9.8 Exercise 28

For an electron in a three-dimensional metal, show that the average energy is given by $\overset{−}{E} = \frac{1}{N} \int_{0}^{E_{\text{F}}} E g \left(E\right) d E = \frac{3}{5} E_{\text{F}} ,$

Where N is the total number electrons in the metal.