I want to calculate the definite integral of inverse of a 5th degree polynomial. The problem is that the inverse of the polynomial cannot be calculated (by using Matlab). However without calculating the inverse of the function, I can calculate $x$ given $y$ so that $f(x)=y$ (by using a solver). Is there a simple logic to calculate the area in analogy.
For example the inverse of $x^5+x^4+x^3+x^2+x+1$
If a function is strictly increasing on an interval, then the integral of the inverse function can be seen by drawing a picture. You will see:
$$\int_{f(a)}^{f(b)} f^{-1}(y)\,dy + \int_{a}^{b} f(x)\,dx = bf(b)-af(a)\tag 1$$
This can be proven (see below), but geometrically, the two areas add up to a rectangle minus a rectangle. This is most obvious if you draw a picture when $0<a<b,f(a),f(b)>0$.
So:
$$\begin{align} \int_{c}^{d} f^{-1}(y)\,dy &= \int_{f(f^{-1}(c))}^{f(f^{-1}(d))} f^{-1}(y)\,dy\\ &=f^{-1}(d)d-f^{-1}(c)c - \int_{f^{-1}(c)}^{f^{-1}(d)}f(x)\,dx\tag{2} \end{align}$$
If $f$ is decreasing, then the picture yields a different-seeming formula. Then:
$$\int_{f(b)}^{f(a)} f^{-1}(y)\,dy - a(f(a)-f(b))= \int_{a}^{b} f(x)\,dx - f(b)(b-a)$$
But re-jiggering that formula gives the same formula $(1)$ above.
So (2) works if you know that $f$ is strictly increasing or decreasing on your region.
That means you only need to compute two specific values of $f^{-1}$ to get the integral, if you know how to integrate $f$.
You can prove this for $f$ and $f^{-1}$ differentiable with a relatively simple substitution.
The chain rule shows that:
$$(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$$
So let $u=f^{-1}(y)$ then $dy=f'(u)\,du$:
$$\int_{f(a)}^{f(b)} f^{-1}(y)\,dy = \int_{a}^{b} uf'(u)\,du$$
Then integration by parts gives that the right side is $bf(b)-af(a) -\int_a^b f(u)\,du$.