Assume that $f: [a, b] \rightarrow [a,b]$ is continuously differentiable bijection and that $f(a) = a$, $f(b) = b$.

Show that

$$\int_a^b g(x) \, dx + \int_a^b f(x) \, dx = b^2-a^2$$

where $g:[a, b] \rightarrow [a, b]$ is the inverse function of $f$

With this one I have no idea where to start. What should we do with the information about $f$ being bijective and $g$ being the inverse of $f$?

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    $\begingroup$ Do you know that you wrote {\int_{{a}}^{{b}}}{{g(x)}}{d}{x}}}} where it would suffice to write \int_a^b g(x)\,dx? $\qquad$ $\endgroup$ – Michael Hardy Feb 16 at 17:33
  • $\begingroup$ Your result holds under the more general condition that $f$ is strictly increasing on $[a, b] $. $\endgroup$ – Paramanand Singh Feb 17 at 2:47
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    $\begingroup$ Does this answer your question? integration of sum of $f$ and its inverse $\endgroup$ – Martin R Feb 17 at 9:09

Let $u=f^{-1}(x)$. Then $x=f(u)$ and $dx=f'(u)\,du$. We will integrate by parts:

$$\int_a^b g(u) h'(u)du = \left[g(u)h(u)\right]_a^b-\int_a^b g'(u)h(u)du$$

by choosing $g(u)=u$ and $h(u)=f(u)$, so:

$$ \begin{aligned} \int_a^b u f'(u)du &= \left[uf(u)\right]_a^b-\int_a^b 1\cdot f(u)du\\ &=bf(b)-af(a)-\int_a^b f(u)\,du \end{aligned} $$

Since $f(b)=b$ and $f(a)=a$, we have:


Chaining all this together:

$$ \begin{aligned} \int_a^b f^{-1}(x)\,dx&=\int_a^b uf'(u)\,du\\ &=\int_a^b u(f(u))'\,du\\ &= \left[uf(u)\right]_a^b-\int_a^b f(u)\,du\\ &=b^2-a^2-\int_a^bf(x)\,dx \end{aligned} $$

from which we deduce the conclusion.

  • $\begingroup$ There's something Im not seeing here. Im used to using this $\int udv = uv - \int v du$ for integrating by parts. What is our $dv$ here? $\endgroup$ – Daniel Feb 16 at 16:32
  • $\begingroup$ @Daniel, that's $f'(u)$. In your notation $(u,v)$ correspond to $(u,f(u))$ in mine. $\endgroup$ – LHF Feb 16 at 16:37
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    $\begingroup$ I changed $f'(u)du$ and $f(x)dx$ to $f'(u)\,du$ and $f(x)\,dx.$ The latter is standard. $\endgroup$ – Michael Hardy Feb 16 at 17:34
  • $\begingroup$ @Atticus Is there any chance you could include a couple more steps to this? Im not sure I see where you applied integration by parts and how did $\left[uf(u)\right]_a^b = b^2 - a^2$. $\endgroup$ – Daniel Feb 17 at 17:06
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    $\begingroup$ @Daniel, I added some detail. I hope it's more clear now. $\endgroup$ – LHF Feb 17 at 17:16

Suppose for convenience $0<a<b.$ Draw a picture. The area between the graph of $f$ and the $x$-axis is $\int_a^b f(x)\, dx.$ Same for $g$ with respect to the $y$-axis. If we add these two areas, we get the area of $[0,b]\times [0,b]$ minus the area of $[0,a]\times [0,a],$ which is $b^2-a^2.$


Here's a graphical hint (not a proof of course): enter image description here

  • $\begingroup$ That's exactly what I was doing. Nice picture though. $\endgroup$ – zhw. Feb 17 at 16:07

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