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Area between two graphs

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the area of the region enclosed between y=4sin(x) and y=2cos(x) from x=0 to x=0.8pi.


    2. Relevant equations
    [itex]\int^{0.8\pi}_0 dx [/itex]

    [itex]
    g(x) = 4\sin(x)
    [/itex]

    [itex]
    f(x) = 2\cos(x)
    [/itex]


    3. The attempt at a solution
    This problem needs to be split up into two parts.

    [itex]
    \int_0 ^n [f(x) - g(x)] dx + \int_n^{0.8\pi} [g(x) - f(x)] dx
    [/itex]

    My major problem is finding n.

    I set:

    [itex]
    f(n) = g(n) \rightarrow 4\sin(n) = 2 \cos(n) \rightarrow 2\sin(n) = cos(n) \rightarrow 2 = \frac{cos(n)}{sin(n)} \rightarrow 2 = \cot(n)
    [/itex]

    I'm having trouble finding that point n. I've worked out that it's near

    [itex]
    \frac{15\pi}{96} = n
    [/itex]


    and with that n I have:
    [itex]
    \int_0^n [2 \cos(x) - 4 \sin(x)] dx + \int_n^{0.8\pi} [2 \cos(x) - 4\sin(x)] dx
    [/itex]

    [itex]
    -2 \sin(x) - (-4) \cos(x)]_0^\frac{15\pi}{96} + [-2 \sin(x) - (-4) \cos(x)]_\frac{15\pi}{96}^{0.8\pi}
    [/itex]
     
    Last edited: Sep 18, 2011
  2. jcsd
  3. Aug 11, 2013 #2
    Looks like you did a very good job. You got cot(n) = 2. If you take the reciprocal of both sides, you will see that it is the same as tan(n) = 1/2. If you are allowed to solve the problem numerically, then you can find arctan(0.5) on your calculator and evaluate the two integrals using the decimal value. If you need an exact answer, then you can try substituting arctan(1/2) as it is and evaluate the sin and cos of arctan(1/2) using your knowledge of trig.

    Junaid Mansuri
     
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