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Math Help - Integration problem

  1. #1
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    Integration problem

    Question: The area bounded by y = x^{3} - 3x^{2} and the line y=-4 is given by the integral ....


    I deduced this:

    \int_{-1}^2 (x^{3} - 3x^{2} + 4) dx

    So it's the int from -1 to 2.

    But the correct answer from my answer key seems to be:

    \int_{-1}^2 (4 - x^{3} + 3x^{2}) dx

    Does this have something to do with the graph being below the x-axis?
    Last edited by Lord Darkin; January 26th 2010 at 06:04 PM.
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  2. #2
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    It is 4-x^3-3x^2, not x^3-3x^2-4 because y=4 is above the graph of x^3-3x^2. Finding the area of a graph is finding the top part, then subtracting the empty spaces to get your area. In this case y=4 is above the graph of y=x^3-3x^2 from -1 to 2.

    BTW, you can put a negative limit to the integral by using {-1}.
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  3. #3
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    Quote Originally Posted by Lord Darkin View Post
    Question: The area bounded by y = x^{3} - 3x^{2} and the line y=4 is given by the integral ....


    I deduced this:

    \int_1^2 (x^{3} - 3x^{2} + 4) dx

    NOTE: That 1 at the integral should be NEGATIVE 1, the latex wouldn't work when I put in -1.

    So it's the int from -1 to 2.

    But the correct answer from my answer key seems to be (again, int from NEGATIVE 1 to 2, NOT 1 to 2):

    \int_1^2 (4 - x^{3} + 3x^{2}) dx

    Does this have something to do with the graph being below the x-axis?

    I think you mean the area between 3x^2-x^3 and y=4,

    as the other one has no bounded region.

    The curve is not below the x-axis from x=-1 to x=2.
    You subtract the curve equation from 4 as the curve lies below the line between the points of intersection.
    In the bounded region, f(x) is always below y=4.

    Integrating the difference between them calculates the area.

    The integral is \int{5-(3x^2-x^3)}dx
    Last edited by Archie Meade; January 26th 2010 at 06:35 PM.
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  4. #4
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    Quote Originally Posted by Lord Darkin View Post
    Question: The area bounded by y = x^{3} - 3x^{2} and the line y=4 is given by the integral ....
    what region? did you leave out any additional problem information?
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  5. #5
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    Oops, sorry

    it was y= NEGATIVE 4

    That's why ...

    Sorry again, I hate it when I do these typos.
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  6. #6
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    No worries,
    it's all clear now.
    From skeeter's diagram, the curve is now above the line,
    hence subtract -4 from the curve equation (ie add 4)
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  7. #7
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    ^I see, so why does the right integral turn otu tohave negative x^3 rather than a positive x^3?
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  8. #8
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    My earlier integral is incorrect because the line is y=-4, rather than y=4.

    If the line was y=4, the curve would need to be 3x^2-x^3

    As the line is y=-4, it's fine to use the equation as you have it... x^3-3x^2

    Your integral now is \int{x^3-3x^2-(-4)}dx

    from x=-1 to x=2

    Are you sure it's y=-4 and not y=4?
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  9. #9
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    Yes, pretty certain.

    This means the answer I had in my original post matches yours. My answer key says that it's the second thing I posted, the one where the y=x^3 - 3x^2 function is multiplied by -1. I assumed that had something to do with it being below the x-axis.
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  10. #10
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    Yes,

    however, the area will give a positive answer if you subtract the line from the curve, since the curve is above the line.

    Can you check the limits of integration for your answer key.
    I would say what you had was correct.
    the answer key would be correct if the bounds x=-1 and x=2 were reversed above and below the integral sign.
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  11. #11
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    Nope, can't find anything like that. I guess that's a mistake.
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  12. #12
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    The answer key gives the area between 3x^2-x^3 and y=4.
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  13. #13
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    Ye sit does. This was a multiple choice question and D was the second thing I posted in my original post. Choice A (what I deduced originally and is first in my post) doesn't seem to be right.
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  14. #14
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    Quote Originally Posted by Lord Darkin View Post
    Question: The area bounded by y = x^{3} - 3x^{2} and the line y=-4 is given by the integral ....


    I deduced this:

    \int_{-1}^2 (x^{3} - 3x^{2} + 4) dx correct for equations given

    So it's the int from -1 to 2.

    But the correct answer from my answer key seems to be:

    \int_{-1}^2 (4 - x^{3} + 3x^{2}) dx correct for y=4 and \color{red}f(x)=3x^2-x^3

    Does this have something to do with the graph being below the x-axis?
    The thing with the graph being below the x-axis is...

    If you wanted to find the area between a curve and the x-axis and the curve was below the x-axis, then integrating would give a negative answer as integration sums f(x) which would all be negative.
    hence f(x) can be negated before integrating or the result can be negated after integration.

    When you are finding the area between two curves, or between a curve and a line, if you subtract the lower function from the upper one, the answer after integrating is always positive and gives you the area since the subtractions always give you a positive value.
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