Answer
$$\int_{0}^{2}\left( t \ \mathbf{i}+e^{t} \mathbf{j}-t e^{t} \mathbf{k}\right) d t =2 \ \mathbf{i}+\left(e^{2}-1\right) \mathbf{j}-\left(e^{2}+1\right) \mathbf{k}$$
Work Step by Step
Given $$\int_{0}^{2}\left(\mathrm{ti}+e^{t} \mathbf{j}-t e^{t} \mathbf{k}\right) d t $$
First, to find $$\int t e^t dt$$
let $u=t, du=dt $ and $dv=e^t dt, v=e^t$
so, we get
$$\int t e^t dt=uv-\int v du=t e^t-\int e^t dt\\=t e^t- e^t =(t -1)e^t$$
Secondly, by integrating on a component-by-component basis produces
\begin{align}
I&=\int_{0}^{2}\left(\mathrm{ti}+e^{t} \mathbf{j}-t e^{t} \mathbf{k}\right) d t\\
&=\left[\frac{t^{2}}{2} \mathbf{i}\right]_{0}^{2}+\left[e^{t} \mathbf{j}\right]_{0}^{2}-\left[(t-1) \mathbf{e}^{t} \mathbf{k}\right]_{0}^{2}\\
&=2 \mathbf{i}+\left(e^{2}-e^0\right) \mathbf{j}-\left(e^{2}-1(-1)\right) \mathbf{k}\\
&=2 \mathbf{i}+\left(e^{2}-1\right) \mathbf{j}-\left(e^{2}+1\right) \mathbf{k}
\end{align}
