Show that $g$ is differentiable and find $g'(x)$, FTOC Let $f: \Bbb{R} \to \Bbb{R}$ be continuous and let $\alpha > 0$. Define $g: \Bbb{R} \to \Bbb{R}$ by $g(x) = \int_{x-\alpha}^{x+\alpha} f(t)dt, x \in \Bbb{R}$. Show that $g$ is differentiable and find $g'(x)$.
So, I was initially going to create two functions, $a(x) = x-\alpha, b(x) = x+\alpha$, which are clearly both differentiable, and use the chain rule to find the derivative of $g$. But that approach didn't make much sense once I put it down on paper, and I don't know how to do this.
 A: Let $x\in\mathbb R$ and choose $M<x-\alpha$. Then by continuity of $f$ we have
$$g(x) = \int_{-M}^{x+\alpha}f(t)\ \mathsf dt - \int_{-M}^{x-\alpha}f(t)\ \mathsf dt, $$
and so the fundamental theorem of calculus yields
$$g'(x) = f(x+\alpha)-f(x-\alpha). $$
A: The easy way to do it is as others in the comments pointed out, let $F(x)=\int_a^x f(t)dt$
Then $$g(x)=\int_{x-\alpha}^{x+\alpha}f(t)dt=F(x+\alpha)-F(x-\alpha)$$
Differentiating gives
$g'(x)=F'(x+\alpha)-F'(x-\alpha)$
By FTC, we know $F'(x)=f(x)$, so
$g'(x)=f(x+\alpha) - f(x-\alpha)$

The rigorous way to differentiate it is a bit messy:
$$g'(x)=\lim_{\Delta x\to 0} {\int_{x+\Delta x-\alpha}^{x+\Delta x +\alpha}f(t)dt - \int_{x-\alpha}^{x+\alpha}f(t)dt \over \Delta x}$$
We can write the numerator as:
$$\int_{x+\alpha}^{x+\alpha+\Delta x}f(t)dt +\int_{x+\Delta x-\alpha}^{x-\alpha}f(t)dt$$
In the first integral above, writing it as $F(x+\alpha+\Delta x) -F(x+\alpha)$, by MVT, we know there exists some $c\in(x+\alpha,x+\Delta x +\alpha)$ such that $F'(c)(\Delta x)=F(x+\alpha+\Delta x)-F(x+\alpha)$
So $$f(c)\Delta x=\int_{x+\alpha}^{x+\Delta x+\alpha}f(t)dt$$
So the first integral in the limit:
$$lim_{\Delta x\to 0} \frac{f(c)\Delta x}{\Delta x}=f(c)$$
Remember the range of $c$ is between $x+\alpha +\Delta x$ and $x+\alpha$, so as $\Delta x$ approaches $0$, the value of $c$ converges to $x+\alpha$, so the limit of the first integral gives $f(x+\alpha)$
Now I would leave the second integral for you to finish.
