An application of the fundamental theorem of calculus Suppose that 
\begin{align*}
f(x) = \int_a^x [t] \text{ d}t, \quad x\in \mathbb{R}
\end{align*}
where $[x]$ is the ceiling function. i.e., 
\begin{align*}
[x] = n, \quad n \leq x < n+1.
\end{align*}
How can I show that $f$ is not differentiable ? I tried this by showing the discontinuity of $f$ but I found that $f$ is continuous for all $x\in \mathbb{R}$. Can anyone help me?. I appreciate any help.
 A: Hint: Assume $a=0$ and fix $k>0$. Let us show that $f$ is not differentiable at $x = 1$. Observe, we have that
\begin{align}
f(1+kh) -f(1-h) = \int^{1+kh}_0 [x]\ dx-\int^{1-h}_0 [x]\ dx = \int^{1+kh}_{1-h} [x]\ dx =  \int^{1+kh}_1 1\ dx = kh
\end{align}
which means
\begin{align}
\frac{f(1+kh)-f(1)}{h} = k.
\end{align}
A: Notice that when $x$ is not an integer then the integral is differentiable and its derivative  is $[x]$. So when $x$ is an integer you get a different value of the derivative when taking limits from above and below.
A: You need to get the Fundamental Theorem of Calculus in slightly general form:
Fundamental Theorem of Calculus: If a function $f:[a, b]\to \mathbb{R} $ is Riemann integrable on $[a, b] $ and $$F(x) =\int_{a}^{x}f(t) \, dt$$ then $F$ is continuous on $[a, b] $. If for any point $c\in[a, b] $ the limit $\lim_{x\to c^{-}}f(x) $ exists then the left hand derivative $F'_{-} (c) $ exists and $$F'_{-} (c) =\lim_{x\to c^{-}} f(x) $$ Similarly if $\lim_{x\to c^{+}} f(x) $ exists then $$F'_{+} (c) =\lim_{x\to c^{+}} f(x) $$
In your question a function $g$ is given by $g(x) =[x] $ whose left hand and right hand limits are unequal at integers. Hence the the function $f$ given by $$f(x) =\int_{a} ^{x} g(t) \, dt$$ has left hand and right hand derivatives unequal at integers. The function $f$ is therefore not differentiable at integer points. 
