I know the originals formula for arc-length is:

$$\int_{a}^b \sqrt{1+{f'(x)}^2}$$

However most of the formulas don't have closed formed solutions, and are unsolvable in terms of this equation.

So far, I tried taking $$\lim_{n\rightarrow\infty}\sum_{i=1}^n \sqrt{1+\left(z\left(a+\frac{a-b}{n}\right)i\right)^2}\left(\frac{a-b}{n}\right)$$

Where $z(x)$ is the equation of the derivative of $f(x)$, $i$ is the number of line segments made around the intervals of the equation, and n is the sub-intervals. The only problem is my Desmos online graphing calculator has trouble computing the arclength as n goes to infinity.

I would like to know some of the techniques for calculating the arc length numerically in a easier way, from the range of $[a,b]$,and maybe an example.

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    $\begingroup$ Oftentimes it is much easier to find a parametrization $f:\mathbb{R} \rightarrow \mathbb{R}^2$ than it is to find an explicit formula $f:\mathbb{R} \rightarrow \mathbb{R}$ for a given curve. In such a case, the arclength formula is instead $\displaystyle \int_{t = t_1}^{t = t_2} \sqrt{\Big(\frac{dx}{dt}\Big)^2 + \Big( \frac{dy}{dt} \Big)^2 }$. $\endgroup$ – Kaj Hansen Jun 30 '14 at 21:13
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    $\begingroup$ Well if you cant find the parameter of something like $x^x+x^2$, and if it doesn't have a indefinite integral, than then the only was is to "numerically" compute it, not analytically. $\endgroup$ – Arbuja Jun 30 '14 at 22:02
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    $\begingroup$ Unfortunately, you are correct. $\endgroup$ – Kaj Hansen Jun 30 '14 at 22:30
  • $\begingroup$ See math.stackexchange.com/a/61796/589. $\endgroup$ – lhf Jul 1 '14 at 0:29

Assuming $f(x)$ is single-valued and differentiable on $(a,b)$, and writing $g(x) \equiv f^\prime(x)$ to avoid confusion, you have $s$ as an integral: $$ s = \int_a^b \sqrt{1+g(x)^2} \, dx $$

The form you gave, while correct in principle, does this integration by a sum of $n$ rectangles; the error term goes down as $1/n$ and this is much too slow. For most will behaved functions $g(x)$, a reasonable numerical integration scheme is Simpson's rule, which will exhibit an error of order $1/n^4$ so it is feasible to get decent accuracy.

Simpson's rule consists of using an even $n$ and weighting the even points at double strength compared to the odd points, and weighting the two endpoint values at half strength.

See Numerical Recipes if you need a more robost method for poorly behaved functions.


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