Answer
$\sqrt{3}-1$
Work Step by Step
$$
g(t)=\frac{t}{\sqrt{3+t^{2}}},[1,3]
$$
The average value of the given function on the given interval is
$$
\begin{split}
g_{\text {ave }}&=\frac{1}{b-a} \int_{a}^{b} g(t) d t \\
&=\frac{1}{3-1} \int_{1}^{3} \frac{t}{\sqrt{3+t^{2}}} d t \\
&=\frac{1}{2}\left[\left(3+t^{2}\right)^{1 / 2}\right]_{1}^{3} \\
&=\frac{1}{2}(2 \sqrt{3}-2) \\
&=\sqrt{3}-1
\end{split}
$$