What is the average value of the function #f(x)=sin^2(3x)+x# on the interval #[0, π]#?

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What is the average value of the function #f(x)=sin^2(3x)+x# on the interval #[0, π]#?

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Answer 1 · 2015-03-07T18:33:31

The average value is given by:
#A_V=(Area)/(Interval)=#

enter image source here

#=1/(pi-0)int_0^pi[sin^2(3x)+x]dx=#
#=1/pi[int_0^pisin^2(3x)dx+int_0^pixdx]=#
You can set in the first integral #3x=t# so that #x=t/3# and #dx=dt/3#
#=1/pi{int_0^pi[sin^2(t)]dt/3+x^2/2]_0^pi}=#
#=1/pi{[1/(3×2)(t-sin(t)cos(t))]+pi^2/2}=#
but #t=3x#;
#=1/pi{1/6(3x-sin(3x)cos(3x)]_0^pi+pi^2/2}=#
And:

#1/pi[pi/2+pi^2/2]=1/2+pi/2#

Please check my math...I was in a hurry!

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What is the average value of the function #f(x)=sin^2(3x)+x# on the interval #[0, π]#?

1 Answer

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