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In Wikipedia is written in one step that for the integral $\int_{x_1}^{x_1 + \Delta x}f(t) dt$ that by mean value theorem there exists a $c$ in $[x_1, x_1 + \Delta x]$ such that


But mean value theorem states that there exists $c$ in $(x_1, x_1 + \Delta x)$. (the open interval)

It seems to be a problem when squeeze theorem is later applied.

Is this mistake in the proof or is the step valid? If it is valid can you please explain me why? Thank you for any help.

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The open interval is contained in the closed interval, so the condition that $c$ is in the closed interval is weaker than the conclusion of the mean value theorem. – Matthew Pressland Apr 23 '13 at 14:33
@MattPressland But if $c \in (x_1, x_1 + \Delta x)$ and by applying squeeze theorem, $c = x_1$ then this is contradiction because $\Delta x \to 0$ and $x_1 \notin (x_1, x_1)$? – blue Apr 23 '13 at 15:03
The squeeze theorem doesn't tell you $c=x_1$, it tells you that $c\to x_1$ as $\Delta x\to 0$. A sequence of points in the open interval can converge to one of the boundary points in the closed interval. – Matthew Pressland Apr 23 '13 at 15:10
@MattPressland I did not know it can converge to boundary point. You can write in answer? If you do I accept. – blue Apr 23 '13 at 15:13
@blue .Lagrage's Mean value theorem for differentiation differ's from first mean value theorem for integration(which you used!) at that point(the interval in which $c$ is expected to be found is closed in the latter and open in the former). – fermesomme Apr 23 '13 at 15:20
up vote 1 down vote accepted

As boywholived points out, a possible statement of the mean value theorem allows for $c$ to lie in the closed interval. In fact this is simply a weaker statement - if $c$ lies in the open interval, then in particular it lies in the closed interval.

This doesn't cause any problems when applying the squeeze theorem. Perhaps a change of notation is useful; for a particular choice of $\Delta x$, the mean value theorem gives you some $x<c<\Delta x$ (assuming the open interval version of the mean value theorem for now). This $c$ depends on $\Delta x$, so we should maybe write $c(\Delta x)$, making it clear that $c$ is really a function whose value depends on $\Delta x$. Then when applying the squeeze theorem, we are interested in the limit of this function as $\Delta x\to 0$. The fact that $c(\Delta x)>x$ for all $\Delta x>0$ only implies that $\lim_{\Delta x\to0}c(\Delta x)\geq x$, not that the inequality is strict. (Compare to the sequence $a_n=1/n$; every term is strictly positive, but $a_n\to 0$ as $n\to\infty$).

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