El valor medio de una función f(x) en un intervalo [a,b] se calcula como
$$\begin{align}&\overline{f(x)} = \frac{1}{b-a}\int_a^bf(x) dx\\&\text{En este caso}\\&\overline{f(x)} = \frac{1}{3-0}\int_0^3 x \sqrt{x^2+16} dx=\\&Sustitución\ u=x^2+16\\&du = 2xdx \to \frac{du}{2}=x dx\\&Retomando\\&\frac{1}{3}\int_0^3 \sqrt{u} \frac{du}{2}=\frac{1}{6}\int_0^3 \sqrt{u} \ du=\\&\frac{1}{6} \frac{u^{3/2}}{3/2}\bigg|_0^3 = \\&\frac{1}{9} (x^2+16)^{3/2}\bigg|_0^3 = \\&\frac{1}{9} \bigg((3^2+16)^{3/2}-(0+16)^{3/2}\bigg)=\\&\frac{1}{9} \bigg(25^{3/2}-16^{3/2}\bigg)=\\&\frac{1}{9} \bigg(125-64\bigg)=\frac{61}{9}\end{align}$$Salu2