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

Jul 14, 2018

$\approx 16.4316$

#### Explanation:

We are using the Formula

${\int}_{a}^{b} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$
Calculating $f ' \left(x\right)$

$f ' \left(x\right) = 2 x {e}^{\frac{1}{x}} + {x}^{2} \cdot {e}^{\frac{1}{x}} \left(- \frac{1}{x} ^ 2\right)$
simplifying we get

$f ' \left(x\right) = {e}^{\frac{1}{x}} \left(2 x - 1\right)$
so we have to compute

${\int}_{1}^{2} \sqrt{1 + {\left({e}^{\frac{1}{x}} \left(2 x - 1\right)\right)}^{2}} \mathrm{dx}$

This integral can not expressed by the known elementary functions.

By a numerical method we get $\approx 16.4316$