Let f : R → R be a function defined by f x = x , x ≤ 2 0 , x > 2 , where [ x ] is the greatest integer less than or equal to x . If I = ∫ - 1 2 x f ( x 2 ) 2 + f ( x + 1 ) d x , then the value of ( 4 I - 1 ) is

Step-by-step solution

f x + 1 = x + 1 , x + 1 ≤ 2 0 , x + 1 > 2 = x + 1 , x ≤ 1 0 , x > 1 = 0 , - 1 ≤ x < 0 1 , 0 ≤ x < 1 0 , 1 < x < 2 f x 2 = x 2 . x 2 ≤ 2 0 , x 2 > 2 = x 2 . | x | ≤ 2 0 , | x | > 2 = 0, - 1 ≤ x < 0 0, 0 ≤ x < 1 1, 1 ≤ x < 2 0, 2 < x < 2 l = ∫ - 1 0 0 d x + ∫ 0 1 0 d x + ∫ 1 2 x d x 2 + 0 + ∫ 2 2 0 d x = 1 4 x 2 1 2 = 1 4 2 - 1 = 1 4 4 l = 1 4 l - 1 = 0