Posted by Anonymous on Thursday, February 9, 2012 at 7:00pm.
Let u = 1/x
du = -dx/x^2
Int(sin(pi/x)/x^2 dx)[1,2]
= Int(-sin(pi*u) du)[1,.5]
= 1/pi cos(pi*u) [1,.5]
= 1/pi(cos pi/2 - cos pi)
= 1/pi(0+1)
= 1/pi
Since the interval length is 1, the average is just 1/pi
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