Area Under Curve: cos x = sin 2x, [0, pi/2]

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SUMMARY

The discussion focuses on calculating the area under the curves defined by the equations y = cos x and y = sin 2x over the interval [0, π/2]. The points of intersection are identified as x = π/6 and x = π/2. The correct formulation for the area is established as the sum of two integrals: [ S(0, π/6) (cos x - sin 2x) ] + [ S(π/6, π/2) (sin 2x - cos x) ]. The importance of ensuring that the area calculation remains positive is emphasized, clarifying that the area should not include negative values unless specified otherwise.

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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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Since they ask for the area, and area is always positive, you don't need the negative multiple in front of your second integral (unless your text has explicitly defined area to mean signed area or the definite integral of a function over an interval).
I do not see why there is a sum instead of a difference in the second integral; it may be a typo.
 
so [ S(0,pi/6) cos x - sin 2x ] + [ S(pi/6,pi/2) sin 2x - cos x ] would be the area?
 
a.a said:
so [ S(0,pi/6) cos x - sin 2x ] + [ S(pi/6,pi/2) sin 2x - cos x ] would be the area?

Yep. :)
 

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