$\int_0^x \left[\int_0^u f(t)dt\right]du$ is equal to:
Step-by-Step Solution
Key Concept: Use the functional equation with substitution $x = y = 0$ to determine initial conditions, then apply the limit definition of derivative with Taylor series expansion.
Given the functional equation $f(x+2y) = f(x)e^{2y} + f(2y)e^x + x^2(1-e^{2y}) + 4y^2(1-e^x) + 4xy$, set $x = y = 0$ to obtain $f(0) = 2f(0)$, so $f(0) = 0$. To find $f'(x)$, compute $\lim_{y \to 0} \frac{f(x+2y) - f(x)}{2y}$ by expanding $e^{2y}$ and $e^x$ as series and simplifying the resulting expression.
Correct Answer: 1,3