Question

What is the arc length of `f(x)= x ^ 3 / 6 + 1 / (2x)` on `x in [1,3]`?

Answer 1

`14/3 approx 4.667`

Explanation:

If `f : x mapsto x^3/6 + 1/(2x)`, the length is
`L=int_1^3 sqrt(1+f'(x)^2) dx`.

Because `f'(x) = x^2/2 - 1/(2x^2) = (x^4-1)/(2x^2)`, you have
`f'(x)^2+1 = (x^8-2x^4+1)/(4x^4) + 1 = (x^8+2x^4+1)/(4x^4) = ((x^4+1)/(2x^2))^2`,
so, you can write
`L=int_1^3 (x^4+1)/(2x^2) dx = int_1^3 (x^2/2 + 1/(2x^2)) dx`
`L = [x^3/6 - 1/(2x)]_1^3 = 14/3`