Integrate on Floors

$\int _{ -1 }^{ 1 }{ \left\lfloor x+\left\lfloor x+\left\lfloor x \right\rfloor \right\rfloor \right\rfloor dx }$

$=\int _{ -1 }^{ 0}{ \left\lfloor x+\left\lfloor x+\left\lfloor x \right\rfloor \right\rfloor \right\rfloor dx }+\int _{ 0}^{ 1 }{ \left\lfloor x+\left\lfloor x+\left\lfloor x \right\rfloor \right\rfloor \right\rfloor dx }$

$=\int _{ -1 }^{ 0 }{ \left\lfloor x+\left\lfloor x-1 \right\rfloor \right\rfloor dx }+\int _{ 0 }^{ 1 }{ \left\lfloor x+\left\lfloor x+0 \right\rfloor \right\rfloor dx }$

$=\int _{ -1 }^{ 0 }{ \left\lfloor x+(-2) \right\rfloor dx }+\int _{ 0 }^{ 1 }{ \left\lfloor x+0\right\rfloor dx }$

$=-3$

Let's analyze the function $f(x)=\left\lfloor x+\left\lfloor x+\left\lfloor x\right\rfloor \right\rfloor\right\rfloor$ .

For $-1\le x<0$ we have $\lfloor x\rfloor = -1$ Also since $-2\le x-1 <-1$ and $-3\le x-2<-2$ we have $\lfloor x-1\rfloor =-2$ and $\lfloor x-2\rfloor = -3$ . So in that range $f(x)=\left\lfloor x+\left\lfloor x-1\right\rfloor\right\rfloor=\left\lfloor x-2\right\rfloor=-3$ .

For $0\le x<1$ we have $\lfloor x\rfloor = 0$ . So in that range $f(x)=\left\lfloor x+\left\lfloor x+0\right\rfloor\right\rfloor=\left\lfloor x\right\rfloor=0$ . And we trivially have $f(1)=3$ .

Therefore, only the interval $[-1,0)$ contributes to the area. The area is $(-3)\times 1=\boxed{-3}$ .

We can simply use the fact that [x+I] can be written as [x]+I where I is an integer. Now as [x] is itself an integer we can remove it repeatedly to final get integral of "-1 to 1" 3 times [x] to get -3. That's a much shorter solution in practice. (Where [.] represents the greatest integer function)