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Area under curve problem again

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the equation of the curve which passes through the point (-1,0) and whose gradient at
    any point (x,y) is 3x2-6x+4. Find the area enclosed by the curve, the axis of x and the ordinates x=1 and x=2.


    . The attempt at a solution

    I integrated and got the equation of the curve to be x3 -3x2 +4x +8

    Now to find the area I'm not sure what to do, I tried to sketch a graph but I couldn't.

    I tried integrating from x=1 to x=2 and I got 2, which isn't the answer in the book :S.

    Help greatly appreciated.
     
  2. jcsd
  3. Feb 9, 2013 #2
    Which equation did you integrate? You are given the gradient of the curve at any point (x,y). You found the curve. Find the area closed by this curve and the x-axis.
     
  4. Feb 9, 2013 #3
    x^3 - 3x^2 +4x +8
     
  5. Feb 9, 2013 #4
    Show the steps you made when you integrated this from x=1 to x=2 to result in the value 2.
     
    Last edited: Feb 9, 2013
  6. Feb 9, 2013 #5
    Integrate it again from x=1 to x=2. If you integrate 3x^2-6x+4 from x=1 to x=2, then you get the value 2. Knowing this, I asked you which equation did you integrate.
     
  7. Feb 9, 2013 #6
    Sorry I intregrated 3x^2-6x+4 for x=1 to x=2 and got 2
     
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