We first find that

\mathbf{f}(t)=\int \mathbf{f}^{\prime}(t) d t=\left(\frac{t^{2}}{2}+C_{1}\right) \mathbf{i}+\left(\frac{t^{4}}{4}+C_{2}\right) \mathbf{j}=\frac{t^{2}}{2} \mathbf{i}+\frac{t^{4}}{4} \mathbf{j}+\mathbf{C}.

To evaluate \mathbf{C}, we substitute the value t = 0 into (10) to find that \mathbf{f}(0)=\mathbf{C}=\mathbf{i}-2 \mathbf{j}, so that

\mathbf{f}(t)=\frac{t^{2}}{2} \mathbf{i}+\frac{t^{4}}{4} \mathbf{j}+\mathbf{i}-2 \mathbf{j}=\left(\frac{t^{2}}{2}+1\right) \mathbf{i}+\left(\frac{t^{4}}{4}-2\right) \mathbf{j}.