Prove that:$\int_{a}^{b} f(x) dx = b \cdot f(b) - a \cdot f(a) - \int_{f(a)}^{f(b)} f^{-1}(x) dx$ I just wanted to ask, if my proof is correct.
I haven't seen the equation before, but I think it's quite useful.
Let $f$ be an bijective differentiable function. Then the inverse function $f^{-1}$ exists and the following equation holds:

$$\int\limits_{a}^{b} f(x) dx = b \cdot f(b) - a \cdot f(a) - \int\limits_{f(a)}^{f(b)} f^{-1}(x) dx$$

Proof.
$f$ is an bijective differentiable function and $F$ is an antiderivative of $f$.
First we need to find an antiderivative of $f^{-1}$.
$\int f^{-1}(x) dx$ with substitution $x = f(y)$ yields:
$$\int y \cdot f'(y) dy = y \cdot f(y) - \int f(y) dy = y \cdot f(y) - F(y)$$
resubstitution yields:
$$\int f^{-1}(x) dx = x \cdot f^{-1}(x) - F(f^{-1}(x))$$
hence $\int\limits_{f(a)}^{f(b)} f^{-1}(x) dx = \left[x \cdot f^{-1}(x) - F(f^{-1}(x)) \right ]_{f(a)}^{f(b)}$
$$=b \cdot f(b) - F(b) - (a \cdot f(a) - F(a)) = F(a) - F(b) + b \cdot f(b) - a \cdot f(a)$$
$$= \int\limits_{b}^{a} f(x)  dx + b \cdot f(b) - a \cdot f(a) = -\int\limits_{a}^{b} f(x)  dx + b \cdot f(b) - a \cdot f(a)$$
All in all:
$$\int\limits_{f(a)}^{f(b)} f^{-1}(x) dx = -\int\limits_{a}^{b} f(x)  dx + b \cdot f(b) - a \cdot f(a)$$ 
which is equal to

$$\int\limits_{a}^{b} f(x) dx = b \cdot f(b) - a \cdot f(a) - \int\limits_{f(a)}^{f(b)} f^{-1}(x) dx$$

q.e.d.
 A: Look at the area represented by the integrals.  For simplicity take the case $0 < a < b$ and $0 < f(a) < f(b)$. The first integral represents the area under the curve, above the $x$-axis, between $x=a$ and $x=b$.  The second integral represents the area to the left of the curve, to the right of the $y$-axis, between $y=f(a)$ and $y=f(b)$. When combined they give the region which could be described as the rectangle with width $b$ and height $f(b)$, less the rectangle with width $a$ and height $f(a)$, both in the first quadrant with one vertex at the origin.
A: We do not make any use of the integration by part nor assume that $f$ is differentiable to derive this formula. We only Assume $f$ to be continuous ad bijective. Hence $f$ is either increasing or decreasing.

We recall that $$ \left|\int_a^bh(x)dx\right|$$ represent the area covered by the between the $x$-axis and the curve of $h$  represented in the interval $[a,b]$



*

*First case: $f$ is an increasing function


*Second case $f$ is a decreasing function
Now,if we consider the case where the function is deceasing then, a drawing looks as the image below. 

One observe that unlike in the first case, the areas formed by $f$ and $f^{-1}$ intercept at $\color{red}{\mathcal{A}}$
Here we evaluate the area $\color{red}{\mathcal{A}}$ in two different ways as follows


*

*Taking into account the rectangle $aQRb$ we get,
$$\color{red}{\mathcal{A}} = \int_{a}^{b} f(x)  dx  - f(b)(b-a)$$

*Taking into account the rectangle $f(b)QPf(a)$ we get,
$$\color{red}{\mathcal{A}} = \int_{f(b)}^{f(a)} f^{-1}(x) dx  - a(f(a)-f(b))$$
finally by equaling both formula we arrive at, 


$$ \int_{a}^{b} f(x)  dx  - f(b)(b-a) = \int_{f(b)}^{f(a)} f^{-1}(x) dx  - a(f(a)-f(b)) $$
that is 
$$ \color{red}{\int_{a}^{b} f(x)  dx +\int_{f(a)}^{f(b)} f^{-1}(x) dx  = bf(b) - af(a).}$$
A: It should be almost obvious that the identity in question is an immediate consequence of the formula for "integration by parts". Using integration by parts we have $$\int_{a} ^{b} f(x) \, dx=bf(b) - af(a) - \int_{a} ^{b} xf'(x) \, dx\tag{1}$$ and then we put $f(x) =t,x=f^{-1}(t)$ in the integral on right to get $$\int_{a}^{b}xf'(x)\,dx=\int_{f(a)}^{f(b)}f^{-1}(t)\,dt\tag{2}$$ Our formula follows from $(1)$ and $(2)$. The proof can also be carried out using Riemann-Stieltjes integral. This has the advantage that it avoids the derivative $f'(x) $ and establishes the identity under much weaker conditions on $f$ (namely that $f$ should be strictly monotone on $[a, b] $).

The identity in question has a nice geometric interpretation (see other answers here) which helps to convince us of the truth of the identity, but it does not constitute a proof (in the same way as geometric arguments can not be used to prove mean value theorem or intermediate value theorem). 
