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Math Help - Inverse function on definite integral

  1. #1
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    Inverse function on definite integral

    The function \displaystyle f and its inverse \displaystyle f^{-1} are continuous. If \displaystyle f(0) = 0, find \displaystyle\int^5_0 f(x)\ dx + \int^{f(5)}_0 f^{-1}(t)\ dt

    I really have no idea how to proceed.

    Answer: 5f(5)
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  2. #2
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    I would interpret the problem geometrically, in terms of area. Obviously,

    \displaystyle\int_{0}^{5}f(x)\,dx

    is the area under the function f(x), if we assume f(x)\ge 0\;\forall\,x\in[0,5], which we will do for now. Now the graph of the function f^{-1} is just the graph of f, but flipped over the line y=x. So the integral

    \displaystyle\int_{0}^{f(5)}f^{-1}(t)\,dt

    is the area under the function f^{-1}. If you plot this up for a few functions, you might notice that if you flip the area represented by

    \displaystyle\int_{0}^{f(5)}f^{-1}(t)\,dt

    over the line y=x, that will just equal the area required to fill the rectangle whose area is 5f(5).

    The result follows for positive functions, at least.

    I should point out that one assumption here, that is justified by the hypothesis that both f and f^{-1} are continuous, is that f^{-1} exists. That means f would be 1-1.
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  3. #3
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    Quote Originally Posted by nvwxgn View Post
    The function \displaystyle f and its inverse \displaystyle f^{-1} are continuous. If \displaystyle f(0) = 0, find \displaystyle\int^5_0 f(x)\ dx + \int^{f(5)}_0 f^{-1}(t)\ dt

    I really have no idea how to proceed.

    Answer: 5f(5)
    Hint in the 2nd integral let

    u=f^{-1}(t) \iff f(u)=t \implies f'(u)du=dt

    and the new limits of integration are

    t=0 \implies u=0 \text{ and } t=f(5) \implies u=5

    then use integration by parts.

    Also this can be seen on a graph.

    The first integral is in blue the inverse is in brown.

    Inverse function on definite integral-plot.bmp

    Edit: Wow I am really slow at posting today
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