Help with derivation on why antibonding orbitals are more antibonding
For the 1s orbital in hydrogen, the wavefunction is

$$\psi_\mathrm{1s} = N\mathrm{e}^{-r}$$

where $N$ is some normalisation constant. The normalisation constant is not unique: if $N$ is a satisfactory normalisation constant such that $\left&lt;\psi_\mathrm{1s}\middle|\psi_\mathrm{1s}\right&gt; = 1$, then for any value of $\phi$, the normalisation constant $N\mathrm{e}^{\mathrm i\phi}$ is also valid.

However you can make the assumption, without any loss of generality, that the wavefunctions of both 1s orbitals in dihydrogen ($\ce{H2}$) are of the same phase (i.e. both normalisation constants have the same value of $N$ and $\phi$). As a simple example we can just let $N$ be a positive real number. With this, the wavefunction of the 1s orbitals always have a positive value, and the overlap integral

$$S_\mathrm{AB} = \left&lt;\psi_\mathrm{1s,A} \middle| \psi_\mathrm{1s,B} \right&gt; = \int \psi_\mathrm{1s,A}\psi_\mathrm{1s,B}\,\mathrm{d}\tau$$

always has a positive integrand (no imaginary components, so I dropped the complex conjugate). The integral is therefore positive.

(Note that even though both 1s orbitals have the same exponential form, $\psi_\mathrm{1s,A}$ is not equal to $\psi_\mathrm{1s,B}$ since the orbitals are centred at different points in space.)

In fact, this choice of phase is required so that you can express the bonding and antibonding MOs as the specific linear combinations presented. The "in-phase" linear combination, i.e. the bonding MO, only has the form $N(\psi_\mathrm{1s,A} + \psi_\mathrm{1s,B})$ if you let the two orbitals have the same phase. If you had a different choice of normalisation constant, the bonding MO would be represented by a different linear combination $N(\psi_\mathrm{1s,A} + z\psi_\mathrm{1s,B})$ where $z$ is some complex number determined by the relative phases of the 1s orbitals.

As such, by virtue of expressing the MOs in that form, your overlap integral is necessarily positive.

Note that the above discussion only really applies to 1s + 1s overlap. With other orbitals, which have both positive and negative regions, you can end up with a negative overlap integral depending on the relative phases that you choose.

For example in $\ce{BeH2}$ if you consider the middle Be 2p orbital to overlap with both hydrogen 1s orbitals, one "overlap" will be formed with the positive lobe of the 2p orbital and the other "overlap" will be formed with the negative lobe of the 2p orbital. Assuming that you phase both 1s orbitals to be positive, then the first overlap integral will be positive (positive 1s times positive 2p gives a positive integrand) and the second will be negative (positive 1s times negative 2p gives a negative integrand).



In general your choice of phase is arbitrary; it does not affect the final physical result, and is simply chosen to make the calculations easier. That is to say, the eigenvalues (MO energies) that you get out of it will still be the same. The eigenfunctions may take on a different form, but that is physically unimportant.