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 $x^5+x^4+x^3+x^2+x+1$

  • $\begingroup$ Presumably, you mean the inverse function, and not, say, $$f(x)=\frac{1}{x^5-x+1}?$$ $\endgroup$ – Thomas Andrews Oct 17 '15 at 14:15
  • $\begingroup$ The inverse function of any 5th degree polynomial like I edited. $\endgroup$ – Dirk Oct 17 '15 at 14:18
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    $\begingroup$ You essentially have a black box method to compute values of the inverse function, which means you can use any of the standard quadrature techniques to estimate the integral. But you also need to be careful, because typically a polynomial is not globally invertible. Your solver may not catch this, so it may switch between pieces of the inverse without raising any exceptions. This will give garbage results. $\endgroup$ – Ian Oct 17 '15 at 14:22
  • $\begingroup$ Sorry, I am a practitioner of mathetatics so I could not get what you mean. For the calculation of the definite integral by using quadrature function in matlab, I have to define the inverse of the polynomial f^-1(x) otherwise there wont be any computation. $\endgroup$ – Dirk Oct 17 '15 at 14:31
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    $\begingroup$ Also, a small suggestion: while it is not "obligatory", you can write your mathematical expressions between "\$\$", so they look better. For instance, for writing "f(x)=y" you can write it as \$f(x)=y\$, which looks like this: $f(x)=y$. $\endgroup$ – Nate River Oct 17 '15 at 14:47

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 $0<a<b,f(a),f(b)>0$.


$$\begin{align} \int_{c}^{d} f^{-1}(y)\,dy &= \int_{f(f^{-1}(c))}^{f(f^{-1}(d))} f^{-1}(y)\,dy\\ &=f^{-1}(d)d-f^{-1}(c)c - \int_{f^{-1}(c)}^{f^{-1}(d)}f(x)\,dx\tag{2} \end{align}$$

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 $bf(b)-af(a) -\int_a^b f(u)\,du$.

  • $\begingroup$ Thx for the answer ! It worked. $\endgroup$ – Dirk Oct 17 '15 at 14:54

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