If I was trapped on a desert island and needed to compute $\log(2)$, the natural logaritm of $2$, one thing I could do is use the equality

$$\log(2) = \int_1^2 \frac{1}{x} \ dx$$

and approximate the definite integral with Simpson's rule. Say I want to partition $[1,2]$ into $10$ subintervals to do this. But, maybe I don't like working with fracftions very much. Well, then I could use instead $$ \log(2) = \int_{10}^{20} \frac{1}{x} \ dx$$ since it is easier to divide $[10,20]$ into $10$ subintervals.

Now, it happens that using Simpon's rule approximation with $n=10$ gives the same approximation to both of these definite integrals. You can try this here; just puch in

f(x): 1/x    From: 1    To: 2    Amount: 10

and then

f(x): 1/x    From: 10   To: 20   Amount: 10

and note the results are the same. My intuition for why this happens is a bit murky. I don't think it would be difficult to justify this with formulas, but I feel as though there is probably something about areas and scale and the geometry of the situation, something which I am missing, which would make this clear. Maybe someone here would like to give a clear explanation? Thanks.


This happens because $x$ is linear in $x$. An approximation of the first integral could be $(1/1 + 1/1.1 + \dots + 1/2) \times 1/11$ where the corresponding approximation of the second integral would be $(1/10+1/11+\dots+1/20) \times 10/11$.

  • $\begingroup$ I should have taken the values at the centers of the intervals to get a better approximation, but the numbers are easier here and the point is the same. $\endgroup$
    – JPi
    Jan 16 '14 at 1:06

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