You can simply use definition of the derivative.
You have $F(x)=\int_{x-1}^{x+1} f(t) dt.$
$F(x+h)-F(x)=\int_{x+h-1}^{x+h+1} f(t) dt-\int_{x-1}^{x+1} f(t) dt= \int_{x+1}^{x+h+1} f(t) dt - \int_{x-1}^{x+h-1} f(t) dt$
$\frac{F(x+h)-F(x)}{h}=\frac{\int_{x+1}^{x+h+1} f(t) dt}{h} - \frac{\int_{x-1}^{x+h-1} f(t) dt}h$
$ \min_{c\in_\langle x+1,x+1+h\rangle} f(c)-\max_{c\in_\langle x-1,x-1+h\rangle} f(c) \le \frac{F(x+h)-F(x)}{h} \le \max_{c\in_\langle x+1,x+1+h\rangle} f(c)-\min_{c\in_\langle x-1,x-1+h\rangle} f(c)$ (From the continuity we know that the minima and maxima exists.)
Now (also from the continuity) both expression converge to $f(x+1)-f(x-1)$.
However, you might want to have a look at a general version of this http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign