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When deriving the minima, the classical approach is to say
$$ \frac d 2 \sin(\theta) = \frac{\lambda}2$$
therefore, $$d \sin(\theta) = \lambda$$ It can then be shown for any integer value of $\lambda$. But why couldn't you say that the 2 point sources that interfere destructively are the full width of the slit apart, so $d\sin(\theta) = \lambda/2$, therefore $d\sin(\theta) = \lambda/2$ for destructive interference, which wouldn't be an integer. Maybe I've misunderstood something.
you really are misunderstanding. You should consider the slit as the source of many or n elementary waves, so every elementary wave of the first half has the Difference $\lambda/2$ to one of the second half, while in your picture only two of the n waves have the difference of 1/2 all the rest have less, so the amplitude will be smaller, but never zero
