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Math Help - Integration Transformations Proof

  1. #1
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    Integration Transformations Proof

    I was told this was a rather easy question, but I'm having troubles sorting it all out....

    Given that f is an integrable function on [a,b], prove

    1) For any c>0, \int_{a}^{b}f(x)dx=c\int_{a/c}^{b/c}f(cx)dx
    2) For any c\in\mathbb{R}, \int_{a}^{b}f(x)dx=\int_{a-c}^{b-c}f(x+c)dx

    The proof is said to be simplest by starting with the direct definitions of integrability.
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  2. #2
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    Quote Originally Posted by DonnMega View Post
    I was told this was a rather easy question, but I'm having troubles sorting it all out....

    Given that f is an integrable function on [a,b], prove

    1) For any c>0, \int_{a}^{b}f(x)dx=c\int_{a/c}^{b/c}f(cx)dx
    2) For any c\in\mathbb{R}, \int_{a}^{b}f(x)dx=\int_{a-c}^{b-c}f(x+c)dx

    The proof is said to be simplest by starting with the direct definitions of integrability.
    What, particularly, are you having trouble with?
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  3. #3
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    This is what I've got so far...

    For 1)
    taking the definition of an integral, we know that the integral is equal to the supremum of the lower sums and the infimum of the upper sums:
    \int_{a}^{b}f(x)dx=sup _P(L(f,P))=inf _P(U(f,P))

    taking this a step farther, we can put this into summation notation:
    \int_{a}^{b}f(x)dx=sup _P(\sum_{1}^{J}m_j(x_j-x_{j-1}))

    Then it's as easy as seeing that since f(x) has become f(cx), we now have
    sup _P(\sum_{1}^{J}m_j(cx_j-cx_{j-1}))

    then
    sup _P(\sum_{1}^{J}m_jc(x_j-x_{j-1}))

    then
    sup _P(c\sum_{1}^{J}m_j(x_j-x_{j-1}))

    then
    c sup _P(\sum_{1}^{J}m_j(x_j-x_{j-1}))=c\int_{a}^{b}f(cx)dx

    so it's obvious where the c comes from, it's the a/c and b/c of the integral that i can't quite explain
    Last edited by DonnMega; December 2nd 2009 at 02:56 PM.
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  4. #4
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    So I've finished the proof! YAY! But...I've come up with a slightly different answer. Have i gone wrong somewhere? Is the question wrong? Any comments will help, thank you!

    pf//
    taking the definition of an integral, we know that the integral is equal to the supremum of the lower sums and the infimum of the upper sums:
    \int f(x)dx=sup _P(L(f,P))=inf _P(U(f,P))

    taking this a step farther, we can put this into summation notation:
    \int f(x)dx=sup _P(\sum_{1}^{J}m_j(x_j-x_{j-1}))

    Now we see that since f(x) becomes f(cx), we essentially have a mapping x->cx and our image is
    \int f(cx)dx= sup _P(\sum_{1}^{J}m_j(cx_j-cx_{j-1}))

    from here we simplify and bring our constant c completely out
    sup _P(\sum_{1}^{J}m_jc(x_j-x_{j-1}))

    sup _P(c\sum_{1}^{J}m_j(x_j-x_{j-1}))

    c sup _P(\sum_{1}^{J}m_j(x_j-x_{j-1}))=c\int f(x)dx

    Equating this to the left side of the equation after mapping, we see that
    \int f(cx)dx=c\int f(x)dx

    \frac {1}{c} \int f(cx)dx=\int f(x)dx

    As for the change in the limits of integration, we first note that our mapping of x -> cx has shifted our interval
    [a,b] -> [ca, cb]

    to obtain our origional interval, we must compose our limits of integration with the inverse mapping function cx -> \frac {cx}{c} so we get
    [a,b] -> [\frac {a}{c},\frac {b}{c}]

    putting this all together, we see that
    \frac {1}{c} \int_{a/c}^{b/c}f(cx)dx=\int_{a}^{b}f(x)dx

    Q.E.D (sort of)

    So.....why am i getting \frac {1}{c} when i "should" be getting just c???
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