Question

(1 point) Let f be a periodic function of period 10 such that f(x)=x2 for −5≤x<5. Then the Fourier series of f is

Answer 1

Answer:

The answer is "$\frac{25}{3}$".

Step-by-step explanation:

please find the complete question in the attached file.

$a_0=\frac{1}{10}\int^{5}_{x=-5} x^2 dx$

   $=\frac{1}{10} (\frac{x^3}{3})^{5}_{-5}\\\\=\frac{1}{30}[5^3+5^3]\\\\=\frac{2}{30}(125)\\\\=\frac{1}{15} \times 125\\\\=\frac{25}{3}$