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.
Look at the area represented by the integrals.  For simplicity take the case $0 &lt; a &lt; b$ and $0 &lt; f(a) &lt; f(b)$. 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.
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).