Question

the rms value of y=x^2 in the interval -pi to pi

Answer 1

Given:

A function has been provided as ;

$y = {x}^{2}$

To find:

Root Mean Square (RMS) value of y in the interval from -π to +π

Calculation:

$\therefore \: \: y_{(rms)} = \sqrt{ \dfrac{ \displaystyle\int_{ - \pi}^{\pi}( {y}^{2} \: dx)}{ \displaystyle \int_{ - \pi}^{\pi}dx} }$

Putting the value of "x" in place of "y" in RHS :

$= > \: \: y_{(rms)} = \sqrt{ \dfrac{ \displaystyle\int_{ - \pi}^{\pi} \{ {( {x}^{2} )}^{2} \: dx \}}{ \displaystyle \int_{ - \pi}^{\pi}dx} }$

$= > \: \: y_{(rms)} = \sqrt{ \dfrac{ \displaystyle\int_{ - \pi}^{\pi} ({x}^{4} \: dx )}{ \displaystyle \int_{ - \pi}^{\pi}dx} }$

Following rules of Integration :

$= > \: \: y_{(rms)} = \sqrt{ \dfrac{ \bigg \{ \dfrac{ {x}^{5} }{5} \bigg \}_{ - \pi}^{\pi} }{\pi - ( - \pi)} }$

$= > \: \: y_{(rms)} = \sqrt{ \dfrac{ \bigg \{ \dfrac{ {x}^{5} }{5} \bigg \}_{ - \pi}^{\pi} }{\pi + \pi} }$

$= > \: \: y_{(rms)} = \sqrt{ \dfrac{ \bigg \{ \dfrac{ {x}^{5} }{5} \bigg \}_{ - \pi}^{\pi} }{2\pi } }$

$= > \: \: y_{(rms)} = \sqrt{ \dfrac{ \bigg \{ {x}^{5} \bigg \}_{ - \pi}^{\pi} }{10\pi } }$

$= > \: \: y_{(rms)} = \sqrt{ \dfrac{ \bigg \{ {\pi}^{5} - {\pi}^{( - 5)} \bigg \} }{10\pi } }$

$= > \: \: y_{(rms)} = \sqrt{ \dfrac{ \bigg \{ {\pi}^{5} - \dfrac{1}{ {\pi}^{5} } \bigg \} }{10\pi } }$

$= > \: \: y_{(rms)} = \sqrt{ \dfrac{ \bigg \{ \dfrac{ {\pi}^{10} - 1}{ {\pi}^{5} } \bigg \} }{10\pi } }$

$= > \: \: y_{(rms)} = \sqrt{ \dfrac{ {\pi}^{10} - 1}{10 {\pi}^{6} } }$

$= > \: \: y_{(rms)} = \dfrac{1}{ {\pi}^{3} } \bigg \{ \sqrt{ \dfrac{ {\pi}^{10} - 1}{10 } } \bigg \}$

So the final answer is :

$\boxed{ \red{ \bold{ \: \: y_{(rms)} = \dfrac{1}{ {\pi}^{3} } \bigg \{ \sqrt{ \dfrac{ {\pi}^{10} - 1}{10 } } \bigg \}}}}$