Show the function is differentiable $F(x) = \int_{x-1}^{x+1}f(t)dt$ for x an element of the reals.
Show that $F$ is differentiable on Reals, and compute $F^\prime$.
I am unsure about how to showing $F$ is differentiable. I know that I need to use the fundamental theorem of calculus, but can someone please explain how to do so?
 A: If $f$ is not continuous, then $F(x) = \int_{x-1}^{x+1} f(t) \text{d}t$ need not be differentiable at the points of discontinuity of $f$.
For instance I believe the following is a counter-example:
$ f(x) = \begin{cases} 0 & x \le 3 \\ 1 & x \gt 3 \end{cases}$
I believe we can show that $F(x) = \int_{x-1}^{x+1} f(t) \text{d}t$ is not differentiable $2$.
However, it is a well known theorem that at any point of continuity $c$ of $f$, the function $G(x) = \int_{a}^{x} f(t) \text{d}t$ is differentiable and $G'(c) = f(c)$.
A: If you haven't done any measure theory , a simple answer would be: 
If f is continuous then it has a primitive (the integral is supposed to be rieman one) If 0 belongs to the domain of f, let us then call G(x)=integral from 0 to x f(t)dt such a primitive function  wich is differentiable (as f is its derivative and f is continuous).
F(x)=G(x+1)-G(x-1).
F'(x)=(x+1)'f(x+1)-(x-1)'f(x-1).
Thus
F'(x)=f(x+1)-f(x-1).
A: 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
