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So I recently came across the formula $$\int_b^a [f(x)+f^{-1}(x)]dx=a^2-b^2$$ which apparently came up in a final of the MIT integration bee, and is true when $a,b$ are fixed points of $f$. I was wondering how one would go about proving this formula? I have no clue how to even make a start!

Geometry Fan
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  • I guess that you in general should use Riemann sums (I'll try to find a link to an old question). If you assume some regularity on $f$, you could probably split the integral in two, and do the substitution $u=f^{-1}(x)$ in the second part. (or even draw a picture, from where the statement is quite convincing) – mickep Sep 16 '16 at 11:09
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    This technique appears at 2013 MIT integration Bee. . see this.https://math.stackexchange.com/questions/906521/show-fa-a-fb-b-implies-int-ab-left-fx-f-1x-right – Refaat M. Sayed Sep 16 '16 at 11:15

2 Answers2

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Use the substitution $x=f(y)$ we get $$ \int_a^bf^{-1}(x)\,dx=\int_a^byf'(y)\,dy=yf(y)|_a^b-\int_a^bf(y)dy=(b^2-a^2)-\int_a^bf(x)dx $$

Jack's wasted life
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  • Very nice idea...but after the first equality the integral's limits should be, I believe, $;\int_{y_a}^{y_b};$ , with $;f(y_a)=a,,f(y_b)=b;$ . They don't remain the same after the change of variable. +1 – DonAntonio Sep 16 '16 at 12:34
  • @DonAntonio I think they do remain the same as a,b are fixed points. Am I missing something? – Jack's wasted life Sep 16 '16 at 12:41
  • @J I think you are. For example, with those limits, you'd get $;yf(y)|_a^b=bf(b)-af(a);$ instead of the correct $;b\cdot b-a\cdot a=b^2-a^2$ – DonAntonio Sep 16 '16 at 14:44
  • @DonAntonio But $f(b)=b,f(a)=a$ so we get the correct answer. – Jack's wasted life Sep 16 '16 at 14:46
  • @J I can't see where, from what you wrote/ is given, could that be deduced. If you do the substitution $;x:=f(y);$ , then $;f^{-1}(x)=y;$ (because from the very wording of the problem it follows $;f;$ is bijective and thus its inverse exists), so that $;x=a\implies y_a=f^{-1}(a);$ ... – DonAntonio Sep 16 '16 at 14:50
  • @DonAntonio a,b are fixed points, it's given in the question. – Jack's wasted life Sep 16 '16 at 14:51
  • @J I know that: so what? When doing integration by substitution the limits must be changed as well according to the new variable by the law you chose. For example, we can (and must) write $$\int_1^\infty\frac{dx}{x^2}=\int_0^1 du=1$$ under the substitution $;u=\cfrac1x;$ . You can not leave the original limits $;1,,\infty;$ , right? – DonAntonio Sep 16 '16 at 14:55
  • @DonAntonio Limits did change but the new values are the old values because a,b are fixed points - what's the problem with that? For example the change of var $x\to\sin x$ does not change the limit $0$ because $0$ is a fixed point of sine. – Jack's wasted life Sep 16 '16 at 14:57
  • With all due respect, and taking into account I liked your idea for solving this problem, but I think you really need to take a peek at your notes in basic integration. You can read here: https://en.wikipedia.org/wiki/Integration_by_substitution , or also here: http://tutorial.math.lamar.edu/Classes/CalcI/SubstitutionRuleDefinite.aspx . In this last, observe how they state that it is impossible to evaluate directly: either you go back to your original variable, which can be extremely cumbersome and practically impossible many times...or change the limits! – DonAntonio Sep 16 '16 at 15:02
  • @Ja And who said that in this case the limit points are fixed points of $;f;$ ?? Perhaps I missed that... – DonAntonio Sep 16 '16 at 15:03
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    @DonAntonio It's in the question "...is true when a,b are fixed points of f..." – Jack's wasted life Sep 16 '16 at 15:04
  • @Ja Then I say no more: you were right, I was wrong. I honestly didn't see that part! Thank you (no need to go over your basic integration notes...) . – DonAntonio Sep 16 '16 at 15:06
  • @DonAntonio I'm very much relieved :) – Jack's wasted life Sep 16 '16 at 15:08
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If $a$ and $b$ are fixed points of $f$, with $0 \le a \le b$, the points $(a, f(a))$ and $(b, f(b))$ lie on the line $y=x$. Then $\int_b^a f(x)$ and $\int_b^a f^{-1}(x)$ are the area of the region $([0, b] \times [0, b]) \setminus ([0, a] \times [0, a])$. The general case of $a \le b$ follows easily.

Reinstate Monica
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