For an electron in a three-dimensional metal, show that the average energy is given by $\stackrel{\text{−}}{E}=\frac{1}{N}{\displaystyle \underset{0}{\overset{E_{\text{F}}}{\int }}Eg(E)dE}=\frac{3}{5}{E}_{\text{F}},$

Where N is the total number electrons in the metal.