If f(x) satisfies the relation f(x)=ex+∫01(y+xex) f(y) dy, then e+f(0) is equal to _______ . [2026]
(2)
f(x)=ex+∫01yf(y)dy+xex∫01f(y)dy
f(x)=ex+A+Bxex
A=∫01yf(y)dy=∫01y(A+ey+Byey)dy
A=A2+0-(-1)+B(e-1)
A2+B(1-e)=1
B=∫01f(y)dy
B=∫01(ey+A+Byey)dy
B=(e-1)+A+B(0-(-1))
B=e-1+A+B⇒A=1-e
f(0)=1+A=1-e+1=2-e
e+f(0)=2