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Interesting Integral

  1. Apr 7, 2004 #1
    Here's a pretty difficult integral that our prof threw at us a couple days ago... left the whole class a little dumbfounded. I wish I had symbols to use (such as the ones that a lot of the pros on this site use)... but I'm sure you'll be able to understand.

    Integrate (from a to b) the following (1-x^2)(f ''(x)) dx

    He went on to say that if we couldnt get that, perhaps we should look at this one:

    Integrate (from -a to a) the following (a^2-x^2)(f ''(x))
    (he suggested IBP)

    I think what threw most people off was how he knew what f''(x) was without identifying it directly.

    Thanks for everyones help in advance!!

  2. jcsd
  3. Apr 7, 2004 #2


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    I can't say I have ever done anything like that. However would it not be fairly easy to say:

    [tex]\int_b^a (1-x^2)f''(x) dx = \int_b^a f''(x) dx - \int_b^a (x^2)f''(x)dx[/tex]

    [tex]\left[ f'(x) \right]_b^a - \int_b^a (x^2)f''(x)dx [/tex]

    And apply by parts on the remaining integral. That's just a guess :confused:
  4. Apr 8, 2004 #3


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    There's no reason to separate the (1- x2).

    Let u= 1-x2 and dv= f"(x)dx
    Then du= -2xdx and v= f '(x)

    so the integral becomes [itex](1-x^2)f'(x)\|_a^b+ 2\int_a^bxf'(x)dx[/itex]

    To do that second integral let u= x, dv= f'(x)dx so
    du= dx and v= f(x). The second integral becomes [itex]xf(x)\|_a^b- 2\int_a^bf(x)dx[/itex].

    The entire integral is [itex](1-b^2)f'(b)-(1-a^2)f'(a)+ 2bf(b)-2af(a)- 2\int_a^bf(x)dx[/itex]

    The "advantage" of using a2-x2 instead of 1- x2 is that the "(1-a2) term is 0 and we have only [itex](a^2-b^2)f'(b)+ 2bf(b)-2af(a)- 2\int_a^bf(x)dx[/itex].

    Of course, we can't evaluate that last integral precisely because we don't know f itself.
    Last edited: Apr 8, 2004
  5. Apr 8, 2004 #4
    ahhh.. thank you both very much for your help!!

    [bows to HallsofIvy]

    very much appreciated :)
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