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Math Help - area between curves.

  1. #1
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    area between curves.

    Can someone please help with this question,



    I have attached the sketch of the curves,

    I am not sure how to integrate, because one part is below the x-axis and the other is above.

    Appreciate if someone can show me how to solve these problems.

    Thanks in advance.
    Attached Thumbnails Attached Thumbnails area between curves.-graph.jpg  
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  2. #2
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    If you subtract x^2-4x from 4x-x^2,

    then you will always have the difference between the curves.
    Integrating this difference gives the area between the points of intersection.

    Or, since the region is symmetrical about the x-axis,
    integrate the top curve and multiply the answer by 2,
    or integrate the region under the x-axis and multiply by -2.
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  3. #3
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    Remember....

    integration sums f(x),
    so if f(x) is negative, integration sums negative values.
    If you want area, you just change the sign,
    though you must be aware of where the graph crosses the x-axis
    to isolate where f(x) is positive or negative.
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  4. #4
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    Quote Originally Posted by Archie Meade View Post
    If you subtract x^2-4x from 4x-x^2,

    then you will always have the difference between the curves.
    Integrating this difference gives the area between the points of intersection.

    Or, since the region is symmetrical about the x-axis,
    integrate the top curve and multiply the answer by 2,
    or integrate the region under the x-axis and multiply by -2.
    Thank you, so should I integrate each area from 4 to 0 separately and than subtract?
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  5. #5
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    No, Tweety,

    When you are dealing with the area between curves as in your case,
    with one below the other, you can subtract the functions first,
    then integrate.
    Subtract the lower curve from the upper curve.

    4x-x^2-(x^2-4x)=8x-2x^2.

    now you can integrate the difference between them
    and that gives the area.

    Notice that 4x-x^2 is x^2-4x upside down.

    Hence the area above the x axis between the points of intersection,
    is always positive there.

    Hence integrating that will give a positive answer.
    Then double your answer since that will give you the area.

    Or, integrate the lower curve, but since that one is negative, your answer
    will be negative.
    So just change the sign and multiply by 2.

    If you do all 3 ways, it would be excellent practice.
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    No, Tweety,

    When you are dealing with the area between curves as in your case,
    with one below the other, you can subtract the functions first,
    then integrate.
    Subtract the lower curve from the upper curve.

    4x-x^2-(x^2-4x)=8x-2x^2.

    now you can integrate the difference between them
    and that gives the area.

    Notice that 4x-x^2 is x^2-4x upside down.

    Hence the area above the x axis between the points of intersection,
    is always positive there.

    Hence integrating that will give a positive answer.
    Then double your answer since that will give you the area.

    Or, integrate the lower curve, but since that one is negative, your answer
    will be negative.
    So just change the sign and multiply by 2.

    If you do all 3 ways, it would be excellent practice.
    Okay I understand now thanks, I will give all three methods ago.
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  7. #7
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    if you subtract the lower function from the upper one,
    the difference between them is always positive.
    Therefore, integrating the difference gives the area,
    you don't have to worry about getting a negative answer.
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