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Area under a curve question

  1. Feb 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the area bounded by the curve y= x2-x-2 and the x-axis
    from x=-2 and x=3.






    The attempt at a solution

    I integrated from x=-2 to x=3 using (x^3/3)-(x^2/2)-2x and I got -4 5/6
    but the answer is -4 1/2 .

    I don't really see where I went wrong.
     
  2. jcsd
  3. Feb 6, 2013 #2

    haruspex

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    I agree with your answer.
     
  4. Feb 6, 2013 #3

    SteamKing

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    Good News! You're both wrong.

    Did you try checking your math again?
     
  5. Feb 6, 2013 #4
    I still get - 4 5/6 :S
     
  6. Feb 6, 2013 #5

    SteamKing

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    What I originally meant was the book answer is also wrong.

    Remember, the integral is F(3) - F(-2)
     
  7. Feb 6, 2013 #6
    When I do that I get -4 5/6!!
    ((3)^3/3)- (9/2)- 2(3) - [(-2)^3/3) - (2) - 2(-2)]
     
  8. Feb 6, 2013 #7
    You guys are all wrong. Look carefully that he said it's also bounded by the x-axis. You guys are counting the NEGATIVE area under the x-axis which you are not supposed to. The area is

    ##\displaystyle\int\limits_{-2}^{-1} x^2 - x - 2\ dx + \int\limits_{1}^{3} x^2 - x - 2\ dx = 5/2##
     
  9. Feb 6, 2013 #8
    5/2? really? The answer in the book is that far off?
     
  10. Feb 6, 2013 #9
    Sorry, a mistake in the limits. It's actually:

    ##\displaystyle\int\limits_{-2}^{-1} x^2 - x - 2\ dx + \int\limits_{2}^{3} x^2 - x - 2\ dx = 11/3##

    See here for a better understanding: http://i.imgur.com/U8iWBwJ.png
     
  11. Feb 6, 2013 #10
    I don't. How is the answer not -5/6 ?
     
  12. Feb 6, 2013 #11
    but shouldn't the 2nd integral be from -2 to 3?
     
  13. Feb 6, 2013 #12
    Why is that? You don't want to count the negative area, do you?
     
  14. Feb 6, 2013 #13
    Oh I see what you did.. so the answer in my book is 100% wrong. so it's 11/3? oh and may I ask what program you used to produce that sketch?
     
  15. Feb 6, 2013 #14
    Actually, thinking about it some more, I realized that you would count the negative area. It says bounded by the curve and x axis and not above the x axis. So yes, the area would be -5/6.

    That particular diagram was from Mathematica 8, but I'm sure you can make one using WolframAlpha.
     
  16. Feb 6, 2013 #15

    haruspex

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    It's one of those awkward 'areas':smile:. I was always taught to treat area under the x-axis as negative. This gives the right result in most practical circumstances, e.g. distance advanced when y=velocity, x=time.
    If you want to count all areas as positive then it becomes ##\int_{-2}^3|x^2-x-2|dx##. I see no basis for simply ignoring the area below the axis. May well have gone wrong in the arithmetic somewhere, but I make that 49/6.
     
  17. Feb 6, 2013 #16
    :S so confused like 3 different answers.
     
  18. Feb 6, 2013 #17

    haruspex

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    Yes, -5/6, thanks.
     
  19. Feb 7, 2013 #18
    I'm confused I don't see how you guys got -5/6 =/
     
  20. Feb 7, 2013 #19
    It's about interpreting the question.

    The area bounded by the curve and the x-axis, if you count below as negative, is -5/6 by a simple integral from -2 to 3.

    If you count the area below the x-axis as positive too, then the answer is 49/6, by integrating between -2 to -1, -1 to 2, and 2 to 3, summing the absolute values of each.

    If you discount any area below the x-axis altogether (bizarre if you ask me), then you just integrate between -2 to -1 and 2 to 3, giving 11/3.
     
  21. Feb 7, 2013 #20

    HallsofIvy

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    That's an impossible answer. Area is always positive.
    From x= -2 to x= -1, the graph is above the x-axis. The integral is positive so that area is just the integral from -2 to -1. From x= -1 to x= 2, the graph is below the x-axis. The integral is negative so that area is the negative of the integral from -1 to 2. Finally, from x= 2 to x= 3, the graph is above the x-axis. The integral is positive so the area is equal to the integral from 2 to 3.

    The area bounded by the x-axis, the graph of [itex]y= x^2- x- 2[/itex], and the vertical lines x= -2 and x= 3, is given by
    [tex]\int_{-2}^{-1} x^2- x- 2 dx- \int_{-1}^{2} x^2- x- 2 dx+ \int_{2}^{3} x^2- x- 2 dx[/tex]
     

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