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)
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.