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Math Help - Check: Area of region of plane between 2 graphs

  1. #1
    Junior Member shepherdm1270's Avatar
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    Check: Area of region of plane between 2 graphs

    y = 9 - x^2 and y = 1 + 2x

    9 - x^2 = 1 + 2x
    x^2 - 2x - 8 = 0
    (x-4)(x+2) = 0
    x = 4 & -2

    Interval from -2 to 4: [(9-x^2) - (1+2x)] dx = -x^3 / 3 - x^2 = F(4) - F(-2)

    F(4) = -(4)^3/3 - (4)^2 = 64/3 - 16 = 16/3

    F(-2) = -(-2)^3 / 3 - (-2)^2 = -8/3 - 12/3 = -20/3


    So F(4) - F(-2) = 16/3 - 20/3 = -4/3


    answer: -4/3


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  2. #2
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    Quote Originally Posted by shepherdm1270 View Post
    y = 9 - x^2 and y = 1 + 2x

    9 - x^2 = 1 + 2x
    x^2 - 2x - 8 = 0 Mr F says: It should be x^2 + 2x - 8 = 0. Here's a tip: Always substitute your answers back into the equations as a check. Ideally you will also draw the curves and shade the required region. I haven't checked past your solutions for the intersection points.

    (x-4)(x+2) = 0
    x = 4 & -2

    Interval from -2 to 4: [(9-x^2) - (1+2x)] dx = -x^3 / 3 - x^2 = F(4) - F(-2)

    F(4) = -(4)^3/3 - (4)^2 = 64/3 - 16 = 16/3

    F(-2) = -(-2)^3 / 3 - (-2)^2 = -8/3 - 12/3 = -20/3


    So F(4) - F(-2) = 16/3 - 20/3 = -4/3


    answer: -4/3


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  3. #3
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    Firstly, you know you made a mistake when you get a negative area.

    You have to know the limits therefore you must calculate the intersections betwen the two curves.

    9-x^2=1+2x

    From this quatdration equation you can deduce using the Wiet's rule for quadratic formulae that intersections are x= -4 and x=2 which are now the limits.

    Then you integrate each of the equations with respect to x from -4 to 2.

    When you get the two results for integrals you must (!) take the absolute value for the difference between them. Then you will obtain positive value and true anwser.
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  4. #4
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    Firstly, you know you made a mistake when you get a negative area.

    You have to know the limits therefore you must calculate the intersections betwen the two curves.

    9-x^2=1+2x

    From this quatdration equation you can deduce using the Wiet's rule for quadratic formulae that intersections are x= -4 and x=2 which are now the limits.

    Then you integrate each of the equations with respect to x from -4 to 2.

    When you get the two results for integrals you must (!) take the absolute value for the difference between them. Then you will obtain positive value and true anwser.
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  5. #5
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    Quote Originally Posted by cigster View Post
    Firstly, you know you made a mistake when you get a negative area.
    Hmm, what about \displaystyle<br />
\int_{-\frac{\pi}{2}}^{0} \sin x \, dx = \displaystyle - \cos x |_{-\frac{\pi}{2}}^{0} = -1 + 0 = -1?
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  6. #6
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    Quote Originally Posted by Isomorphism View Post
    Hmm, what about \displaystyle<br />
\int_{-\frac{\pi}{2}}^{0} \sin x \, dx = \displaystyle - \cos x |_{-\frac{\pi}{2}}^{0} = -1 + 0 = -1?
    The area between two curves is always positive. It's found by finding int ((upper curve) - (lower curve).

    The example you've given is a definite integral - it does not give the area between two curves.

    The area between the sine curve and the x-axis from x = -pi/2 to x = 0 is given by int (0) - (sin x) ......
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