Finding the number of holes in the valence band for a semiconductor

Given that: $$p=D_{v}\int_{VB} (1-F(E))dE=D_{v}\int_{VB}(1-\frac{1}{e^{(E-E_{f})/KT}+1})dE$$ We can make the approximation $(1+x)^{-1}=1-x$ for $x<<1$ in this case(for problem 1). $$D_{v}\int_{VB}[1-(\frac{1}{e^{(E-E_{f})/KT}}+1)]dE=D_{v}\int_{VB}[e^{-(E-E_{f})/KT}]dE$$ $$=D_{v}\int_{E_{v}}^{E_{v}+B_{v}} [e^{-(E-E_{f})/KT}]dE$$ Approximating $E_{v}+B_{v}\rightarrow\infty$: $$p=KTD_{v}e^{-(E_{v}-E_{f})/KT}$$ I hope this is helpful! Please point out any errors or bad approximations, if you see any.