Let f : ℝ → ℝ be a differentiable function such that its derivative f ' is continuous and f π = - 6 . If F : 0 , π → ℝ is defined by F x = ∫ 0 x f t d t , and if ∫ 0 π f ' x + F x cos x d x = 2 , then the value of f 0 is_______

Step-by-step solution

I = ∫ 0 π f ' x · cos   x + F x · cos   x d x = 2 = ∫ 0 π f ' x · cos   x · d x + ∫ 0 π F x cos   x · d x = 2 = cos x · f ( x ) 0 π - ∫ 0 π - sin x · f x d x + ∫ 0 π F x · cos   x · d x = 2 ⇒    cos π . f π - cos 0 . f 0 + ∫ 0 π sin x · f x d x + ∫ 0 π F x · cos   x d x = 2 ⇒    - 1 · - 6 - f 0 +