Let f : R → R be defined as f x = e - x sin x . If F : 0 , 1 → R is a differentiable function such that F x = ∫ 0 x f t d t , then the value of ∫ 0 1 F ' ( x ) + f ( x ) e x d x lies in the interval

Step-by-step solution

Given f ( x ) = e - x sin x Now, F x = ∫ 0 x f t d t Using Newton Leibnitz rule i.e. d d x ∫ u x v x f t d t = f v x · v ' x - f u x · u ' x , ⇒ F ' ( x ) = f ( x ) Now, I = ∫ 0 1 F ' ( x ) + f ( x ) e x dx ⇒ I = ∫ 0 1 ( f ( x ) + f ( x ) ) · e x dx ⇒ I = 2 ∫ 0 1 f ( x ) · e x d x ⇒ I = 2 ∫ 0 1 e - x sin x · e x d x ⇒ I = 2 ∫ 0 1 sin x d x ⇒ I = 2 - cos x 0 1 ⇒ I = 2 ( 1 - cos 1 ) Using the e