Given a differentiable real-valued function f, the arclength of its graph on [a,b] is given by

∫ba√1+(f′(x))2dx

For many choices of f this can be a tricky integral to evaluate, especially for calculus students first learning integration. I’ve found a few choices of f that make the computation pretty easy:

- Letting f be linear is super easy, but then you don’t even need the formula.
- Taking f of the form (stuff)32
mightwork out nicely if stuff is chosen carefully.- Calculating it for f(x)=√1−x2 is alright if you remember that ∫1x2+1dx is arctan(x)+C.
- Letting f(x)=ln(sec(x)) results in ∫sec(x)dx, which classically sucks.
But it looks like most choices of f suggest

at leasta trig substitution f′(x)↦tan(θ), and will be computationally intensive, and unreasonable to ask a student to do. Are there other examples of a function f such that computing the arclength of the graph of f won’t be too arduous to ask a calculus student to do?

**Answer**

Ferdinands, in his short note “Finding Curves with Computable Arc Length”, also comments on the difficulty of coming up with suitable examples of curves with easily-computable arclengths. In particular, he gives a simple recipe for coming up with examples: let

f(x)=12∫(g(x)−1g(x))dx

for some suitably differentiable g(x) over the desired integration interval for the arclength. The arclength over [a,b] is then given by

12∫ba(g(x)+1g(x))dx

g(x)=x10 and g(x)=tanx are some of the example functions given in the article that are amenable to this recipe.

**Attribution***Source : Link , Question Author : Mike Pierce , Answer Author : J. M. ain’t a mathematician*