Calculate the average over the given interval. f(x)=sin(pi/x)/x^2, [1,2]

Answer 1

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