Answer
$$\frac{{5\sqrt 5 - 1}}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\left\| {t{\bf{i}} + {t^2}{\bf{j}}} \right\|} dt \cr
& {\text{Calculate }}\left\| {t{\bf{i}} + {t^2}{\bf{j}}} \right\| \cr
& \left\| {t{\bf{i}} + {t^2}{\bf{j}}} \right\| = \sqrt {{{\left( t \right)}^2} + {{\left( {{t^2}} \right)}^2}} \cr
& \left\| {t{\bf{i}} + {t^2}{\bf{j}}} \right\| = \sqrt {{t^2} + {t^4}} \cr
& {\text{Therefore,}} \cr
& \int_0^2 {\left\| {t{\bf{i}} + {t^2}{\bf{j}}} \right\|} dt = \int_0^2 {\sqrt {{t^2} + {t^4}} } dt \cr
& = \int_0^2 {\sqrt {{t^2}\left( {1 + {t^2}} \right)} } dt \cr
& = \int_0^2 {t\sqrt {1 + {t^2}} } dt \cr
& = \frac{1}{2}\int_0^2 {2t\sqrt {1 + {t^2}} } dt \cr
& {\text{Integrate}} \cr
& = \frac{1}{2}\left[ {\frac{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}{{3/2}}} \right]_0^2 \cr
& = \frac{1}{3}\left[ {{{\left( {1 + {t^2}} \right)}^{3/2}}} \right]_0^2 \cr
& = \frac{1}{3}\left[ {{{\left( {1 + {{\left( 2 \right)}^2}} \right)}^{3/2}} - {{\left( {1 + {{\left( 0 \right)}^2}} \right)}^{3/2}}} \right] \cr
& = \frac{1}{3}\left[ {{{\left( 5 \right)}^{3/2}} - {{\left( 1 \right)}^{3/2}}} \right] \cr
& = \frac{{5\sqrt 5 - 1}}{3} \cr} $$
