Answer
$\int_1^4\langle2t^{\frac{3}{2}},0,(t+1)\sqrt{t}\rangle dt=\langle\frac{124}{5},0,\frac{256}{15}\rangle$.
Work Step by Step
$\int_1^4\langle2t^{\frac{3}{2}},0,(t+1)\sqrt{t}\rangle dt=\int_1^42t^{\frac{3}{2}}dt+\int_1^40dt+\int_1^4t^\frac{3}{2}+t^{\frac{1}{2}}dt=(\frac{4}{5}t^\frac{5}{2})\big|_1^4+(0)\big|_1^4+(\frac{2}{5}t^\frac{5}{2}+\frac{2}{3}t^\frac{3}{2})\big|_1^4=\langle\frac{124}{5},0,\frac{256}{15}\rangle$
