Chapter 5 - Applications of Integration - 5.5 Average Value of a Function - 5.5 Exercises - Page 391: 4

$$\sqrt{3}-1$$

Given $$g(t)=\frac{t}{\sqrt{3+t^2}},\ \ \ [1,3] $$ Then \begin{align*} g_{\text{ave}}&=\frac{1}{b-a}\int_{a}^{b}g(t)dt \\ &=\frac{1}{3-1}\int_{1}^{3}\frac{t}{\sqrt{3+t^2}}dt\\ &=\frac{1}{4} \int_{1}^{3}2t(3+t^2)^{-1/2}dt\\ &=\frac{1}{2}(3+t^2)^{1/2}\bigg|_{1}^{3}\\ &=\frac{1}{2}(2\sqrt{3}-2)\\ &=\sqrt{3}-1 \end{align*}