Question 1070645
Integrate the function over one period and divide by the length,
{{{int((25+10sin(6pi*t)),dt)=25t-(5/(3pi))cos(6pi*t))+C}}}
Integrating from {{{t=0}}} to {{{t=(2pi)/(6pi)=1/3}}},
{{{A=25(1/3-0)-(5/(3pi))(cos(2pi)-cos(0))=25/3}}}
{{{f[ave]=A/(1/3-0)=(25/3)/(1/3)=25/3}}}
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Integrate the square of the function over one period and divide by the length,
then take the square root,
{{{int((25+10sin(6pi*t)^2),dt)=int((100sin^2(6pi*t)+500sin(6pi*t)+625),dt)=675t-(25/(6pi))sin(12pi*t)-(250/3)cos(6pi*t)+C}}}
Integrating from {{{t=0}}} to {{{t=1/3}}},
{{{A=675(1/3-0)-(25/(6pi))(sin(4pi)-sin(0))-(250/3)(cos(2pi)-cos(0))=225}}}
{{{f[RMS]=sqrt(225/(1/3))=sqrt(675)=15sqrt(3)}}}