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Math Help - Proof

  1. #1
    Senior Member slevvio's Avatar
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    Proof

    Prove \int^{a}_{0} f(x)dx = \int^{a}_{0}f(a-x)dx

    I managed to prove that LHS = F(a) -F(0) and RHS = F(0) - F(a).

    Any help would be appreciated!
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by slevvio View Post
    Prove \int^{a}_{0} f(x)dx = \int^{a}_{0}f(a-x)dx

    I managed to prove that LHS = F(a) -F(0) and RHS = F(0) - F(a).

    Any help would be appreciated!
    Hi,

    you have nothing else to do than use the definitions:
    LHS:

    \int_0^a f(x)dx = F(a) - F(0)

    RHS:
    \int_0^a f(a-x) dx = F(a-a) - F(a-0) = F(0) - F(a)
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  3. #3
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    Quote Originally Posted by earboth View Post
    Hi,

    you have nothing else to do than use the definitions:
    LHS:

    \int_0^a f(x)dx = F(a) - F(0)

    RHS:
    \int_0^a f(a-x) dx = F(a-a) - F(a-0) = F(0) - F(a)
    Not quite, EB.

    Unfortunately, your RHS = - LHS ...... a consequence of the small matter of f(a-x) = f(-[x - a]) being got from f(x) by a translation (a units from the y-axis) and a reflection (in y-axis) .... The definite integral is invariant under the translation, but not under the reflection.

    Therefore (correction in red):

    \int_0^a f(a-x) dx = - [F(a-a) - F(a-0)] = - F(0) + F(a)

    = LHS.


    Personally, rather than explicitly using transformations, I'd implicitly use them by making the substitution u = a - x. Then:

    du = -dx.
    x = 0 => u = a.
    x = a => u = 0.

    Therefore:

    \int_0^a f(a-x) \, dx = \int_a^0 f(u) \, (- du) = - \int_a^0 f(u) \, du = \int_0^a f(u) \, du = F(a) - F(0) = \, LHS

    Note: When reversing the integral limits you multiply by a negative. So the resulting double negative becomes a positive.
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  4. #4
    Senior Member slevvio's Avatar
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    thank you!
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