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Area for cosx on interval [0,2pi]

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data
    f(x)=cosx and the x-axis on the interval [0,2pi]

    A) Set up definite integral that represents area above
    B) Find area using the fundamental theorem

    2. Relevant equations



    3. The attempt at a solution
    cosxdx [0,2pi]
    = sinx [0,2pi]
    = sin(2pi)-sin(0)
    = 0

    Area= (cosxdx [0,pi/2]) - (cosxdx [pi/2,3pi/2]) + (cosxdx [3pi/2,2pi])

    Area= (sinx [0,pi/2]) - (sinx [pi/2,3pi/2]) + (sinx [3pi/2,2pi])

    Area= (1-0) - (-1-1) + (0-1)
    Area= (1) -(-2) + (1)
    Area= 4 square units.

    So.... What do you think? Is it right? And what exactly does the question mean by "Set up definite integral that represents area above"?
     
  2. jcsd
  3. Apr 22, 2010 #2
    When it says set up the definite integral it means to do this.

    [tex]\int^{2\pi}_{0}cosx dx[/tex]

    now you solve it.
     
  4. Apr 22, 2010 #3
    I did that, but I didn't know how to type it into the forums. So I made it the cosxdx [0,2pi] and solved from there. What do you think of my answer for the area?
     
  5. Apr 22, 2010 #4
    You have the integral correct, check this again though
    [tex] sin(2\pi)-sin(0)[/tex]
     
  6. Apr 22, 2010 #5

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    As you have noted, the area in question is not given by

    [tex]\int_0^{2\pi} \cos x\ dx[/tex]

    because that gives zero. That formula only works for non-negative integrands. To express it as a single integral you might write:

    [tex]\int_0^{2\pi} |\cos x|\ dx[/tex]

    which you have properly evaluated by breaking it up to subtract the negative parts.
     
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