Answer
$$s = \frac{1}{{27}}\left[ {{{13}^{3/2}} - 8} \right]$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = {t^3}{\bf{i}} + {t^2}{\bf{j}},{\text{ }}\left[ {0,1} \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{t^3}{\bf{i}} + {t^2}{\bf{j}}} \right] \cr
& {\bf{r}}'\left( t \right) = 3{t^2}{\bf{i}} + 2t{\bf{j}} \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( {3{t^2}} \right)}^2} + {{\left( {2t} \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {9{t^4} + 4{t^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = t\sqrt {9{t^2} + 4} \cr
& {\text{Find the arc length }}s{\text{ using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& s = \int_0^1 {t\sqrt {9{t^2} + 4} } dt \cr
& s = \frac{1}{{18}}\int_0^1 {18t\sqrt {9{t^2} + 4} } dt \cr
& {\text{Integrate}} \cr
& s = \frac{1}{{18}}\left[ {\frac{{{{\left( {9{t^2} + 4} \right)}^{3/2}}}}{{3/2}}} \right]_0^1 \cr
& s = \frac{1}{{27}}\left[ {{{\left( {9{t^2} + 4} \right)}^{3/2}}} \right]_0^1 \cr
& {\text{Simplifying}} \cr
& s = \frac{1}{{27}}\left[ {{{\left( {9 + 4} \right)}^{3/2}} - {{\left( {0 + 4} \right)}^{3/2}}} \right] \cr
& s = \frac{1}{{27}}\left[ {{{13}^{3/2}} - 8} \right] \cr
& {\text{Graph}} \cr} $$
