If $f$ has integral on $[a,b]$ and continuous at $x_0∈[a,b]$ then $F$ has derivative at point $x_0$ and $F'(x_0)=f(x_0)$.

Question

Suppose that $f$ is integrable on $[a,b]$. Let

$$F(x)=\int_{a}^xf(t)dt, \quad a\leq x \leq b.$$

If $f$ is continuous at $x_0 \in [a,b]$ then $F$ has derivative at $x_0$ and $F^\prime(x_0)=f(x_0)$.

This is the proof from my book: $$ F(x_0 + \Delta x) - F(x_0 ) = \int_a^{x_0 + \Delta x} {f(t)dt} - \int_a^{x_0 } {f(t)dt} = \int_{x_0 }^{x_0 + \Delta x} {f(t)dt} . $$ From this $$\frac{F(x_0+\Delta x)-F(x_0)}{\Delta x}-f(x_0)=\frac{1}{\Delta x}\int_{x_0}^{x_0+\Delta x}[{f(t)-f(x_0)}]dt.$$

Can you please help how we got this? Shouldn't it be $ \frac{1}{\Delta x}\int_{x_0}^{x_0+\Delta x}[{f(t)-f(x_0)\Delta x}]dt$?