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Math Help - Integration by parts problem.

  1. #1
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    Integration by parts problem.

    Ok so I don't understand what I'm supposed to be doing in this problem:

    if f(0) = g(0) = 0 and f" and g" are continuous, show that

    \int_0^a f(x)g''(x) \, dx = f(a)g'(a) - f'(a)g(a) + \int_0^a f''(x)g(x) \, dx
    Last edited by mr fantastic; September 27th 2011 at 09:24 PM. Reason: Title, added integral terminals using latex.
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  2. #2
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    Re: Integration by parts problem, please help?

    Quote Originally Posted by ourdemise View Post
    Ok so I don't understand what I'm supposed to be doing in this problem:

    if f(0) = g(0) = 0 and f" and g" are continuous, show that

    \int(from 0 to a) f(x)g"(x)dx = f(a)g'(a) - f'(a)g(a) + \int(from 0 to a)f"(x)g(x)dx
    \int_0^a f(x) \cdot g''(x) \, dx

    u = f(x) , dv = g''(x) \, dx

    du = f'(x) \, dx , v = g'(x)

    \left[f(x) \cdot g'(x)\right]_0^a - \int_0^a g'(x) \cdot f'(x) \, dx

    do parts again w/ the last integral
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Integration by parts problem, please help?

    You have to prove that:
    \int_{0}^{a} f(x)g''(x)dx=f(a)g'(a)-f'(a)g(a)+\int_{0}^{a}f''(x)g(x)dx
    If we use integration by parts on this integral: \int_{0}^{a}f(x)g''(x)dx then, let f(x)=u with dv=g''(x) \Rightarrow v=g'(x)
    Therefore we get:
    \int_{0}^{a}f(x)d[g'(x)]=[f(x)\cdot g'(x)]_{0}^{a}-\int_{0}^{a} f'(x)g'(x)dx
    Now use integration by parts again to compute \int_{0}^{a}f'(x)g'(x)dx

    EDIT: sorry, I didn't see skeeter's post
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