Calculus

posted by Anna

I'm supposed to find the average value of the function over the given interval.

f(x) = sin(nx), interval from 0 to pi/n, where n is a positive integer.

I know the average value formula, and I know that the integral of that function would be (-1/n)cos(nx), but I keep getting zero for my final answer, which is wrong. Can someone help me?

  1. Anonymous

    u = n x

    d u = n dx Divide both sides by n

    du / n = dx

    d x = du / n


    integ sin ( n x ) dx =

    integ sin u du / n =

    ( 1 / n ) integ sin u du =

    ( 1 / n ) ( - cos u ) + C =

    - cos ( n x ) / n + C


    average = [ 1 / ( b - a ) ] definite integral sin( n x ) dx from x = 0 to pi /n

    average = [ 1 / ( pi / n - 0 ) ] * [ - cos ( pi * x / n) - cos ( 0 * x ) ]

    average = [ 1 / ( pi / n ) ] * [ - cos ( pi * x / n) - 1 ]

    average = n / pi * [ - cos ( pi * x / n ) - 1 ]

  2. Mutlaq

    This answer is wrong people

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