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April 1, 2015

April 1, 2015

Posted by **zoe** on Monday, October 31, 2011 at 6:48am.

- calculus -
**Steve**, Monday, October 31, 2011 at 3:43pmDouble check my math, because it gets messy...

Break it into two intervals: [0,pi/4] and [pi/4,pi/2]

Int(sin x)[0,pi/4] + Int(cos x)[pi/4,pi/2]

= (-cos x)[0,pi/4] + (sin x)[pi/4,pi/2]

= 2-√2

This is the denominator in the formulas for xbar and ybar: D = 2-√2

Now for the numerators:

xbarn = Int(x sin x)[0,pi/4] + Int(x cos x)[pi/4,pi/2]

Recall that using integration by parts,

Int(x sin x) = -x cos x + sin x)

Int(x cos x) = x sin x + cos x)

If my math is right, xbarn = pi/4 * (2-√2)

So, xbar = xbarn/D = pi/4

Makes sense, since the area is symmetric about the line x = pi/4.

Now for ybar

ybarn = 1/2 Int(sin^2 x)[0,pi/4] + 1/2 Int(cos^2 x)[pi/4,pi/2]

Recall that

sin^2 x = (1 - cos(2x))/2

cos^2 x = (1 + cos(2x))/2

ybarn = 1/4 Int(1 - cos 2x)[0,pi/4] + 1/4 Int(1 + cos 2x)[pi/4,pi/2]

= 1/4 (x - 1/2 sin 2x)[0,pi/4] + 1/4 (x + 1/2 sin 2x)[pi/4,pi/2]

= 1/8 (pi-2)

So, ybar = ybarn/D = (pi-2)/(8*(2-√2))

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