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April 16, 2014

April 16, 2014

Posted by **Michelle** on Thursday, October 27, 2011 at 3:24pm.

Thanks for your help :)

- Calculus -
**Steve**, Thursday, October 27, 2011 at 5:04pmSince cos x > sin x on the interval,

Integrate 2cos x - 2sin x on [0,pi/4]

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

at x=pi/4, 2(1/√2 + 1/√2) = 2*2/√2 = 2√2

at x=0, 2*0 + 2*1 = 2

area = 2√2 - 2

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