Let f: [1, ) R be a differentiable function defined as f(x) = ₁^x f(t) ,dt + (1-x)( _e x - 1) + e . Then the value of f(f(1)) is :

Step-by-step solution

Given f(x) = ₁^x f(t) ,dt + (1-x)( _e x - 1) + e Substituting x = 1 in the given equation: f(1) = ₁^1 f(t) ,dt + (1-1)( _e 1 - 1) + e f(1) = 0 + 0 + e = e Differentiating the given equation with respect to x using the Leibniz rule: f'(x) = f(x) + d dx [(1-x)( _e x - 1)] f'(x) = f(x) + (-1)( _e x - 1) + (1-x) ( 1 x ) f'(x) = f(x) - _e x + 1 + 1 x - 1 f'(x) - f(x) = 1 x - _e x This is a linear differential equation of the form dy dx + Py = Q , where P = -1 and Q = 1 x - _e x . Integrating Factor (IF) = e^ -1 ,dx = e^