# Why does $\frac{1}{\Delta x} \int^{x+\Delta x}_x f(u) du$ tend to $f(x)$ as $\Delta x \to 0$? (From proof of Fundamental Theorem of Calculus)

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}

• This is true provided $f$ is continuous at $x$. I assume that was stated before these calculations. Aug 27, 2021 at 15:00
• 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 $\Delta x\to 0^-$. Aug 28, 2021 at 14:20

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

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 $$\Delta x \rightarrow 0$$

• Of course the mean value theorem for integrals requires $f$ to be continuous. Aug 27, 2021 at 15:01

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.

• (+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 $f(x)$?
– Joe
Aug 27, 2021 at 14:24
• @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. Aug 27, 2021 at 14:25
• 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.
– Joe
Aug 27, 2021 at 14:31