Question

If $f(x) = \begin{cases} -\pi, & \text{in } -\pi<x<0 \\ x, & \text{in } 0<x<\pi \end{cases}$ then the constant term of the Fourier series is

• A. \( -\frac{\pi}{4} \)
• B. \( -\frac{\pi}{2} \)
• C. \( -\frac{4}{\pi} \)
• D. \( -\frac{2}{\pi} \)

Hint:

The constant term $a_0$ in a Fourier series represents the average value of the function over one period. It is simply the mean value of the function. Remember to correctly identify the period ($2L$) and then use the appropriate integral formula for $a_0$. For piecewise functions, split the integral at the points where the function definition changes.

Answer 1

The Correct Option is A

The constant term of the Fourier series is given by $a_0$, which is calculated as: $$ a_0 = \frac{1}{2L} \int_{-L}^{L} f(x) \, dx $$ For a function defined on the interval $(-\pi, \pi)$, $L = \pi$. So, the formula becomes: $$ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx $$ Given the function: $$ f(x) = \begin{cases} -\pi, & \text{in } -\pi < x < 0 \\x, & \text{in } 0 < x < \pi \end{cases} $$
Step 1: Split the integral according to the definition of $f(x)$. $$ a_0 = \frac{1}{2\pi} \left( \int_{-\pi}^{0} (-\pi) \, dx + \int_{0}^{\pi} x \, dx \right) $$ 
Step 2: Evaluate the first integral. $$ \int_{-\pi}^{0} (-\pi) \, dx = [-\pi x]_{-\pi}^{0} = (-\pi \cdot 0) - (-\pi \cdot (-\pi)) = 0 - \pi^2 = -\pi^2 $$ 
Step 3: Evaluate the second integral. $$ \int_{0}^{\pi} x \, dx = \left[\frac{x^2}{2}\right]_{0}^{\pi} = \frac{\pi^2}{2} - \frac{0^2}{2} = \frac{\pi^2}{2} $$ 
Step 4: Substitute the evaluated integrals back into the expression for $a_0$. $$ a_0 = \frac{1}{2\pi} \left( -\pi^2 + \frac{\pi^2}{2} \right) $$ 
Step 5: Simplify the expression for $a_0$. $$ a_0 = \frac{1}{2\pi} \left( -\frac{2\pi^2}{2} + \frac{\pi^2}{2} \right) = \frac{1}{2\pi} \left( -\frac{\pi^2}{2} \right) $$ $$ a_0 = -\frac{\pi^2}{4\pi} = -\frac{\pi}{4} $$ Thus, the constant term of the Fourier series is $-\frac{\pi}{4}$.