Let f x = ∫ 0 x e t f t d t + e x be a differentiable function for all x ∈ R . Then f x equals :

Step-by-step solution

Given f x = ∫ 0 x e t f t d t + e x ⇒ f 0 = 1 Differentiating with respect to x we get, f ' x = e x f x + e x f ' x = e x f x + 1 f ' x f x + 1 = e x Integrate both the sides we get, ∫ 0 x f ' x f x + 1 d x = ∫ 0 x e x d x ln f x + 1 0 x = e x 0 x ln f x + 1 - ln f 0 + 1 = e x - 1 ln f x + 1 2 = e x - 1 as f 0 = 1 f x = 2 e e x - 1 - 1