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Math Help - Area of the curve between 2 functions

  1. #1
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    Area of the curve between 2 functions

    I need to find the area of the curve between 2 functions:

    y= x^2 and y= 2x-x^2

    Taking the integral of both is easy:
    \frac{1}{3}x^3 and x^2-\frac{1}{3}x^3

    At this point I know I need to find the intersects of the functions so I can then find the definitive integral.

    I begin the process like so:
    x^2=2x-x^2

    Then:
    2x-x^4

    Then:
    2x(1-x^3)

    At this point I think I am just having a brain spaz as I don't know how to get the 2 points I need so I can plug them into the functions. Am I doing something wrong and in general, do I have the correct idea?
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  2. #2
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    Quote Originally Posted by x0ne View Post
    I need to find the area of the curve between 2 functions:

    y= x^2 and y= 2x-x^2

    Taking the integral of both is easy:
    \frac{1}{3}x^3 and x^2-\frac{1}{3}x^3

    At this point I know I need to find the intersects of the functions so I can then find the definitive integral.

    I begin the process like so:
    x^2=2x-x^2

    Then:
    2x-x^4

    Then:
    2x(1-x^3)

    At this point I think I am just having a brain spaz as I don't know how to get the 2 points I need so I can plug them into the functions. Am I doing something wrong and in general, do I have the correct idea?
    How did you get from x^2=2x-x^2 to 2x-x^4

    I get x^2-x=0 which gives x=0 and x=1 which will be your limits
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    How did you get from x^2=2x-x^2 to 2x-x^4

    I get x^2-x=0 which gives x=0 and x=1 which will be your limits
    I subtracted the x^2 to the other side and that gave me x^4. Could you break out exactly what you did? For some reason I am not seeing it.
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by x0ne View Post
    I subtracted the x^2 to the other side and that gave me x^4. Could you break out exactly what you did? For some reason I am not seeing it.
    I added x^2 to both sides. When adding you change the coefficient rather than the exponent. x^4=(x^2)(x^2) whereas 2x^2 = x^2+x^2
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  5. #5
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    Quote Originally Posted by e^(i*pi) View Post
    I added x^2 to both sides. When adding you change the coefficient rather than the exponent. x^4=(x^2)(x^2) whereas 2x^2 = x^2+x^2
    Got it!

    Now to be sure I am on the right path:

    I thought I would have ended up with some room between the points. I now need to do the definite integral of both the above functions, but I only have points that go from 0 to 1. How would I proceed now?
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  6. #6
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by x0ne View Post
    Got it!

    Now to be sure I am on the right path:

    I thought I would have ended up with some room between the points. I now need to do the definite integral of both the above functions, but I only have points that go from 0 to 1.

    \frac{1}{3}x^3 + x^2-\frac{1}{3}x^3

    How would I proceed now?
    Normally you would do one function subtract the other one:


    \frac{1}{3}x^3 <br />
(\frac x^2-\frac{1}{3}x^3) - (\frac{1}{3}x^3)<br />

     = [1 - \frac{1}{3} - \frac{1}{3}] - 0 = \frac{1}{3}

    Then evaluate for a number.
    Last edited by e^(i*pi); May 10th 2009 at 01:42 PM.
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