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Area under the curve

  1. May 5, 2009 #1

    a.a

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    1. The problem statement, all variables and given/known data

    Find the area under bounded by y = cos x and y = sin 2x on the intervacl [0, pi/2]



    3. 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?
     

    Attached Files:

  2. jcsd
  3. May 5, 2009 #2
    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.
     
  4. May 5, 2009 #3

    a.a

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    so [ S(0,pi/6) cos x - sin 2x ] + [ S(pi/6,pi/2) sin 2x - cos x ] would be the area?
     
  5. May 5, 2009 #4
    Yep. :)
     
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