Consider the function $f(x)=\int_0^x[t]dt$ where $x>0$ and $[t]$ denotes the largest integer less than or equal to $t$. Then
(A) $f(x)$ is not defined for $x=1,2,3,\dots$
(B) $f(x)$ is defined for all $x>0$ but is not continuous at $x=1,2,3,\dots$
(C) $f(x)$ is continuous at all $x>0$ but is not differentiable at $x=1,2,3,\dots$
(D) $f(x)$ is differentiable at all $x>0$
I have edited the question. I am able to show that the function is continuous for all $x>0$. I am facing problem in checking the differentiability. I tried to do it using the definition of left hand and right hand derivatives but got stuck after a few steps. The correct answer is (C). But until now I have been able to eliminate (A) and (B). So, the answer can be either (C) or (D). How do I deal with the differentiability part?
Some pointers-
Consult a graph: convince yourself that this is correct:
This graph is showing in particular that the $\color{red}{\text{area under the graph}}$ of $\color{blue}{y=[x]}$ from $x=0$ to $4$ is $\color{orange}6$. In general the graph of the $\color{green}{\text{integral as a function of $x$}}$ looks continuous. Does that feel correct?
Simplify your formula. Why write $\int_{[i]}^{[i]+1}[i] dt$? Isn't this just equal to $i$? You should be able to simplify the formula into $$\int_0^x [t] dt = f\left(x\right)=\frac{[x]([x]-1)}2+\left(x-[x]\right)[x]=x[x]-\frac{[x]}2-\frac{[x]^2}2.$$ This matches the above graph: if you check $x=4$, you indeed get $\frac{4^2}2 - \frac42=6$.
Verify continuity. This isn't so hard with the above formula, but its not enough to just study $[x]$ as in the previous version of the question.
Verify that the derivative changes at the integers. The graph shows its smooth outside the integers; the formula should make this obvious. Study the derivative at the left and right of an integer. Recall that a derivative exists iff the left and right derivatives exist and are equal.