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

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Homework Help Overview

The problem involves finding the area bounded by the curves y = cos x and y = sin 2x over the interval [0, pi/2]. Participants are discussing the setup of the integrals needed to calculate this area.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set up the integrals based on the points of intersection found at x = pi/6 and x = pi/2. There is confusion regarding the correct formulation of the integrals, particularly whether to use a sum or a difference in the second integral.

Discussion Status

Participants are actively questioning the formulation of the integrals and discussing the implications of area being positive. There is a suggestion that a potential typo may exist in the original setup, and some participants express agreement on the revised integral formulation.

Contextual Notes

There is an emphasis on the definition of area in the context of the problem, with participants noting that area is typically considered positive. The discussion also reflects uncertainty about the specific definitions used in their text regarding signed area.

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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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