If f(x) satisfies the relation f(x)=e^ x + ₀¹ (y+x e^ x ) f(y) d y , then e+f(0) is equal to _ _ _ _ .

Step-by-step solution

f(x) = e^x + ₀^1 (y + xe^x)f(y) , dy Splitting the integral: f(x) = e^x + ₀^1 yf(y) , dy + xe^x ₀^1 f(y) , dy Let A = ₀^1 yf(y) , dy and B = ₀^1 f(y) , dy f(x) = (1 + Bx)e^x + A At x = 0 : f(0) = 1 + A Now compute B using the expression for f(y) : B = ₀^1 [(1 + By)e^y + A ] dy B = ₀^1 e^y , dy + B ₀^1 ye^y , dy + ₀^1 A , dy Evaluating each integral: ₀^1 e^y , dy = e - 1 ₀^1 ye^y , dy = [ye^y]₀^1 - ₀^1 e^y , dy = e - (e - 1) = 1 ₀^1 A , dy = A B = (e - 1) + B(1) + A 0 = e - 1 + A A = 1 - e f(0) = 1 + A = 1 + (1 - e)