Answer
$${\mathbf{r}(t)= \mathbf{i}+\left(t^{3}+2\right) \mathbf{j}}+4 t^{\frac{3}{2}} \mathbf{k}$$
Work Step by Step
Given$$ \mathbf{r}^{\prime}(t)=\left(3 t^{2} \mathbf{j}+6 \sqrt{t} \mathbf{k}\right), \ \ \ \mathbf{r}(0)=\mathbf{i}+2 \mathbf{j}$$
As we have $$\mathbf{r}(t)=\int \mathbf{r}^{\prime}(t) dt$$
So, integrating on a component-by-component basis produces
\begin{array}{l}{\mathbf{r}(t)=\int \mathbf{r}^{\prime}(t)(t)dt=\int\left(3 t^{2} \mathbf{j}+6 \sqrt{t} \mathbf{k}\right) d t= t^{3 } \mathbf{j}+4 t^{\frac{3}{2}} \mathbf{k}+\mathbf{C}} \\
\text{since} \ \mathbf{r}(0)=\mathbf{i}+2 \mathbf{j}, \quad \text{we get}
\\ {\mathbf{r}(0)= 0 \mathbf{j}+ 0\mathbf{k}+\mathbf{C}=\mathbf{i}+2 \mathbf{j}}
\\ { \Rightarrow \mathbf{C}=\mathbf{i}+2 \mathbf{j}} \\
{\text {so we get }}\\
{\mathbf{r}(t)=t^{3 } \mathbf{j}+4 t^{\frac{3}{2}} \mathbf{k}+\mathbf{i}+2 \mathbf{j}= \mathbf{i}+\left(t^{3}+2\right) \mathbf{j}}+4 t^{\frac{3}{2}} \mathbf{k}\end{array}
