In the context of proving integral rule (here), I'm very confused to why the following statement is true:
A form of the mean value theorem, $\int_a^b f(x)dx = (b-a)f(\xi)$ where $a < \xi < b$, may be applied to the first and last integrals of the fomula for $\Delta \psi$ above, resulting in $$ \Delta \psi = -\Delta a f(\xi_1, \alpha + \Delta \alpha) + \int_a^b [f(x,\alpha + \Delta \alpha) - f(x,\alpha)] dx + \Delta b f(\xi_2, \alpha + \Delta \alpha)$$ Divide by $\Delta \alpha$ and let $\Delta \alpha \to 0$. Notice $\xi_1 \to a$ and $\xi_2 \to b$.
It's the last part that is puzzling. Why do $\xi_1 \to a$ and $\xi_2 \to b$?
The $\xi_1$ is from the interval $(a,a+\Delta a)$ corresponding to an integral $\int_a^{a+\Delta a}$. The mean value theorem is only applied on this small interval, not on $(a,b)$. As $\Delta a\to0$, this interval shrinks and therefore $\xi_1\to a$. Similarly, $\xi_2\in(b,b+\Delta b)$.
It might be that $\Delta a$ or $\Delta b$ is negative, but the proof works in that case as well with suitable reinterpretation.
In the calculation both $a$ and $b$ depend on $\alpha$ in a differentiable way. The calculation (which is insufficiently explained in Wikipedia) we have $\Delta a=a(\alpha+\Delta\alpha)-a(\alpha)$. Therefore $\Delta a\to0$ as $\Delta\alpha\to0$ and similarly for $b$.