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Homework Statement
Find the area under bounded by y = cos x and y = sin 2x on the intervacl [0, pi/2]
The Attempt at a Solution
cos x = sin 2x
2sin x cos x  cos x = 0
x = pi/6 and pi/2
shouldn't the intergral be: [ S(0,pi/6) cos x  sin 2x ]  [ S(pi/6,pi/2) sin 2x  cos x ]
I have [ S(0,pi/6) cos x  sin 2x ]  [ S(pi/6,pi/2) sin 2x + cos x ]
why is it sin 2x + cos x?
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