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Math Help - integrating sine functions

  1. #1
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    integrating sine functions

    Posted, it here as the homework has no help was given in the other thread, and it has become a matter which quite urgently must be solved.

    I am to determine the area between these two curves. Although I am familiar to the process I fail to recognize and grasp what one should do to these specific types of functions. If possible could someone assist me and post a solution along with work so I can process it.

    To the question.

    We are to find the area between these two curves
    y = 8.5sin(0.32x-0.12)+0.56
    y = 4.7sin(0.42x+0.5)-4.3

    From x=-10 to x=10

    Thanks in advance
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  2. #2
    MHF Contributor
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    A diagram would be very helpful, but I don't know how to graph in graphing calculators or computers so I cannot see the actual, precise figure. But I can "see" that the
    dA = (difference in y's)*dx

    As for the areas that are "negatives", if there are, I cannot see those without the figure or a sketch. Although I can do the graphs roughly by sketching them on paper here, I will not do that because the idea is to show you only the general way to do your question here.

    dA is vertical. height is (y2 -y1). width is dx.

    dA = [(8.5sin(0.32x -0.12) +0.56) -(4.7sin(0.42x +0.5) -4.3)]*dx
    dA = [8.5sin(0.32x -0.12) -4.7sin(0.42x +0.5) +4.86)]*dx

    A = INT.(-10 to 10)[8.5sin(0.32x -0.12) -4.7sin(0.42x +0.5) +4.86)]*dx

    A = [(8.5/0.32){-cos(0.32x -0.12)} -(4.7/0.42){-cos(0.42x +0.5)} +(4.86)x]|(-10 to 10)

    A = [-26.5625cos(0.32(10) -0.12) +11.1905cos(0.42(10) +0.5) +4.86(10)]
    -[-26.5625cos(0.32(-10) -0.12) +11.1905cos(0.42(-10) +0.5) +4.86(-10)]

    A = 74.9735 -(-31.9498) = 106.9233 sq.units -------answer.
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  3. #3
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    I thank you for your response, however the functions do in fact cross the x axis. Is there any specific method to find that area too? Or should one derive a function by means of stationary points etc.?
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  4. #4
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    Quote Originally Posted by Mouseman View Post
    I thank you for your response, however the functions do in fact cross the x axis. Is there any specific method to find that area too? Or should one derive a function by means of stationary points etc.?
    Deriving a single function from the two given functions will give another sine curve only. You then cannot get the area between the two given functions.

    By "negative" areas, I meant the areas if the upper sine curve goes below the lower sine curve.
    Are there intervals, in the given domain, where the [y2 = 8.5sin(.....] curve goes below the [y1 = 4.7sin(.....] curve? Because in those intervals, (y2 -y1) will be negative....so the dA will be "negative".

    As for the two given curves going down the x-axis, the (y2 -y1) height of the dA will remain positive as long as y2 is above y1 even if both curves or one curve is below the x-axis.
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  5. #5
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    I thank you again to clear up the confusion evoked in my response. There are no such areas, and I feel confident I could apply the method proposed in other exercises.
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