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    The curve with the equation  y= 12 + 4x - x^2 cuts the y-axis at A, the positive x-axis at B and the other negative x-axis at C as shown in the diagram. The point O is the origin. The shaded region R is bounded by the line AB and the curve. The point B has the coordinates (6,0), show that the triangle OAB and the region R have equal areas.

    So I got that A = 12 (by making x = 0)
    by factorising the curve:
     x^2 - 4x + 12 = 0

(x-6)(x+2) = 0

x = 6, x = -2  so C is -2

    So the area of triangle OAB should be \frac{1}{2}6x12 which is 36.

    When I integrate the curve:
     \displaystyle\int^6_0 12+4x-x^2 \ dx
     \left[12x+2x^2-\frac{1}{3}x^2] _0^6
    I get 72 so I'm not sure where I've gone wrong, any help please. Thanks!
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    Haven't read it all, but with your integration at the end it should be 1/3x^3 not squared, if that helps?
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    (Original post by Secret.)

    The curve with the equation  y= 12 + 4x - x^2 cuts the y-axis at A, the positive x-axis at B and the other negative x-axis at C as shown in the diagram. The point O is the origin. The shaded region R is bounded by the line AB and the curve. The point B has the coordinates (6,0), show that the triangle OAB and the region R have equal areas.

    So I got that A = 12 (by making x = 0)
    by factorising the curve:
     x^2 - 4x + 12 = 0

(x-6)(x+2) = 0

x = 6, x = -2  so C is -2

    So the area of triangle OAB should be \frac{1}{2}6x12 which is 36.

    When I integrate the curve:
     \displaystyle\int^6_0 12+4x-x^2 \ dx
     \left[12x+2x^2-\frac{1}{3}x^2] _0^6
    I get 72 so I'm not sure where I've gone wrong, any help please. Thanks!
    area of shaded bit = area under curve - area under triangle.

    You have the correct area under the curve, but you've missed something obvious with the triangle

    (Original post by .JJ)
    Haven't read it all, but with your integration at the end it should be 1/3x^3 not squared, if that helps?
    72 is what you get if you treat it as such.
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    (Original post by Secret.)

    I'm not sure where I've gone wrong
    Nowhere

    You just have not answered the question yet
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    (Original post by .JJ)
    Haven't read it all, but with your integration at the end it should be 1/3x^3 not squared, if that helps?
    Yeah it is sorry

    (Original post by TenOfThem)
    Nowhere

    You just have not answered the question yet
    But doesn't that mean the area of R is 72 :O?
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    (Original post by Indeterminate)
    area of shaded bit = area under curve - area under triangle.

    You have the correct area under the curve, but you've missed something obvious with the triangle



    72 is what you get if you treat it as such.
    Any hints please? Can't see what I've missed
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    (Original post by Secret.)
    But doesn't that mean the area of R is 72 :O?
    No ... it means that the area under the curve ... ie between the curve and the x-axis is 72
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    (Original post by TenOfThem)
    No ... it means that the area under the curve ... ie between the curve and the x-axis is 72
    This. Look intently @ R. You need to minus an area of a regular shape from the area under curve.
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    (Original post by TenOfThem)
    No ... it means that the area under the curve ... ie between the curve and the x-axis is 72
    Aah! How could I mix them up! Thank you!
 
 
 
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