Is $$ f(x)= \begin{cases} 0,\quad &x=0,\\ x\sin x,&x>0, \end{cases} $$ integrable, in the Darboux sense, on $[0,1]$?
I know the Darboux integral has to do with the upper sum and lower sum resulting in an equivalence between them. But I'm just a bit confused on how the partition is chosen?