Question

Fourier series of \(x^2,\;0<x<2\) is \[ f(x)=\frac{a_0}{2} +\sum_{n=1}^{\infty}a_n\cos(n\pi x) +\sum_{n=1}^{\infty}b_n\sin(n\pi x), \] then \(b_2=\)

• A. Zero since \(x^2\) is an even function
• B. \[ \int_{0}^{2}x^2\sin2x\,dx \]
• C. \[ \int_{0}^{2}x^2\cos2\pi x\,dx =-\frac{2}{\pi} \]
• D. \[ \int_{0}^{2}x^2\sin2\pi x\,dx =-\frac{2}{\pi} \]

Hint:

For Fourier series on \[ 0<x<2L, \] \[ \boxed{ b_n = \frac1L \int_{0}^{2L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx. } \] Here, \[ L=1, \] so \[ b_n = \int_{0}^{2} f(x)\sin(n\pi x)\,dx. \]

Answer 1

The Correct Option is D

Step 1: Recall the Fourier coefficient. For the interval \[ 0<x<2, \] the sine coefficient is \[ \boxed{ b_n=\int_{0}^{2}f(x)\sin(n\pi x)\,dx. } \] Since \[ f(x)=x^2, \] we obtain \[ b_2 = \int_{0}^{2}x^2\sin(2\pi x)\,dx. \]

Step 2: Evaluate the integral. Using integration by parts, \[ \int_{0}^{2}x^2\sin(2\pi x)\,dx = -\frac{2}{\pi}. \] Hence, \[ \boxed{ b_2=-\frac{2}{\pi}. } \] Therefore, \[ \boxed{(D)} \] is the correct answer.