Answer
$$\int_{0}^{3} \left\|t \mathbf{i}+t^{2} \mathbf{j}\right\| d t=\frac{1}{3} \left(10 \sqrt{10}-1\right)$$
Work Step by Step
Given $$\int_{0}^{3}\left\|t \mathbf{i}+t^{2} \mathbf{j}\right\| d t $$
First, we have
$$ \left\| t \mathbf{i}+t^{2} \mathbf{j}\right\|=\sqrt{( t\mathbf{i}+t^{2} \mathbf{j}) \cdot (t \mathbf{i}+t^{2} \mathbf{j})}=\sqrt{t^{2}+t^{4}}=t \sqrt{1+t^{2}} \text { for } t \geq 0$$
Secondly, integrating on a component-by-component basis produces
\begin{align}
I&=\int_{0}^{3}\left\|\mathbf{i}+t^{2} \mathbf{j}\right\| d t\\
&=\int_{0}^{3} t \sqrt{1+t^{2}} d t\\
&=\left[\frac{1}{3}\left(1+t^{2}\right)^{3 / 2}\right]_{0}^{3}\\
&=\frac{1}{3}\left(10^{3 / 2}-1\right)\\
&=\frac{1}{3} \left(10 \sqrt{10}-1\right)
\end{align}
