Find the average value of the function defined by

Question

f(x)= x^2 sin 3 x^3

on the interval [0, 3rd Root (pi)]

Answer 1

let u = 3x^3
du = 9x^2 dx
so you have

∫[0,∛π] x^2 sin(3x^3) dx
= 1/9 ∫[0,3π] sin(u) du
= 1/9 (-cos(u))[0,3π]
= 1/9 (-cos3π + cos0)
= 1/9 (1+1)
= 2/9

so, the average value is 2/(9∛π)