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