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    The diagram shows the finite region, R, bounded by the curve with equation y=x4+x), the line with equation y=12 and the y-axis. A) find coordinate A B) Find area R I have the coordinate as (2,12). I am having trouble with how to find the area. Any ideas?
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    (Original post by SophieL1996)
    The diagram shows the finite region, R, bounded by the curve with equation y=x4+x), the line with equation y=12 and the y-axis. A) find coordinate A B) Find area R I have the coordinate as (2,12). I am having trouble with how to find the area. Any ideas?
    Have you tried drawing a sketch?
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    (Original post by SophieL1996)
    The diagram shows the finite region, R, bounded by the curve with equation y=x4+x), the line with equation y=12 and the y-axis. A) find coordinate A B) Find area R I have the coordinate as (2,12). I am having trouble with how to find the area. Any ideas?
    What is the equation?
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    (Original post by Mr M)
    What is the equation?
    The equation for the curve is x(4+x) and for the line y=12
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    As it's quadratic there will be two answers, of which -6 is the other. You may find it easier with two coordinates. If I remember correctly, I think then you integrate and sub the x values in and subtract them from each other. Hope this helps, sorry my memory isn't too good but c2 was last year for me!


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    (Original post by SophieL1996)
    The equation for the curve is x(4+x) and for the line y=12
    Ok that isn't what you put in your opening post!

    Either change the subject and integrate \int x \, dy or find \int y \, dx and subtract your answer from the area of a rectangle.
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    Sorry that is no help to me... The parabola is positive with the horizontal line through it, the area is between A (point of intersection) the y axis and (0,0) What do I take away from what ?
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    (Original post by SophieL1996)
    Sorry that is no help to me... The parabola is positive with the horizontal line through it, the area is between A (point of intersection) the y axis and (0,0) What do I take away from what ?
    Who are you talking to?

    Do you want to integrate with respect to x or to y. Either will work so you choose?

    Edit: As this is C2 I would strongly recommend x.
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    (Original post by SophieL1996)
    Sorry that is no help to me... The parabola is positive with the horizontal line through it, the area is between A (point of intersection) the y axis and (0,0) What do I take away from what ?
    You can split it up into three rectangles, can you work it out from that?
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    (Original post by Jackabc)
    You can split it up into three rectangles, can you work it out from that?
    No, as I have never been taught that
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    (Original post by SophieL1996)
    The diagram shows the finite region, R, bounded by the curve with equation y=x4+x), the line with equation y=12 and the y-axis. A) find coordinate A B) Find area R I have the coordinate as (2,12). I am having trouble with how to find the area. Any ideas?
    could you attach the diagram, it will make it easier for me to help
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    (Original post by jadecross)
    could you attach the diagram, it will make this easier to help
    Yes I have been trying to but it won't upload
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    (Original post by SophieL1996)
    No, as I have never been taught that
    Me neither.

    :confused:
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    so intergrate between 0 and 2 for equation of curve and minus that from (2x12) ??
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    The shaded area in this diagram is \int_0^2 x(4+x) \, dx

    Can you see a rectangle that you could subtract this area from to find the area you are interested in?

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    (Original post by SophieL1996)
    so intergrate between 0 and 2 for equation of curve and minus that from (2x12) ??
    yes, it's as simple as that.
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    (Original post by jadecross)
    yes, it's as simple as that.
    ok thanks I shall have a go
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    Thanks I got the right answer of 40/3 thanks for all your help, especially the diagram (helped a lot!)
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    (Original post by Mr M)
    Me neither.

    :confused:
    (Original post by SophieL1996)
    No, as I have never been taught that
    Isn't it bounded by all these graphs twice? So if you worked out the area between -4 and 0, added it to the area under -6 and -4 and then added it to the area between 0 and 2. Or was you just on about one side?

    Edit: lol you could still split it up into three rectangles though if you added the areas between 0 and 1, 1 and 1.5, and 1.5 and 2 but that might take a bit longer.
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    (Original post by Jackabc)
    Isn't it bounded by all these graphs twice? So if you worked out the area between -4 and 0, added it to the area under -6 and -4 and then added it to the area between 0 and 2. Or was you just on about one side?

    Edit: lol you could still split it up into three rectangles though if you added the areas between 0 and 1, 1 and 1.5, and 1.5 and 2 but that might take a bit longer.
    You are making this WAY too difficult.
 
 
 
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