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Area problem

  1. Oct 9, 2006 #1
    Consider the region bounded by y = x^2, y = 5x+6, and the negative x-axis

    Compute the area of this region.

    Im somewhat confused by what they mean by the negative x-axis?

    The points of intersection between the two functions are [-1,1] and [6,36]

    A = Integral -1 to 6 (5x-6-x^2)?

    I'd appreciate if someone evaluated this problem for me step by step for I have spent several hours on it now without any clue.. Thank You!!
    Last edited: Oct 9, 2006
  2. jcsd
  3. Oct 9, 2006 #2


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    It should be the parabola is on the top right of the region, the line is on the top left of the region, and the x-axis is on the bottom of the region
  4. Oct 9, 2006 #3


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    Draw the graphs! The "negative x-axis" is the part of the x-axis left of (0,0), that corresponds to negative values of x. The integral you give is incorrect. It (with 5x+ 6, not 6x- 6) gives the area of the region between y= x2 and y= 5x- 6. The area you want has upper boundry the straight line y= 5x- 6, from the point where it crosses the x-axis to (-1, 1) where it meets the parabola y= x2, then that parabola to (0,0). The lower boundary is the x-axis, y= 0.

    You will have to do it as to separate integrals.
    Last edited by a moderator: Oct 9, 2006
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