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