Answer
$$s = \int_{{e^{ - 2}}}^1 {\sqrt {1 + \frac{1}{{{x^2}}}} } dx \approx 2.221418$$
Work Step by Step
$$\eqalign{
& x = {e^{ - y}},{\text{ }}0 \leqslant y \leqslant 2 \cr
& {\text{Solve for }}y \cr
& \ln x = - y \cr
& y = - \ln x \cr
& 0 \leqslant y \leqslant 2 \to {e^{ - 2}} \leqslant x \leqslant 1 \cr
& \cr
& \left( {\text{a}} \right){\text{Graph below}} \cr
& \cr
& {\text{Use the formula for arc length}} \cr
& s = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ { - \ln x} \right] \cr
& \frac{{dy}}{{dx}} = - \frac{1}{x} \cr
& s = \int_{{e^{ - 1}}}^1 {\sqrt {1 + {{\left( { - \frac{1}{x}} \right)}^2}} } dx \cr
& s = \int_{{e^{ - 2}}}^1 {\sqrt {1 + \frac{1}{{{x^2}}}} } dx \cr
& \cr
& \left( {\text{c}} \right){\text{Integrate by using a CAS or graphing utility}} \cr
& s = \int_{{e^{ - 2}}}^1 {\sqrt {1 + \frac{1}{{{x^2}}}} } dx \approx 2.221418 \cr
& \cr
& {\text{Graph}} \cr} $$
