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Math Help - [SOLVED] area under the curve

  1. #1
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    [SOLVED] area under the curve

    I'm having trouble proving an equation.

    Take f(x) to be a continuous strictly increasing function with f(0)=0 and let {f}^{-1}(x) be its inverse. Then:

    \int_{0}^{a}f(x)dx+\int_{0}^{f(a)}{f}^{-1}(y)dy=af(a)

    This should be visually obvious. The graph of f(x) divides the rectangle with vertices (0,0) and (a,f(a)) in two.

    In the proof, I would like not to assume differentiability (since the equation is obviously true non-smooth functions too)

    Thanks.
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  2. #2
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    Here's what I mean by visually obvious:

    File:Young.png - Wikipedia, the free encyclopedia

    Just put b=f(a) in that figure.

    any ideas?
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  3. #3
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    \int_0^{f(a)}f^{-1}(y)dy =<br />

    <br />
( \: y=f(t) \: \: t=f^{-1}(y) \: )<br />

    <br />
= \: \int_0^{f^{-1}(f(a))}f^{-1}(f(t))df(t) \:= \: \int_0^{a}tdf(t) \: = \: tf(t)| _0^a - \int_0^a f(t)dt<br /> <br />
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  4. #4
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    I forgot that using Stieltjes integrals one is free to do change of variable and integration by parts without assuming differentiability.

    Thanks.
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