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March 31, 2015

March 31, 2015

Posted by **Anonymous** on Thursday, February 9, 2012 at 7:00pm.

- College Calculus -
**Steve**, Thursday, February 9, 2012 at 9:23pmLet 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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