Use the WKB approximation to find the allowed energies ($E_n$) of an infinite square well with a "shelf", of height $V_0$ extending half-way across:

$$V(x)=V_0 \quad , \text{if} \quad 0<x<a/2$$ $$V(x)=0 \quad , \text{if} \quad a/2<x<a$$ $$V(x)=\infty \quad , \text{otherwise}$$

This is what I did:

For the region $0<x<a/2$: $$\phi (x)=\frac{1}{\hbar}\int_0^{a/2}p(x)dx=n\pi$$ $$\frac{ap}{2}=n\pi \hbar$$

$p=\sqrt{2m(E-V_0)}$, so solving for $E$ yields:

$$E=\frac{2n^2\pi ^2 \hbar ^2}{ma^2}+V_0$$

For the region $a/2<x<a$: $$E=\frac{2n^2\pi ^2 \hbar ^2}{ma^2}$$

So then I said that we can't have 2 different allowed energies defining the entire potential, so I summed them up.

$$E_n = \frac{4n^2\pi ^2 \hbar ^2}{ma^2} + V_0$$

$$=8E_n^0 + V_0$$

where $E_n^0 = \frac{n^2\pi ^2 \hbar ^2}{2ma^2}$

...but the given answer is

$$E_n = E_n^0 + \frac{V_0}{2} + \frac{V_0^2}{16E_n^0}$$

Why isn't it correct to simply add the energies like I did?