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Math Help - Calculating Surfaces

  1. #1
    Junior Member shinhidora's Avatar
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    Calculating Surfaces

    I'm back and stuck once again...

    Calculate the surface between y = (1/2)x^2+1 and y = -x+4 between [0,2.5]

    It's supposed to be 2.059

    Though I don't get there...

    I did, using my TI-84 Plus Calculator:

    y1 = (1/2)x^2 + 1

    2nd + Trace --> 7) f(x)dx
    Lower Limit: 0
    Upper Limit: 2.5

    Gives 5.1041667

    y2 = -x+4

    2nd + Trace --> 7) f(x)dx
    Lower Limit: 0
    Upper Limit: 2.5

    Gives 6.875

    6.875 - 5.104... = 1.4408333

    Anyone got an idea?
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by shinhidora View Post
    I'm back and stuck once again...

    Calculate the surface between y = (1/2)x^2+1 and y = -x+4 between [0,2.5]

    It's supposed to be 2.059

    Though I don't get there...

    I did, using my TI-84 Plus Calculator:

    y1 = (1/2)x^2 + 1

    2nd + Trace --> 7) f(x)dx
    Lower Limit: 0
    Upper Limit: 2.5

    Gives 5.1041667

    y2 = -x+4

    2nd + Trace --> 7) f(x)dx
    Lower Limit: 0
    Upper Limit: 2.5

    Gives 6.875

    6.875 - 5.104... = 1.4408333

    Anyone got an idea?
    Draw a diagram. Then you'll see that the area is:

    \int_{0}^{-1 + \sqrt{7}} (-x + 4) - \left(\frac{x^2}{2} + 1\right) \, dx + \int_{-1 + \sqrt{7}}^{5/2} \left(\frac{x^2}{2} + 1\right) - (-x + 4) \, dx.

    Obviously you can simplify this a little bit ......

    Note: -1 + \sqrt{7} is the x-coordinate of the intersection point of the line with the parabola that lies between x = 0 and x = 2.5.

    As good practice in basic algebraic skills you should get the exact value by hand before using the calculator.
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  3. #3
    Junior Member shinhidora's Avatar
    Joined
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    How can I forget about the intersection.... all this math makes me dumb
    Gonna work it out now!

    Edit: Got it Thank you very much!
    Last edited by shinhidora; June 8th 2008 at 05:32 AM.
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