Answer
$$s = \int_{ - 2}^1 {\sqrt {1 + {{\left( {2x + 1} \right)}^2}} } dx \approx 5.65263$$
Work Step by Step
$$\eqalign{
& y = {x^2} + x - 2,{\text{ }} - 2 \leqslant x \leqslant 1 \cr
& \cr
& \left( {\text{a}} \right){\text{Graph below}} \cr
& \cr
& {\text{Use the formula for arc length}} \cr
& s = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^2} + x - 2} \right] \cr
& \frac{{dy}}{{dx}} = 2x + 1 \cr
& s = \int_{ - 2}^1 {\sqrt {1 + {{\left( {2x + 1} \right)}^2}} } dx \cr
& \cr
& \left( {\text{c}} \right){\text{Integrate by using a CAS or graphing utility}} \cr
& s = \int_{ - 2}^1 {\sqrt {1 + {{\left( {2x + 1} \right)}^2}} } dx \approx 5.65263 \cr
& \cr
& {\text{Graph}} \cr} $$
