quantum mechanics - Expectation values of the position operator is equal to zero in case of even potentials?

Assuming the eigenvalue of position operator $\hat x$ equal to $k$, can I not write:

$$\begin{align} \langle\psi_n|x|\psi_m\rangle &= \langle x\psi_n|\psi_m\rangle \\ &=\langle k\psi_n|\psi_m\rangle \\ &=k\langle\psi_n|\psi_m\rangle \\ &=k\delta_{nm} \end{align}$$

But I know that $\langle x \rangle =0$ in case of even potentials (I don't know how that happens) and what I have written above is wrong, at least in case of even potentials.

Taking the example of infinite 1D square well, the states are :

$$\psi_{n} \left(x\right)=A\sin\left(\frac{nx\pi}{L}\right)dx$$ then, $$\langle\psi_n|x|\psi_m\rangle =A\int_{-\infty}^{\infty}\sin\left(\frac{mx\pi}{L}\right)x\sin\left(\frac{nx\pi}{L}\right)dx$$ If m=n=1, $$ =A\int_{-\infty}^{\infty}x\sin^{2}\left(\frac{x\pi}{L}\right)dx $$if we apply an even potential then the equation gets reduced to$$ =\frac{1}{L}\int_{-L}^{L}x\sin^{2}\left(\frac{x\pi}{L}\right)dx=0 $$while in case of a potential(neither even nor odd), the equation leads to$$ =\frac{2}{L}\int_{0}^{L}x\sin^{2}\left(\frac{x\pi}{L}\right)dx=L/2 ~? $$ Here $n=m=1$ but $ =0 $for even potentials, which is confusing me! It should be$k$ right?