calculus - 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.