Is $\int_a^xf(t)dt$ differentiable iff $f$ is continuous? Let $a,b\in\mathbb R$, $a<b$ and $f:[a,b]\rightarrow\mathbb R$ Riemann integrable. Define $F(x):=\int_a^xf(t)dt$ for $x\in[a,b]$. Is it true that $F$ is differentiable if and only if $f$ is continuous? If so, how do you prove this?
 A: No. For simplicity, let $a = 0$. Now, let $f(x) = 2x\sin(1/x) - \cos(1/x)$ for $x \neq 0$, and $f(0) = 0$. Then, $F(x) = x^2\sin(1/x)$ for $x \neq 0$, and $F(0) = 0$. Then, $F$ is differentiable, but $f$ is not continuous.
e: Of course, the other direction is just (a weaker case of) the fundamental theorem of calculus.
e2: This function also provides a counter example to the statement "if $f$ is differentiable, then $f'$ is continuous".
A: Notice that $F$ does not notice changes to isolated points in intervals where $f$ is continuous, so the answer is no.
For instance, let $f(x)=1$ except at some point $y\in[a,b]$, where it is any other value.
A: Assume $f $ continuous at $x=c\in [a,b]. $
let
$$\Delta=\frac {\int_a^{c+h}f-\int_a^cf}{h}-f (c) $$
$$=\frac {1}{h}\int_c^{c+h} (f (t)-f (c))dt $$
Given $\epsilon>0$, 
$f $ continuous at $x=c  \implies
\exists \eta>0 \;:$
$|h|<\eta \implies |f (c+h)-f (c)|<\epsilon $
$\implies | \int_c^{c+h} (f (t)-f (c))dt|\le $
$\int_c^{c+h}|f (c+t-c)-f (c)|dt\le h\epsilon $
$\implies \Delta <\epsilon $.
We proved that
$$\lim_{h\to 0}\frac {F (c+h)-F (c)}{h}=f (c)=F'(c) $$
