1}[/tex]

    Well that's
    [tex]x^{n}e^{x} - nI_{n-1}[/tex]
    is it not?

    So now put the limits in

    [tex][x^{n}e^{x} - nI_{n-1}]^{2}_{0}[/tex]

    [tex][2^{n}e^{2} - nI_{n-1}] - [ - nI_{n-1}][/tex]
    but that's isn't right because I'm getting no [tex] nI_{n-1}[/tex] because they cancel!

    Where have I gone wrong?

    Last edited: Apr 9, 2010
  2. jcsd
  3. Apr 9, 2010 #2

    Gib Z

    User Avatar
    Homework Helper

    The step is correct but neither of them represent [itex]I_n[/itex]. The Integration by parts formula for definite integrals is [tex]\int^b_a u(x) \frac{dv(x)}{dx} dx = u(b)v(b)-u(a)v(a) - \int^b_a v(x) \frac{du(x)}{dx}[/tex].

    You have put in the limits of integration for the second term, but not the first.
  4. Apr 9, 2010 #3


    Staff: Mentor

    You omitted part of the statement above. Show that 2ne2 - nIn - 1 equals what or does what?
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