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Integration problem

  1. May 8, 2005 #1
    Problem:

    (a) If [tex]f[/tex] is one-to-one and [tex]f^{\prime}[/tex] is continuous, prove that

    [tex]\int _a ^b f(x) \: dx = bf(b) - af(a) - \int _{f(a)} ^{f(b)} f ^{-1} (y) \: dy[/tex]

    (b) In the case where [tex]f[/tex] is a positive function and [tex]b > a > 0[/tex], draw a diagram to give a geometric interpretation of part (a).

    My work:

    (a) [tex]\int _a ^b f(x) \: dx[/tex]

    Integrating by parts gives

    [tex]u = f(x) \Rightarrow \frac{du}{dx} = f ^{\prime} (x) \Rightarrow du = f ^{\prime} (x) \: dx[/tex]
    [tex]dv = dx \Rightarrow v = x[/tex]

    [tex]\int _a ^b f(x) \: dx = \left. xf(x) \right] _a ^b - \int _a ^b x f ^{\prime} (x) \: dx[/tex]
    [tex]\int _a ^b f(x) \: dx = bf(b) - af(a) - \int _a ^b x f ^{\prime} (x) \: dx[/tex]

    Applying the Substitution Rule gives

    [tex] y = f(x) \Leftrightarrow x = f^{-1} (y) \Rightarrow \frac{dy}{dx} = f ^{\prime} (x) \Rightarrow dx = \frac{dy}{f ^{\prime} (x)}[/tex]

    [tex]y(b) = f(b)[/tex]
    [tex]y(a) = f(a)[/tex]

    [tex]\int _a ^b f(x) \: dx = bf(b) - af(a) - \int _{f(a)} ^{f(b)} f ^{-1} (y) \: dy[/tex]

    (b) I'm not sure how I should handle this one. The left-hand side is quite easy to visualize: it corresponds to a generic integral from a to b. The right-hand side does not seem to be that simple, and I need some help.

    Any help is highly appreciated.
     
  2. jcsd
  3. May 8, 2005 #2

    arildno

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    Dearly Missed

    1) Draw orthogonal x-and y-axes on a piece of paper
    2) Mark the interval [a,b] on the x-axis, and draw f(x) over it, so that it ranges between f(a) and f(b); mark f(a) and f(b) on the y-axis.
    3) What rectangle can you naturally construct whose area is a*f(a)?
    4) What rectangle can you naturally construct whose area is b*f(b)?
    5) Look and behold, and see if you find an easy geometric interpretation of the equality..
     
    Last edited: May 8, 2005
  4. May 8, 2005 #3

    saltydog

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    You know Thiago, I don't claim to know much math, no more than 1% in fact, so I'm not surprised your relation is news for me and I'm sure you will follow Arildno so I don't think I'm giving anything away by posting the attached plot for the function [itex]y(x)=0.2 x^2 [/itex] from 2 to 4.
     

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    Last edited: May 8, 2005
  5. May 8, 2005 #4

    arildno

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    I met this relation in my first analysis course, and it was, to me, one of those "Wow, math is really cool!"-experiences..
     
  6. May 8, 2005 #5
    Thank you guys! I can now see what you're talking about. It's pretty straight-forward. I was a bit confused by a "big fat" generic equation. :)
     
  7. May 8, 2005 #6

    saltydog

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    Just want to correct my statement above: The function [itex]y(x)=0.2x^2[/itex] is NOT one-to-one and only because it's so in the first quadrant would it qualify for the relation above with a and b in the same quadrant.
     
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