Word Count : 440

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The given function is ( f(x) = \pi^2 - x^2 ) with period ( 2\pi ). Since ( f(x) ) is even (( f(-x) = f(x) )), its Fourier series contains only cosine terms: [ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx), ] where [ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} (\pi^2 - x^2) , dx, \quad a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (\pi^2 - x^2) \cos(nx) , dx. ] Step 1: Compute ( a_0 ) [ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} (\pi^2 - x^2) , dx = \frac{1}{\pi} \left[ \int_{-\pi}^{\pi} \pi^2 dx ________ ___ ________ __________ __________ _________ ______ _______ ____ ____ ______ __________.
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