Asked by Abby
Find the average value of the function defined by
f(x)= x^2 sin 3 x^3
on the interval [0, 3rd Root (pi)]
f(x)= x^2 sin 3 x^3
on the interval [0, 3rd Root (pi)]
Answers
Answered by
Steve
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)
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)
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