What is the value of the Fourier coefficient $a_0$ for the function given below? $f(x) = \begin{cases} \sin 2x, & -\pi < x < -\frac{\pi}{2} \\ 0, & -\frac{\pi}{2} \le x \le 0 \\ \sin 2x, & 0 < x \le \pi \end{cases}$
The Fourier coefficient $a_0$ represents the average value of the function $f(x)$ over the interval $[-\pi, \pi]$. It is calculated using the formula:
$a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx$
The function $f(x)$ is defined piecewise. To calculate the integral $\int_{-\pi}^{\pi} f(x) dx$, we split the integral according to the function's definition:
The total integral becomes:
$ \int_{-\pi}^{\pi} f(x) dx = \int_{-\pi}^{-\pi/2} \sin(2x) \, dx + \int_{-\pi/2}^{0} 0 \, dx + \int_{0}^{\pi} \sin(2x) \, dx $
$ \int_{-\pi}^{-\pi/2} \sin(2x) \, dx = \left[ -\frac{\cos(2x)}{2} \right]_{-\pi}^{-\pi/2} $
Evaluate at the limits:
$ = \left(-\frac{\cos(2 \times (-\pi/2))}{2}\right) - \left(-\frac{\cos(2 \times (-\pi))}{2}\right) $
$ = \left(-\frac{\cos(-\pi)}{2}\right) - \left(-\frac{\cos(-2\pi)}{2}\right) = \left(-\frac{-1}{2}\right) - \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{2} = 1 $
The function is $0$ in this interval:
$ \int_{-\pi/2}^{0} 0 \, dx = 0 $
$ \int_{0}^{\pi} \sin(2x) \, dx = \left[ -\frac{\cos(2x)}{2} \right]_{0}^{\pi} $
Evaluate at the limits:
$ = \left(-\frac{\cos(2\pi)}{2}\right) - \left(-\frac{\cos(2 \times 0)}{2}\right) $
$ = \left(-\frac{1}{2}\right) - \left(-\frac{1}{2}\right) = -\frac{1}{2} + \frac{1}{2} = 0 $
Sum the results from the individual intervals:
$ \int_{-\pi}^{\pi} f(x) dx = 1 + 0 + 0 = 1 $
Now substitute the value of the total integral back into the formula for $a_0$:
$ a_0 = \frac{1}{2\pi} \times (1) = \frac{1}{2\pi} $
The value of the Fourier coefficient $a_0$ is $\frac{1}{2\pi}$.
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