1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Mentallic

    User Avatar
    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

    HallsofIvy

    User Avatar
    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook