more functional equation

Differentiating the above functional equation with respect to x yields $f''(x) = f'(x)$ since the last integration term is a real constant. We ultimately end up with the general solution $f(x) = A + Be^{x},$ which after substituting back into the functional equation produces:

$Be^{x} = (A + Be^{x}) + \int_{0}^{2} A + Be^{x} dx \Rightarrow 0 = A + (Ax + Be^{x})|^{2}_{0} \Rightarrow 0 = 3A + B(e^2 - 1) \Rightarrow A = (\frac{1-e^{2}}{3})B.$

Taking our given boundary condition, $f(0) = \frac{4-e^{2}}{3},$ we now obtain:

$f(0) = A + Be^{0} = (\frac{1-e^{2}}{3})B + B = \frac{4-e^{2}}{3};$

which yields: $A = \frac{1-e^{2}}{3}, B = 1$ , or $\boxed{f(x) = e^{x} + \frac{1-e^{2}}{3}}.$ Finally, $\int_{0}^{1} f(x) dx = \int_{0}^{1} e^{x} + \frac{1-e^2}{3} dx = e^x + (\frac{1-e^2}{3})x|^{1}_{0} = \boxed{(e-1) + \frac{1-e^2}{3}}.$