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 ,t h e nt h ev a l u eof(4I-1)i s`

Answer

0

Explanation

Here , ` f (x) = {{:([x]_(,),xle2),(0_(,),xgt2):}`
`:. I= int_(-1)^(2) ( xf (x^(2)))/(2+f (x+1))dx`
`=int _(-1)^(0)(x f(x^(2)))/(2+f (x+1))dx + int _(0)^(1) (xf (x^(2)))/(2+f(x=1))dx`
`+int_(1)^(sqrt(2))(x f (x^(2)))/(2+f(x+1))dx + int _(sqrt(2))^(sqrt(3))(xf (x^(2)))/(2+f(x+1))dx+ int_(sqrt(3))^(2)(xf (x^(2)))/(2+f(x+1))dx`
`=int_(-1)^(0)dx+int_(0)^(1) 0 dx + int_(1)^(sqrt(2))(x*1)/(2+0)dx+int_(sqrt(2))^(sqrt(3))0 dx + int_(sqrt(3))^(2)0 dx`
`[{:(because-1ltxltrArr0 ltx^(2)lt 1 rArr [x^(2)]=0_(,),),(0 lt x lt 1 rArr 0 lt x^(2) lt 1 rArr [x^(2)] = 0 _(,),),(1 ltx lt sqrt(2)rArr{{:(1ltx^(2)lt2 rArr [x^(2)]=1,),(2ltx+1lt1+sqrt(2)rArrf(x+1)=0_(,),):}:},),(sqrt(2)ltxltsqrt(3)rArr 2 lt x^(2)lt3 rArr f(x^(2))=0_(,),),(andsqrt(3)lt xlt 2 rArr 3 lt x^(2)lt 4 rArr f (x^(2))=0,):}]`
`rArr I = int_(1)^(sqrt(2))(x)/(2)dx = [(x^(2))/(4)]_(1)^(sqrt(2))=(1)/(4)(2-1)=(1)/(4)`
`:. 4I =1 rArr 4 I -1=0`