Question

What is the arclength of `f(x)=x^2e^(1/x)` on `x in [0,1]`?

Answer 1

`oo`

Explanation:

Without calculation, we can see easily that `lim_(x->0^+)x^2e^(1/x)=oo`, meaning `f(x)` has a vertical asymptote at `x=0`, and so the arclength must be infinite on `[0, 1]`.