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Suppose $f$ is a continuous, strictly increasing function defined on a closed interval $[a,b]$ such that $f^{-1}$ is the inverse function of $f$. Prove that, $$\int_{a}^bf(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=bf(b)-af(a)$$

A high school student or a Calculus first year student will simply, possibly, apply change of variable technique, then integration by parts and he/she will arrive at the answer without giving much thought into the process. A smarter student would probably compare the integrals with areas and conclude that the equality is immediate.

However, I am an undergrad student of Analysis and I would want to solve the problem "carefully". That is, I wouldn't want to forget my definitions, and the conditions of each technique. For example, while applying change of variables technique, I cannot apply it blindly; I must be prudent enough to realize that the criterion to apply it includes continuous differentiability of a function. Simply with $f$ continuous, I cannot apply change of variables technique.

Is there any method to solve this problem rigorously? One may apply the techniques of integration (by parts, change of variables, etc.) only after proper justification.

The reason I am not adding any work of mine is simply that I could not proceed even one line since I am not given $f$ is differentiable. However, this seems to hold for non-differentiable functions also.

I would really want some help. Pictorial proofs and/or area arguments are invalid.

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    $\begingroup$ Draw a picture and you'll see why this is so. $\endgroup$
    – mickep
    Jan 22, 2015 at 16:35
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    $\begingroup$ You should write out the Riemann sums that those integrals correspond to. $\endgroup$ Jan 22, 2015 at 16:39
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    $\begingroup$ @yedaynara Many people will take a pictorial proof as a proof, if it is decent enough. $\endgroup$
    – Pedro
    Jan 22, 2015 at 17:04
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    $\begingroup$ @yedaynara I would note accurate-pictoral proofs are rigorous, they just illustrate the words more succinctly (eg. an accurate Venn diagram). If you don't want them, fine, but you shouldn't say they are universally non-rigorous. $\endgroup$ Jan 22, 2015 at 17:41
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    $\begingroup$ Possible duplicate of $\int_{a}^{b} f(x) dx = b \cdot f(b) - a \cdot f(a) - \int_{f(a)}^{f(b)} f^{-1}(x) dx$ proof $\endgroup$
    – Guy Fsone
    Nov 7, 2017 at 19:11

6 Answers 6

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Let $\{x_0,x_1,\dots,x_N\}$ be a partition of $[a,b]$. Then $\{f(x_0),f(x_1),\dots,f(x_N)\}$ is a partition of $[f(a),f(b)]$. The following equality holds: $$ \sum_{i=0}^{N-1}f(x_i)(x_{i+1}-x_i)+\sum_{i=0}^{N-1}x_i(f(x_{i+1})-f(x_i))+\sum_{i=0}^{N-1}(x_{i+1}-x_i)(f(x_{i+1})-f(x_i))=b\,f(b)-a\,f(a). $$ The first two sums are Riemann sums for $\int_a^bf$ and $\int_{f(a)}^{f(b)}f^{-1}$ respectively. The third sum converges to $0$ as the size of the partition goes to $0$.

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  • $\begingroup$ Fabulous proof +1 $\endgroup$
    – Vim
    Jan 23, 2015 at 2:55
  • $\begingroup$ Simplest proof. It assumes that $f$ is strictly monotone and nothing more. +1 $\endgroup$
    – Paramanand Singh
    Nov 7, 2017 at 19:28
  • $\begingroup$ Although this makes pictographic sense, that third sum seems to resist crude estimates. $N$ could grow extremely large as the mesh stays constant. And the best bounds for the mesh we can garner are $(b-a)/n\leq \text{mesh}\leq b-a$. Do you have a trick for that last sum? $\endgroup$
    – user123641
    Nov 7, 2017 at 23:31
  • $\begingroup$ @ParamanandSingh We still need continuity of $f$ (in the form of uniform continuity) in order to conclude that the mesh of $\{f(x_0), f(x_1), \ldots, f(x_N)\}$ actually tends to zero as well so that the second sum is guaranteed to converge to the integral of $f^{-1}$. $\endgroup$
    – user123641
    Nov 8, 2017 at 1:22
  • $\begingroup$ @Robert: Well if $\int f\, dx$ exists then $\int x\, df$ also exists (Riemann-Stieltjes integral). So the second sum also converges. We only need monotone nature of $f$. $\endgroup$
    – Paramanand Singh
    Nov 8, 2017 at 4:11
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I think it is very natural from a geometrical point of view. It's just about the addition of two areas, which make up a big rectangle substracting a small one. See the graph below:enter image description here


Now, obviously, in the case shown in my graph $$S_1=\int_{a}^{b}f(x)dx$$ and $$S_2=\int_{f(a)}^{f(b)} f^{-1}(x)dx$$ Geometrically, we have $$S_1+S_2=S_{big}-S_{small}$$ where $S_{big}$ and $S_{small}$ respectively denotes the area of the big rectangle and the small one in the graph. Therefore $$\int_{a}^{b}f(x)dx+\int_{f(a)}^{f(b)} f^{-1}(x)dx=bf(b)-af(a)$$

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    $\begingroup$ For a technical proof, I think a substitution $x=f(t)$, $t\in[a,b]$ will suffice, but I have not worked on it yet. $\endgroup$
    – Vim
    Jan 22, 2015 at 16:53
  • $\begingroup$ Hehe, you hit me by 1 min. I just made the same picture in Mathematica, but I was too slow. I think the picture says it all. $\endgroup$
    – mickep
    Jan 22, 2015 at 16:53
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    $\begingroup$ Geometrically such a proof is totally acceptable, but perhaps a technical proof is more welcome to the OP, uhm? And by the way, the reason I'm faster than you is I don't know how to use Mathematica and the only thing I'm able to use for making a graph is MS Paint XXD $\endgroup$
    – Vim
    Jan 22, 2015 at 16:59
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    $\begingroup$ @yedaynara Well, I'm going to add my technical proof to my post, just a minute. $\endgroup$
    – Vim
    Jan 22, 2015 at 17:06
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    $\begingroup$ @yedaynara Oh I think I underestimated this problem. It seemed to me that writing a technical proof will take some more time. But I'm sorry that it's very late in my time-zone and I need to get up early tomorrow. I gotta go to sleep. I will review this problem later. Good night. $\endgroup$
    – Vim
    Jan 22, 2015 at 17:16
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Let $C$ be the graph of $y = f(x)$ over the interval $[a,b]$. Then $\int_a^b f(x)\, dx$ is the line integral $\int_C y\, dx$, and $\int_{f(a)}^{f(b)} f^{-1}(y)\, dy$ is the line integral $\int_C x\, dy$. Thus $$\int_a^b f(x)\, dx + \int_{f(a)}^{f(b)} f^{-1}(y)\, dy = \int_C x\, dy + y\, dx = \int_C d(xy) = bf(b) - af(a).$$

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  • $\begingroup$ I am not really sure if I understand line integral right now. So sorry for that. Actually I have just covered one semester in Real Analysis and line integral will be taught to us in our 3rd semester. But your approach seems fascinating!! At least you have not made any ungiven assumption. But could you kindly think of something a bit more elementary for someone who is learning the Riemann-Stieltjes Integral now? $\endgroup$ Jan 22, 2015 at 17:20
  • $\begingroup$ Is there any chance that you were taught line integrals in calculus, but don't remember? $\endgroup$
    – kobe
    Jan 22, 2015 at 17:25
  • $\begingroup$ In Calculus, yes, but in Analysis, things get restricted, you know. We are not so independent it seems. So before writing some thing down, I need to ponder 10 times whether I am adhering to definitions. In fact, I think I encountered line integrals only in my high school physics, not in Mathematics. $\endgroup$ Jan 22, 2015 at 17:28
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For completeness, the case where $f$ is differentiable is handled this way $$\begin{split} I&=\int_a^bf(x)\mathrm dx+\int_{f(a)}^{f(b)}f^{-1}(x)\mathrm dx\\ & = \int_a^bf(x)\mathrm dx+\int_a^bf^{-1}\big(f(u)\big)f'(u)\mathrm du \quad \text{(by substitution $x=f(u)$)}\\&= \int_a^bf(x)\mathrm dx+\int_a^bxf'(x)\mathrm dx \qquad\text{(simply renaming $u$ as $x$)}\\&= \int_a^b\left[f(x)+xf'(x)\right]\mathrm dx\qquad\text{(merging of the integrals)}\\&= \big[xf(x)\big]_a^b=bf(b)-af(a)\qquad\text{(recognizing the derivative of $x\mapsto xf(x)$)} \end{split}$$

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Let $P' = \{f(x_0=a), f(x_1), \cdots, f(x_n=b)\}$ be a partition of $f^{-1}$ on $[f(a), f(b)]$. Then we know $P = \{x_0, \cdots, x_n\}$ is a partition of $f$ on $[a,b]$. Observe that \begin{align*} L(f,P) + U(f^{-1}, P') &= \sum_{k=0}^{n-1}f(x_k) (x_{k+1} - x_k) + \sum_{k=0}^{n-1}x_{k+1}(f(x_{k+1}) - f(x_k))\\ &= \sum_{k=0}^{n-1} x_{k+1}f(x_{k+1}) - x_kf(x_k)\\ &= bf(b) - af(a) \end{align*} Similarly we have \begin{align*} U(f,P) + L(f^{-1}, P') = bf(b) - af(a) \end{align*} Thus, for all $P'$ we have \begin{align*} bf(b) - af(a) - U(f^{-1}, P') = L(f,P) \le \int_a^b f(x)dx \le U(f,P) = bf(b) - af(a) - L(f^{-1}, P') \end{align*} Thus $\int_a^b f(x) dx = bf(b) - af(a) - \int_{f(a)}^{f(b)} f^{-1}(x)dx$.

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I think you want to start by proving three things:

  1. The theorem holds for linear functions.
  2. The theorem holds for piecewise-defined functions, if it holds for each individual piece and everything remains monotone and continuous.
  3. The trapezoidal rule can be used to approximate the integral of any continuous function $f$ with the integral of a piecewise-linear function (which will be continuous and monotone if $f$ is).

Numbers 1. and 2. basically don't require any analysis at all. Number 3. follows pretty quickly from the intermediate value theorem and the definition of the Riemann integral.

Now, given any $\epsilon > 0$, choose a mesh so that the trapezoidal rule approximates $f$ to within $\epsilon/2$, and another mesh so it approximates $f^{-1}$ to within $\epsilon/2$. Any mesh on $f$ yields a mesh on $f^{-1}$ and vice versa, so we can pass to a common refinement of these two meshes and consider the piecewise linear function $\tilde f$ given by approximating $f$ on that common refinement. Then the sum of the two integrals for $\tilde f$ will:

  • Be exactly equal to $bf(b)-af(a)$
  • Be within $\epsilon$ of the sum of the two integrals for $f$.

Since $\epsilon$ was arbitrary, we are done.

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