Show rigorously that the sum of integrals of $f$ and of its inverse is $bf(b)-af(a)$ 
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.
 A: 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 $$\int_a^b f(x)\, dx + \int_{f(a)}^{f(b)} f^{-1}(y)\, dy = \int_C x\, dy + y\, dx = \int_C d(xy) =  bf(b) - af(a).$$
A: For completeness, the case where $f$ is differentiable is handled this way
$$\begin{split} I&=\int_a^bf(x)\mathrm dx+\int_{f(a)}^{f(b)}f^{-1}(x)\mathrm dx\\
& = \int_a^bf(x)\mathrm dx+\int_a^bf^{-1}\big(f(u)\big)f'(u)\mathrm du \quad
\text{(by substitution $x=f(u)$)}\\&=
\int_a^bf(x)\mathrm dx+\int_a^bxf'(x)\mathrm dx \qquad\text{(simply renaming $u$ as $x$)}\\&=
\int_a^b\left[f(x)+xf'(x)\right]\mathrm dx\qquad\text{(merging of the integrals)}\\&=
\big[xf(x)\big]_a^b=bf(b)-af(a)\qquad\text{(recognizing the derivative of $x\mapsto xf(x)$)}
\end{split}$$
A: 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 $\int_a^b f(x) dx = bf(b) - af(a) - \int_{f(a)}^{f(b)} f^{-1}(x)dx$.
A: 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 $0$.
A: I think you want to start by proving three things:


*

*The theorem holds for linear functions.

*The theorem holds for piecewise-defined functions, if it holds for each individual piece and everything remains monotone and continuous.

*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.
A: 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:

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