Let f:R→R be a function defined by fx=x,x≤20,x>2 where x  is the greatest integer less than or equal to x . If I=∫−12xfx22+fx+1dx, then the value of 4I−1 is

We havefx2=x2x2≤20x2>2and fx+1=x+1x+1≤2,x≤10x+1>2,x>1Now−2≤x<−1⇒x2=1−1≤x<0⇒x2=00≤x<1⇒x2=01≤x<2⇒x2=1−1≤x<0⇒fx+1=00≤x<1⇒fx=11≤x<2⇒fx+1=0∴I=∫-12xfx22+f(x+1)dx      =∫-10xfx22+f(x+1)dx+∫01xfx22+f(x+1)dx+∫12xfx22+f(x+1)dx+∫22xfx22+f(x+1)dx      =0+0+∫12x2dx+0       =12·x2212=14(2-1)=14⇒4I-1=0