Answer
$\mathop \smallint \limits_0^3 \left( {4t + 3,{t^2}, - 4{t^3}} \right){\rm{d}}t = \left( {27,9, - 81} \right)$
Work Step by Step
$\mathop \smallint \limits_0^3 \left( {4t + 3,{t^2}, - 4{t^3}} \right){\rm{d}}t = \left( {\mathop \smallint \limits_0^3 \left( {4t + 3} \right){\rm{d}}t,\mathop \smallint \limits_0^3 {t^2}{\rm{d}}t, - \mathop \smallint \limits_0^3 4{t^3}{\rm{d}}t} \right)$
Evaluate $\mathop \smallint \limits_0^3 \left( {4t + 3} \right){\rm{d}}t$
$\mathop \smallint \limits_0^3 \left( {4t + 3} \right){\rm{d}}t = \left( {2{t^2} + 3t} \right)|_0^3 = 27$
Evaluate $\mathop \smallint \limits_0^3 {t^2}{\rm{d}}t$
$\mathop \smallint \limits_0^3 {t^2}{\rm{d}}t = \frac{1}{3}{t^3}|_0^3 = 9$
Evaluate $ - \mathop \smallint \limits_0^3 4{t^3}{\rm{d}}t$
$ - \mathop \smallint \limits_0^3 4{t^3}{\rm{d}}t = - {t^4}|_0^3 = - 81$
Thus,
$\mathop \smallint \limits_0^3 \left( {4t + 3,{t^2}, - 4{t^3}} \right){\rm{d}}t = \left( {27,9, - 81} \right)$
