The value of the integral _ -2 ² ² x [ x ]+ 1 2 d x (where [x] denotes the greatest integer less than or equal to x) is

Step-by-step solution

Let f(x)= ² x [ x ]+ 1 2 . array l So, f(-x)= ²(-x) [ -x ]+ 1 2 [-x]=-1-[x] f(-x)= ² x -1- [ x ]+ 1 2 = ² x - 1 2 - [ x ] =-f(x) array f(x) is odd function Hence, _ -2 ² f(x) d x=0