The Fourier series for a function $f(x)$ with period $T=2\pi$ is given by:
$f(x) = A_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx))$
The coefficient $A_0$ is calculated using the formula:
$A_0 = \frac{1}{2\pi} \int_{0}^{2\pi} f(x) dx$
Given the function $f(x) = x^2$ and the interval $0 \le x \le 2\pi$. Substitute $f(x)$ into the formula for $A_0$:
$A_0 = \frac{1}{2\pi} \int_{0}^{2\pi} x^2 dx$
First, find the definite integral of $x^2$ from $0$ to $2\pi$:
$\int_{0}^{2\pi} x^2 dx = \left[ \frac{x^3}{3} \right]_{0}^{2\pi}$
Evaluate the integral at the limits:
= $\frac{(2\pi)^3}{3} - \frac{(0)^3}{3}$
= $\frac{8\pi^3}{3} - 0$
= $\frac{8\pi^3}{3}$
Now, substitute this result back into the formula for $A_0$:
$A_0 = \frac{1}{2\pi} \times \frac{8\pi^3}{3}$
Simplify the expression:
$A_0 = \frac{8\pi^3}{6\pi}$
$A_0 = \frac{4\pi^2}{3}$
Therefore, the value of the Fourier coefficient $A_0$ for the function $f(x) = x^2$ is $\frac{4\pi^2}{3}$.
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