Chapter 13 - Section 13.2 - Derivatives and Integrals of Vector Functions - 13.2 Exercise - Page 861: 35

$\int_0^2\langle t,-t^3,3t^5\rangle dt=\langle2,-4,32\rangle$

$\int_0^2\langle t,-t^3,3t^5\rangle dt=\int_0^2tdt-\int_0^2t^3dt+\int_0^23t^5dt=(\frac{t^2}{2})\big|_0^2-(\frac{t^4}{4})\big|_0^2+(\frac{t^6}{2})\big|_0^2=\langle2,-4,32\rangle$