Answer
$$\frac{1}{3}{\bf{i}} + \frac{1}{4}{\bf{j}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\left( {{t^2}{\bf{i}} + {t^3}{\bf{j}}} \right)} dt \cr
& {\text{Integrating the components yields}} \cr
& \int_0^1 {\left( {{t^2}{\bf{i}} + {t^3}{\bf{j}}} \right)} dt = \left[ {\frac{{{t^3}}}{3}} \right]_0^1{\bf{i}} + \left[ {\frac{{{t^4}}}{4}} \right]_0^1{\bf{j}} \cr
& {\text{Use the vector form of the Fundamental Theorem of Calculus}} \cr
& \int_a^b {{\bf{r}}\left( t \right)} dt = \left[ {{\bf{R}}\left( t \right)} \right]_a^b = {\bf{R}}\left( b \right) - {\bf{R}}\left( a \right) \cr
& {\text{then}} \cr
& \int_0^1 {\left( {{t^2}{\bf{i}} + {t^3}{\bf{j}}} \right)} dt = \left[ {\frac{{{{\left( 1 \right)}^3}}}{3} - \frac{{{{\left( 0 \right)}^3}}}{3}} \right]{\bf{i}} + \left[ {\frac{{{{\left( 1 \right)}^4}}}{4} - \frac{{{{\left( 0 \right)}^4}}}{4}} \right]{\bf{j}} \cr
& {\text{simplify}} \cr
& \int_0^1 {\left( {{t^2}{\bf{i}} + {t^3}{\bf{j}}} \right)} dt = \frac{1}{3}{\bf{i}} + \frac{1}{4}{\bf{j}} \cr} $$
