Consider the function $ \int_0^x [t]dt $ where $ x>0 $ and [t] denotes largest integer less than or equal to t. What is the nature of the function?

Question

I have done the following:
$ \int_0^x [t]dt $ can be divided as
$ \int_0^1 [t]dt + \int_1^2 [t]dt + \int_2^3 [t]dt + ... + \int_{x-1}^x [t]dt$
= $ \int_0^1 0dt +\int_1^2 1dt +....$
= $ 1 + 2 + 3 + ....{(x-1)} $
= $ \frac{x(x-1)}{2} $
Therefore, $ f(x) = \frac{x(x-1)}{2} $
If we use this expression to determine the continuity or differentiability of the f(x) then it is continuous and differentiable in [0,x]
However, if we use the original expression $ f(x)= \int_0^x [t]dt $ to determine the nature of the function with the fundamental formulae, then the results show that the function is continuous but not differentiable. Where is the fallacy?

Answer 1

Your division into several integrals appears to assume $x$ is an integer

Answer 2

An important issue is your argument assumes $x$ is integral, when that is not necessarily the case. This is why your $f\left(x\right) = \frac{x\left(x - 1\right)}{2}$ is not true in all cases. In particular, CY Aries's question comment states your summation is missing a last term of $\left(x - \lfloor x \rfloor\right)\lfloor x \rfloor$ (which is $0$ if $x$ is integral). This means the correct function expression for all $x \gt 0$ is

$$f(x) = \frac{x\left(x - 1\right)}{2} + \left(x - \lfloor x \rfloor\right)\lfloor x \rfloor \tag{1}\label{eq1A}$$

Answer 3

Your function is shown in the graph below. Between two naturals $n$ and $n+1$ the graph has slope $n$, so it is differentiable. At the naturals it is not differentiable because the slope is different on the two sides. You have calculated the values at the naturals correctly, but it has values in between.