Floor in Floor in Floor

A property of Greatest integer function can make it easier.

$\lfloor a+b \rfloor=a+\lfloor b \rfloor$ if a is an integer So the given integrand $\lfloor x+\lfloor x+\lfloor x \rfloor \rfloor \rfloor=\lfloor x+\lfloor x \rfloor \rfloor + \lfloor x \rfloor= \lfloor x \rfloor +\lfloor x \rfloor +\lfloor x \rfloor=3\lfloor x \rfloor$

Now we can easily plot $f(x)=3\lfloor x \rfloor$ .

f(x)=0 for x=0 to 1 and f(x)=3 for x=1 to 2 . So the area will be $(2-1) \times 3=\boxed{3}$

clearly for 0<x<1

the equation is 0,, and hence we can stop worrying about the integral from 0 to 1,,

now from 1 to 2,,, we have 1<x<2 so box(x) =1

also box(x+n) = box(x) + n (n is natural number)

so we have box(x+box(x+box(x))) = box(x+box(x+1))=box(x+box(x)+1) = box(x+2) = box(x) + 2 = 3

so the integral is simply

$\int _{ 1 }^{ 2 }{ 3 } dx$

= 3 (answer)

(note- box(x) means greatest integer function of x)