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

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.
