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=\int_0^{1} \lt t^2, t \cos \pi t, \sin \pi t \gt dt;\\ I=[\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 \\I=\lt \dfrac{1^3}{3}-\dfrac{0}{3}, [\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=\lt \dfrac{1}{3}, [\dfrac{ \sin \pi}{\pi}+\dfrac{ \cos \pi }{\pi^2}]- \dfrac{ \cos \pi (0)}{\pi^2}],-\dfrac{\cos \pi}{\pi} -(\dfrac{\cos 0}{\pi}) \gt =\lt \dfrac{1}{3}, -\dfrac{1}{\pi^2}-\dfrac{ 1}{\pi^2}, \dfrac{1}{\pi}+\dfrac{1}{\pi} \gt $
Hence, we get $\int_0^{1} r(t) dt=\dfrac{i}{3} -\dfrac{2j}{\pi^2}+\dfrac{2k}{\pi}$
