Answer
$\left\langle1,2,-\frac{1}{3 } \sin 3\right\rangle.$
Work Step by Step
We have
\begin{align}
\int_{0}^{1}\left\langle 2t, 4t, -\cos 3t\right\rangle dt &=\left\langle t^2, 2t^2, -\frac{1}{3} \sin 3t
\right\rangle|_{0}^{1}\\
&=\left\langle1,2,-\frac{1}{3 } \sin 3\right\rangle-\left\langle0,0,0\right\rangle\\
&=\left\langle1,2,-\frac{1}{3 } \sin 3\right\rangle.
\end{align}
