Question

find a0 of the function f(x) = 1/4 (π-x)^​

Answer 1

Given:

 $f(x) =\frac{1}{4}(\pi -x)^{2}$

$a_0=\frac{1}{\pi }\int_{0}^{2\pi} \frac{1}{4}(\pi-x)^{2}$

     $=$ from $0$ to $2\pi$

      $= \frac{1}{2}(\frac{1}{3}\pi^{2})$

      $=\frac{1}{6} \pi^2$

      $= \frac{\pi^2}{6}$

   

Answer 2

Answer:

The correct answer will be =

The value of $a_0 = \frac{\pi ^2}{6}$.

Step-by-step explanation:

Correction in the question -

We have to find a₀ of the function $f(x) = \frac{1}{4}(\pi - x)^2$

In Fourier transform $a_0$ corresponds to the average of f(x) in one period and it is given by -

$a_0 = \frac{1}{\pi} \int\limits^\pi_{-\pi} {f(x)} \, dx$

so calculating for the given function -

$a_0 = \frac{1}{\pi} \int\limits^{\pi} _{-\pi} {\frac{1}{4}(\pi - x)^2} \, dx$

Integrating from -π to π we will get -

$a_0 = \frac{1}{4\pi} \int\limits^\pi_{-\pi} ({\pi^2+x^2-2\pi x}) \, dx \\\\a_0 = \frac{1}{4\pi} (2\pi^3+ \frac{8}{3} \pi^3 - 4\pi^3)$

$a_0 = \frac{1}{4\pi}(\frac{2\pi ^3}{3})\\ \\ a_0 = \frac{\pi ^2}{6}$

So the value of the $a_0 = \frac{\pi ^2}{6}$ .

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