Question

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

Answer 1

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

`=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!