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

##### 1 Answer
Sep 18, 2016

$\infty$

#### Explanation:

Without calculation, we can see easily that ${\lim}_{x \to {0}^{+}} {x}^{2} {e}^{\frac{1}{x}} = \infty$, meaning $f \left(x\right)$ has a vertical asymptote at $x = 0$, and so the arclength must be infinite on $\left[0 , 1\right]$.