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Homework Help: Definite integration (area between curves)

  1. Sep 4, 2010 #1
    1. The problem statement, all variables and given/known data

    Calculate the area between the curves y = sin 2x and y = cos x from x = to x = 90 degrees

    2. Relevant equations

    Using the difference function: ∫(upper = 90) (lower = 0) [sin2x - cos x]

    3. The attempt at a solution

    (sin 2x) - (cos x)

    = ∫ sin 2x - cos x
    = ∫-0.5 cos 2x - sin x
    = [(-0.5 cos 2 x 90) - (sin 90)] - [(-0.5 cos 2 x 0) - (sin 0)
    = (0.5) - (-0.5)
    = 1

    The answer is 1/2
  2. jcsd
  3. Sep 4, 2010 #2


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    Homework Helper

    I was sitting there scratching my head for a bit between these two lines, trying to figure out what you had done:

    You already took the integral on the second line so don't put in the integral sign again.

    -0.5cos(180o) = 0.5
    sin(90o) = 1

    By the way, I'm getting an answer of 0.
    Last edited: Sep 4, 2010
  4. Sep 5, 2010 #3


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    Science Advisor

    By the way, you should understand that the calculus properties of sine and cosine, specifically that [itex]d sin(x)/dx= cos(x)[/itex], [itex]d cos(x)/dx= -sin(x)[/itex],[itex]\int sin(x) dx= -cos(x)+ C[/itex], and [itex]\int cos(x)= sin(x)+ C[/itex] are true only if x is in radians.

    Here, because you are taking sine and cosine of the limits of integration, it doesn't matter whether they are in degrees or radians but if you had something like, say,
    [tex]\int x sin(x) dx[/itex]
    from 0 to 90 degrees, where we have an "x" outside the trig function, you would have to convert to radians to get the correct answer.
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