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For bijector functions, I noticed that there is some weird relation between the representation of the function in a plane and the integration by parts.

By "integration by parts" I mean "u.v = ∫udv + ∫vdu", which is better than ∫udv = u.v - ∫vdu for the representation.

If the function is bijector in a plane in which x is u and f(x) is v, while du and dv represent somehow dx and dy, the area ∫vdu, which is under the graphic, plus the area ∫udv to the left of the graphic turn out to be a rectangle of area u.v.

If not bijector... Well, if there is an exactly u.v area, I can't read it.

http://pt-br.tinypic.com/view.php?pic=rjjfog&s=8#.U929fuNdX9w

However (where my question finally arrives), I have no idea if there is actually an official proccess in which it can be made, or if it is actually related to any propriety of integration, or even if it is right at all, once I can't understand what would ∫udv mean if the function were not bijector.

By the way, I am aware we are working with undefined integrals, not defined ones. And therefore, the areas can't start somewhere and end somewhere else. It's the representation of a defined case for being able to understand the proprieties.

Question(Finally!): Is there any way I can actually do it correctly? Is there any relation between integration by parts and that rectangle?

Thank you, sorry for my terrible English and Math knowledge...

Fábio Aguiar
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I think what you are noticing is that if $f$ is a positive strictly increasing function of sufficient regularity then:

$$ab=\int_{0}^{a}f(x)dx+\int_{0}^{b}f^{-1}(x)dx$$

provided that $b=f(a)$. I am guessing this from the drawing you put up in your question. This can be shown through the drawing you have or the following:

$$\int_{0}^{b}f^{-1}(x)dx=\int_{0}^{a}f^{-1}(f(u))f'(u)du=\int_{0}^{a}uf'(u)du=uf(u)\vert_{0}^{a}-\int_{0}^{a}f(u)du$$

$$=af(a)-\int_{0}^{a}f(u)du=ab-\int_{0}^{a}f(u)du$$

This equality is related to a version of Young's inequality and can be demonstrated under less hypotheses. More information can be found here:

  1. https://proofwiki.org/wiki/Young's_Inequality_for_Increasing_Functions/Equality
  2. Young inequality
  3. http://www.emis.de/journals/JIPAM/images/289_06_JIPAM/289_06_www.pdf

The interpretation of the above is as you described. The area of the rectangle of side lengths $a$ and $b$ can be decomposed into the area under $f$ and $f^{-1}$. Notice that the graph of the inverse is flip of the graph of $f$ across the line $y=x$. I believe this idea is studied in Spivak's book on calculus.

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