The value of ∫0^4 x lfloor xrfloor dx, where lfloor xrfloor denotes the greatest integer not exceeding x, is

On each interval $[k,k+1)$ the greatest integer function equals $k$, so split the integral at $x=1,2,3$: $\int_0^4x\lfloor x\rfloor dx=\sum_{k=0}^{3}k\int_k^{k+1}x\,dx=\sum_{k=0}^3k\cdot\dfrac{(k+1)^2-k^2}{2}=\sum_{k=0}^3\dfrac{k(2k+1)}2$. The terms are $0,\ \dfrac32,\ 5,\ \dfrac{21}2$ for $k=0,1,2,3$. Their sum is $0+1.5+5+10.5=17$. The integrand is discontinuous at the integers but bounded, so the integral exists and is computed piecewise. The answer can be confirmed by graphing the step function $\lfloor x\rfloor$ multiplied by $x$ and summing trapezoid areas.