1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook