If f(x)= cases x-[x] x-2 , & x>2 b, & x=2 |x^2-x-2 | a (2+x-x^2 ) , & -1 < x 2 2 a-b, & x -1 cases is continuous on R , then _ x 0 ^2 a x+x b x x^2 =

Step-by-step solution

Given f(x)= array cc x-[x] x-2 & x>2 b & x=2 |x^2-x-2 | a (2+x-x^2 ) & -1 < x 2 2 a-b & x -1 array . Since f(x) is continuous on R So, _ x 2⁺ f(x)=f(2) _ x 2⁺ x-[x] x-2 =b b=1 aligned & and _ x 2⁻ f(x)=f(2) _ x 2⁻ |(x-2)(x+1)| a(2-x)(x+1) =1 & _ x 2⁻ -(x-2)(x+1) -a(x-2)) x+1) =1 a=1 aligned Now, _ x 0 ^2 a x+x b x x^2 = _ x 0 ^2 x+x x x^2 _ x 0 ( ^2 x x^2 + x x )=1+1=2