Answer
$$ \mathbf{r}(t)=2 e^{2 t} \mathbf{i}+3\left(e^{t}-1\right) \mathbf{j}$$
Work Step by Step
Given$$ \mathbf{ \acute{r}}(t) =4 e^{2 t} \mathbf{i}+3 e^{t} \mathbf{j}, \ \ \ \mathbf{r}(0)=2 \mathbf{i}$$
As we have $$\mathbf{r}(t)=\int \mathbf{ \acute{r}}(t)dt$$
So, by integrating on a component-by-component basis we find
\begin{array}{l}{\mathbf{r}(t)=\int \mathbf{ \acute{r}}(t)dt=\int\left(4 e^{2 t} \mathbf{i}+3 e^{t} \mathbf{j}\right) d t=2 e^{2 t} \mathbf{i}+3 e^{t} \mathbf{j}+\mathbf{C}} \\
\text{since} \ \mathbf{r}(0)=2 \mathbf{i}, \quad \text{we get}
\\ {\mathbf{r}(0)=2 e^0\mathbf{i}+3 e^0\mathbf{j}+\mathbf{C}=2 \mathbf{i} }
\\ {\mathbf{r}(0)=2 \mathbf{i}+3 \mathbf{j}+\mathbf{C}=2 \mathbf{i} \Rightarrow \mathbf{C}=-3 \mathbf{j}} \\
{\text {so we get }}\\
{\mathbf{r}(t)=2 e^{2 t} \mathbf{i}+3 e^{t} \mathbf{j}-3 \mathbf{j}=2 e^{2 t} \mathbf{i}+3\left(e^{t}-1\right) \mathbf{j}}\end{array}
