Question

Root mean square of f(x)=x^2in (0,π)

Answer 1

The mean square of a function $f(x)$ in the interval (a, b) or [a, b] is given by,

• $\overline{x^2}=\dfrac{\displaystyle\int\limits_a^b[f(x)]^2\ dx}{\displaystyle\int\limits_a^bdx}$

And, well, the root mean square is given by,

• $x_{rms}=\sqrt{\overline{x^2}}$

Here the mean square of the function $f(x)=x^2$ in the interval (0, π) will be,

$\longrightarrow\overline{x^2}=\dfrac{\displaystyle\int\limits_0^\pi[f\left(x\right)]^2\ dx}{\displaystyle\int\limits_0^\pi dx}$

$\longrightarrow\overline{x^2}=\dfrac{\displaystyle\int\limits_0^\pi\left(x^2\right)^2\ dx}{\displaystyle\int\limits_0^\pi dx}$

$\longrightarrow\overline{x^2}=\dfrac{\displaystyle\int\limits_0^\pi x^4\ dx}{\displaystyle\int\limits_0^\pi dx}$

$\longrightarrow\overline{x^2}=\dfrac{\dfrac{1}{5}\left[x^5\right]_0^\pi}{\big[x\big]_0^\pi}$

$\longrightarrow\overline{x^2}=\dfrac{\dfrac{1}{5}\cdot\pi^5}{\pi}$

$\longrightarrow\overline{x^2}=\dfrac{\pi^4}{5}$

Hence the root mean square will be,

$\longrightarrow x_{rms}=\sqrt{\overline{x^2}}$

$\longrightarrow x_{rms}=\sqrt{\dfrac{\pi^4}{5}}$

$\longrightarrow\underline{\underline{x_{rms}=\dfrac{\pi^2}{\sqrt5}}}$