Integral of a Floor Function (and Graph)

Question

Draw the graph of $f(x)$ over the interval $[0,3]$, where $$f(x)=\int_0^x\lfloor t\rfloor^2\mathrm dt.$$

I don't quite know if I need to use this formula $$\int_0^n\lfloor t\rfloor^2\mathrm dt=\frac{n(n-1)(2n-1)}{6}.$$

Answer 1

Hint: For $0 \le x < 1$, $$f(x) = \int_0^{x} \lfloor t\rfloor^2 dt = \int_0^{x} 0^2 dt = 0.$$ For $1 \le x < 2$, $$f(x) = \int_0^{x} \lfloor t\rfloor^2 dt = \int_0^{1} 0^2 dt + \int_1^{x} 1^2 dt = x-1.$$ Can you continue from here?