Let f : ℝ → ℝ be a function defined as f x = a sin π x 2 + 2 - x , a ∈ ℝ , where t is the greatest integer less than or equal to t . If lim x → - 1 f x exists, then the value of ∫ 0 4 f x d x is equal to

Step-by-step solution

Given, f x = a sin π x 2 + 2 - x , a ∈ ℝ Now given lim x → - 1 f x exists, So lim x → - 1 + a sin π x 2 + 2 - x = - a + 2 And lim x → - 1 - asin π x 2 + 2 - x = 0 + 3 = 3 So, lim x → - 1 f x exist when - a + 2 = 3 ⇒ a = - 1 Now, ∫ 0 4 f x d x = ∫ 0 1 f x d x + ∫ 1 2 f x d x + ∫ 2 3 f x d x + ∫ 3 4 f x d x ⇒ ∫ 0 4 f x d x = ∫ 0 1 - sin π x 2 + 2 - x d x + ∫ 1 2 - sin π x 2 + 2 - x d x + ∫ 2