integrate frm -1 to 1:
[x+[x+[x]]]dx

Don't you know that you can cut this integration to two parts?

ie, _-1∫¹ f(x).dx = _-1∫⁰ f(x).dx + ₀∫¹ f(x).dx

Let f(x) = [x]

By the defenition of [x]; we have,

f(x) = x, x>0 and f(x) = -x, x<0

Then, come to your problem,

Let g(x) = [x+[x+[x]]].

When x>0; g(x) = x+x+x = 3x;

When x<0; g(x) = -x; (Why?? Do it yourself!!)

Then your problem reduces to:

_-1∫¹ [x+[x+[x]]].dx

= _-1∫⁰ (-x).dx + ₀∫¹ 3x.dx

= -(0/2) - (-1/2) + (3*1/2) - 0

= +(1/2) + (3/2)

= 2.

Is this the correct answer??

Sorry Sorry! I misunderstood your question! I'm extremely sorry!

I thought that [x] is modulus function! I thought of "greatest integer function" just after clicking "POST REPLY" button!

I couldn't undo it!

Again, extremely sorry for wasting your time!!