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I was reading a proof of the fundamental theorem of calculus in my textbook and one of the lines states that $$\lim_{\Delta x \to 0} \frac{1}{\Delta x} \int^{x+\Delta x}_x f(u) du = f(x)$$ but it didn't give any explanation for this. The section on limits is later in the books so I assume this can be understood with only basic knowledge of limits.

For context, here is the poof up to this step:

$$ \begin{align} F(x) &= \int^x_a f(u) du \\ F(x+\Delta x) &= \int^x_a f(u) du + \int^{x+\Delta x}_x f(u) du\\ &= F(x) + \int^{x + \Delta x}_x f(u) du \\ \frac{F(x+\Delta x)-F(x)}{\Delta x} &= \frac{1}{\Delta x} \int^{x+\Delta x}_x f(u) du\\ \Delta x &\to 0 \\ &\therefore \\ \frac{dF}{dx} &= f(x) \end{align}$$

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    $\begingroup$ This is true provided $f$ is continuous at $x$. I assume that was stated before these calculations. $\endgroup$
    – GEdgar
    Aug 27, 2021 at 15:00
  • $\begingroup$ This is what one calls the first part of Fundamental Theorem of Calculus. And it holds more generally. If $f(u) \to A$ as $u\to x^+$ then the expression in question tends to $A$ as $\Delta x\to 0^+$ and a similar relation holds for $\Delta x\to 0^-$. $\endgroup$
    – Paramanand Singh
    Aug 28, 2021 at 14:20

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$$ \frac{1}{\Delta x}\int^{x+\Delta x}_x f(u) du $$

Using the mean value theorem for integrals:

$$\Longrightarrow \frac{1}{\Delta x}f(c)(\Delta x +x-x) $$ For some constant $c$ inside the interval $[x,x+\Delta x]$

$$ \Longrightarrow f(c) $$

Since $\Delta x\rightarrow 0$, $c$ also approaches $x$ (as it is squeezed between the said interval)

Hence $$\frac{1}{\Delta x}\int^{x+\Delta x}_x f(u) du = f(x) $$ as $\Delta x \rightarrow 0$

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    $\begingroup$ Of course the mean value theorem for integrals requires $f$ to be continuous. $\endgroup$
    – GEdgar
    Aug 27, 2021 at 15:01
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Here is another way to think about it: Let $M_{\Delta x}$ and $m_{\Delta x}$ be the supremum and infimum of $f$ over the interval $[x,x + \Delta x]$. We have $$ m_{\Delta x} \cdot \Delta x \leq \int_{x}^{x + \Delta x} f(u)\,du \leq M_{\Delta x} \cdot \Delta x $$ So $$ m_{\Delta x} \leq \frac{1}{\Delta x} \int_{x}^{x + \Delta x} f(u)\,du \leq M_{\Delta x} $$ Since $f$ is continuous at $x$, $\lim_{\Delta x \to 0} M_{\Delta x} = f(x)$ and similarly for $m_{\Delta x}$. So by the squeeze theorem, the term in the middle tends to $f(x)$ too.

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    $\begingroup$ (+1), but shouldn't $M_{\Delta x}$ and $m_{\Delta x}$ be the supremum and infimum of $f$ on the interval $[x,x+\Delta x]$, not $f(x)$? $\endgroup$
    – Joe
    Aug 27, 2021 at 14:24
  • $\begingroup$ @Joe yes, absolutely. I'm more used to thinking of $x$ as a variable ranging over an interval than an endpoint of that interval, and in my haste I was doing both. Let me edit. $\endgroup$ Aug 27, 2021 at 14:25
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    $\begingroup$ Great! I slightly prefer this approach to using the Mean Value Theorem, because the inequality you used follows directly from how integrals are defined in the first place. $\endgroup$
    – Joe
    Aug 27, 2021 at 14:31

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