Find the value of the real number 𝑎0 of the Fourier series if f(x) = 𝑥2 for 0 ≤ x ≤ 2𝜋.Question 1Select one:-8𝜋23-4𝜋234𝜋238𝜋23

The Fourier series of a function f(x) is given by:

f(x) = a0/2 + Σ [an cos(nx) + bn sin(nx)]

where the coefficients a0, an, and bn are given by:

a0 = (1/π) ∫ from 0 to 2π f(x) dx an = (1/π) ∫ from 0 to 2π f(x) cos(nx) dx bn = (1/π) ∫ from 0 to 2π f(x) sin(nx) dx

In this case, f(x) = x^2 and we want to find a0. So we need to compute the integral:

a0 = (1/π) ∫ from 0 to 2π x^2 dx

This is a simple power rule integral, so we get:

a0 = (1/π) * [x^3/3] from 0 to 2π

Evaluating at the limits gives:

a0 = (1/π) * [(2π)^3/3 - 0] a0 = (1/π) * [8π^3/3] a0 = 8π^2/3

So, the correct answer is 8π^2/3. However, none of the options you provided match this result. There might be a mistake in the question or the provided options.