Answer
$$\int_{-1}^{1}\left(t \ \mathbf{i}+t^{3} \mathbf{j}+\sqrt[3]{t}
\ \mathbf{k}\right) d t=0$$
Work Step by Step
Given $$\int_{-1}^{1}\left(t \ \mathbf{i}+t^{3} \mathbf{j}+\sqrt[3]{t} \ \mathbf{k}\right) d t $$
Integrating on a component-by-component basis produces
\begin{align}
\int_{-1}^{1}\left(t \ \mathbf{i}+t^{3} \mathbf{j}+\sqrt[3]{t} \mathbf{k}\right) d t&=\left[\frac{t^{2}}{2} \mathbf{i}\right]_{-1}^{1}+\left[\frac{t^{4}}{4} \mathbf{j}\right]_{-1}^{1}+\left[\frac{3}{4} t^{4 / 3} \mathbf{k}\right]_{-1}^{1}\\
&= (\frac{1}{2} -\frac{1}{2})\mathbf{i} + (\frac{1}{4} -\frac{1}{4})\mathbf{j} + (\frac{3}{4} -\frac{3}{4}) \mathbf{k} \\
&=0 \end{align}
