Let a and b be real constants such that the function f defined by f x = x 2 + 3 x + a , x ≤ 1 b x + 2 , x > 1 be differentiable on R . Then, the value of ∫ - 2 2 f x d x equals

Step-by-step solution

Given: f x = x 2 + 3 x + a , x ≤ 1 b x + 2 , x > 1 Now, for function to be continuous, ⇒ 1 2 + 3 × 1 + a = b × 1 + 2 ⇒ 4 + a = b + 2 ⇒ a - b = - 2 . . . i Now, differentiating and putting the value of x , ⇒ 2 x + 3 = b ⇒ 2 1 + 3 = b ⇒ b = 5 ⇒ a = 3 ⇒ ∫ - 2 2 f x d x = ∫ - 2 1 x 2 + 3 x + a d x + ∫ 1 2 b x + 2 d x ⇒ ∫ - 2 2 f x d x = ∫ - 2 1 x 2 + 3 x + 3 d x + ∫ 1 2 5 x + 2 d x ⇒ ∫ - 2 2 f x d x = x 3 3 + 3 x 2 2 + 3 x - 2 1 + 5 x 2 2 + 2 x 1 2 ⇒ ∫ - 2 2 f x d x = 1 3 + 3 2 + 3 - - 8 3 + 6 - 6 + 5 × 4 2 + 2 × 2 - 5