Answer
$$\int_{0}^{1}(8 t \mathbf{i}+t \mathbf{j}-\mathbf{k}) d t =4 \mathbf{i}+\frac{1}{2} \mathbf{j}-\mathbf{k}$$
Work Step by Step
Given $$\int_{0}^{1}(8 t \mathbf{i}+t \mathbf{j}-\mathbf{k}) d t $$
Integrating on a component-by-component basis produces
\begin{align}
\int_{0}^{1}(8 \mathfrak{t} \mathfrak{i}+t \mathbf{j}-\mathbf{k}) d t&=\left[4 t^{2} \mathbf{i}\right]_{0}^{1}+\left[\frac{t^{2}}{2} \mathbf{j}\right]_{0}^{1}-[t \mathbf{k}]_{0}^{1}\\
&=4 \mathbf{i}+\frac{1}{2} \mathbf{j}-\mathbf{k}
\end{align}
