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?

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 ] 
Mutlaq
This answer is wrong people
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