Prove that:$\int_{a}^{b} f(x) dx = b \cdot f(b) - a \cdot f(a) - \int_{f(a)}^{f(b)} f^{-1}(x) dx$ [duplicate]

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.

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

1. Taking into account the rectangle $aQRb$ we get, $$\color{red}{\mathcal{A}} = \int_{a}^{b} f(x) dx - f(b)(b-a)$$
2. 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 $$ \color{red}{\int_{a}^{b} f(x) dx +\int_{f(a)}^{f(b)} f^{-1}(x) dx = bf(b) - af(a).}$$