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Math Help - Area Within a Curve

  1. #1
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    Area Within a Curve

    Hey everybody, newbie here, looking for some quick help. I promised a good friend of mine help with her calculus homework, but I've gotten rusty, and only have a fleeting glimpse of what I should be doing. PLEASE HELP
    Here's the functions, of which I've got to find the area within the curve.

    y=x^3 - 5x + 2
    y=Sqrt(2-x)
    y=-Sqrt(2-x)

    The graph kinda looks like a horizontal parabola & a vertical snake...

    The relevant points of intersection I've found are:
    (.12666695, 2.0567), (2,0), & (.69576957, -1.1420)

    Of course the x-values are the only ones needed.
    Thus
    the upper limits for Int1 are .12666695, and .69576957

    The upper limts for Int2 are .69576957, and 2



    This is where I am stuck.
    Do I plug them into Sqrt(2-x) - ((x^3) - 5x + 2), or have I missed a step?

    It is beyond late right now, so if I could get some assistance while I sleep, I'll be eternally grateful.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jughead5267 View Post
    Hey everybody, newbie here, looking for some quick help. I promised a good friend of mine help with her calculus homework, but I've gotten rusty, and only have a fleeting glimpse of what I should be doing. PLEASE HELP
    Here's the functions, of which I've got to find the area within the curve.

    y=x^3 - 5x + 2
    y=Sqrt(2-x)
    y=-Sqrt(2-x)

    The graph kinda looks like a horizontal parabola & a vertical snake...

    The relevant points of intersection I've found are:
    (.12666695, 2.0567), (2,0), & (.69576957, -1.1420)

    Of course the x-values are the only ones needed.
    Thus
    the upper limits for Int1 are .12666695, and .69576957

    The upper limts for Int2 are .69576957, and 2



    This is where I am stuck.
    Do I plug them into Sqrt(2-x) - ((x^3) - 5x + 2), or have I missed a step?

    It is beyond late right now, so if I could get some assistance while I sleep, I'll be eternally grateful.
    This is an irritating problem. Who came up with this anyway?

    Notice that there are actually 3 integrals you need to do:
    Integrate from x = 0.126666947 to x = 0.695769572
    Integrate from x = 0.695769572 to x = 1.978833423
    Integrate from x = 1.978833423 to x = 2

    (See the two graphs below. The second is a blown up view of the region around x = 2.)

    When you find the area you take the difference between the two functions that are creating the area. So
    A = \int_{0.127}^{0.696}( (\sqrt{2 - x}) - (x^3 - 5x + 2) )~dx + \int_{0.696}^{1.979}( (\sqrt{2 - x}) - (-\sqrt{2 - x}) )~dx   + \int_{1.979}^2 ( (x^3 - 5x + 2) - (-\sqrt{2 - x}) )~dx

    -Dan
    Attached Thumbnails Attached Thumbnails Area Within a Curve-graph1.jpg   Area Within a Curve-graph2.jpg  
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