If f(x) satisfies the relation $$f(x)=e^{x}+\int_{0}^{1}\left(y+xe^{x}\right)f(y)dy,$$ then e + f(0) is equal to ______.

The equation becomes:

$$f(x) = e^x + B + x e^x A = e^x(1 + Ax) + B$$

Now, substitute $$f(x)$$ back into the definitions of $$A$$ and $$B$$:

$$A = \int_0^1 [e^y(1 + Ay) + B] dy$$

$$B = \int_0^1 y[e^y(1 + Ay) + B] dy$$
Solving this system of linear equations for $$A$$ and $$B$$:

$$f(0) = e^0(1 + 0) + B = 1 + B$$.
After integrating and solving, we find that $$B = 1 - e$$.
Therefore, $$f(0) = 1 + (1 - e) = 2 - e$$.
Answer $$e + f(0) = e + (2 - e) = 2$$