Answer
$\dfrac{1}{3} i -\dfrac{2}{\pi^2}j+\dfrac{2}{\pi}k$
Work Step by Step
Let us consider that $I=\int_0^{1} r(t) dt$
and $I=\int_0^{1} \lt t^2, t \cos \pi t, \sin \pi t \gt dt$
This implies that $I=[\lt \dfrac{t^3}{3}, \dfrac{t \sin \pi t}{\pi}+\dfrac{\cos \pi t}{\pi^2},-\dfrac{\cos \pi t}{\pi} \gt ]_0^1$
$\implies \lt \dfrac{1}{3}-0, [\dfrac{(1) \times \sin \pi (1)}{\pi}+\dfrac{ \cos \pi (1)}{\pi^2}]- [0+\dfrac{ \cos \pi (0)}{\pi^2}],-\dfrac{\cos \pi (1)}{\pi} -(\dfrac{\cos 0}{\pi}) \gt $
or, $I= \lt \dfrac{1}{3}, -\dfrac{1}{\pi^2}-\dfrac{ 1}{\pi^2}, \dfrac{1}{\pi}+\dfrac{1}{\pi} \gt $
Hence, $\int_0^{1} r(t) dt=\dfrac{1}{3} i -\dfrac{2}{\pi^2}j+\dfrac{2}{\pi}k$
