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Integration By Parts

  1. Feb 9, 2006 #1
    Okay, so here is the problem I have, which I am getting tripped up on for some reason:
    a) Use integration by parts to show that
    [tex]\int_{a}^{b} f(x) dx = bf(b) - af(a) - \int_{a}^{b} xf'(x) dx[/tex]
    this was pretty easy, just regular old integration by parts with limits of integration.

    b) Use the result in part (a) to show that if y = f(x) then
    [tex] \int_{a}^{b} f(x) dx = bf(b) - af(a) - \int_{f(a)}^{f(b)} f^{-1}(y) dy[/tex]
    I know that the inverse function of f(x) will leave me with x, which gets me to
    [tex] \int_{a}^{b} f(x) dx = bf(b) - af(a) - \int_{a}^{b} f^{-1}(y) f'(x)[/tex]
    but then what happens with f'(x)? For some reason I'm just totally spacing this out.
  2. jcsd
  3. Feb 9, 2006 #2


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    You want the f'(x). It appears in your first formula. Just simplify f-1(y).
  4. Feb 10, 2006 #3
    I think you are working the opposite way. Basically what I'm trying to do is get from
    [tex]\int_a^b f(x)dx[/tex]

    to (if we let [tex] y = f(x)[/tex])

    [tex]\int_{a}^{b} f(x) dx = bf(b) - af(a) - \int_{f(a)}^{f(b)} f^{-1}(y) dy[/tex]

    to (if we let [tex]\alpha = f(a)[/tex] and [tex]\beta = f(b)[/tex])

    [tex]\int_{\alpha}^{\beta} f^{-1}(x)dx = \beta f^{-1}(\beta) - \alpha f^{-1}(\alpha) - \int_{f^{-1}(\alpha)}^{f^{-1}(\beta)} f(x)dx[/tex]
    Last edited: Feb 10, 2006
  5. Feb 10, 2006 #4


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    From part a it should be obvious that it's sufficient to show that the two integrals are equal.

    So, what happens if you try to integrate
    [tex]\int_a^b xf'(x) dx[/tex]
    by subsitution using
    Last edited: Feb 10, 2006
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