Let `f: RvecR` be a function defined by `f(x)={[x],(xlt=2) \ \ (0,x >2)` where `[x]` is the greatest integer less than or equal to `xdot` If `I=int_(-1)^2(xf(x^2))/(2+f(x+1))dx , then \ the \ value \ of \ (4I-1)i s`

Answer

0

Explanation

`I=int_(-1)^(2)(x[x^(2)])/(2+[x+1])dx=int_(1)^(2)(x[x^(2)])/(3+[x])dx`
`=int_(-1)^(0)0/(3-1)dx+int_(0)^(1)0/(3+1)dx+int_(1)^(sqrt(2))(x.1)/(3+1)dx+0`
`=1/4[(x^(2))/2]_(1)^(sqrt(2))=(2-1)/8=1/4`
`:. 4I-1=0`