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Math Help - Proving a Theorem on Integrals

  1. #1
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    Proving a Theorem on Integrals

    I'm clueless when it comes to proving theorems. It's frustrating that I don't really know what I'm expected to do. I can draw diagrams, and test the theorem for specific functions, but I don't know where to begin on proving the theorem in general.
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    Theorem

    If f is continuous on \Re, prove that

    \int_{a}^{b}f(-x)dx=\int_{-b}^{-a}f(x)dx

    for the case that f(x)\geq 0 and 0<a<b.

    ---------------------------------------
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  2. #2
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    Quote Originally Posted by adkinsjr View Post
    I'm clueless when it comes to proving theorems. It's frustrating that I don't really know what I'm expected to do. I can draw diagrams, and test the theorem for specific functions, but I don't know where to begin on proving the theorem in general.
    ---------------------------------------
    Theorem

    If f is continuous on \Re, prove that

    \int_{a}^{b}f(-x)dx=\int_{-b}^{-a}f(x)dx

    for the case that f(x)\geq 0 and 0<a<b.

    ---------------------------------------

    Just make the substitution u = -x... Of course, you can prove using Riemann Sums as well, but it's longer.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Just make the substitution u = -x... Of course, you can prove using Riemann Sums as well, but it's longer.

    Tonio
    Ok, I have \int_{a}^{b}f(-x)dx=-\int_{a}^{b}f(u)du=\int_{b}^{a}f(u)du

    The only thing I can think of next, is to claim that \int_{b}^{a}f(u)du=\int_{-b}^{-a}f(-u)du which is just \int_{-b}^{-a}f(x)dx since u=-x.

    Is this valid?
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    Quote Originally Posted by adkinsjr View Post
    Ok, I have \int_{a}^{b}f(-x)dx=-\int_{a}^{b}f(u)du=\int_{b}^{a}f(u)du

    The only thing I can think of next, is to claim that \int_{b}^{a}f(u)du=\int_{-b}^{-a}f(-u)du which is just \int_{-b}^{-a}f(x)dx since u=-x.

    Is this valid?
    Of course...because it actually is!

    Tonio
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