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An areas question?

  1. Mar 23, 2007 #1
    i added 2 files with the question and the way i tried to solve it
    it messes up and nothing come out

    if my handwriting is problematic to you
    the question is:

    parabula y=x^2 +b*x+c cuts the X axes in two points
    one of them is (1,0)
    the area between the parabula the X and Y axes equals to the area between the parabula and theX axes

    i showed it all in the file it includes a graph

    plz help

    Attached Files:

    • 1.GIF
      File size:
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    • 2.jpg
      File size:
      30.6 KB
  2. jcsd
  3. Mar 23, 2007 #2


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    Staff Emeritus
    Science Advisor

    What, exactly, are you asking?
  4. Mar 23, 2007 #3
    I am asking how can isolve this question
    and find the b, c parameters?
  5. Mar 24, 2007 #4


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    Staff Emeritus
    Science Advisor

    Okay, I didn't see your last question on "page 2".

    I notice that you are integrating from 0 to 1. Why? Clearly the "other" point at which the parabola crosses the x-axis is NOT "0". You are told that one x-intercept is x= 1 so you know that the function factors as (x-1)(x- a) for some number a. a is the other x-intercept and the area is the integral from a to 1 or 1 to a, depending upon which is positive.

    (x-1)(x- a)= x2-(1+a)x+ a= x2+ bx+ c so b= -1-a and c= a.

    The "area between the parabola and the x and y axes" is either
    [tex]/int_0^1 x^2- (1+a)x+ a dx[/tex]
    if 1< a or
    [tex]/int_0^a x^2- (1+a)x+ a dx[/tex]
    if a< 1.

    Similarly, the area between the parabola is either
    [tex]\int_1^a x2-(1+a)x+ a dx[/itex]
    if 1< a or
    [tex]\int_a^1 x2-(1+a)x+ a dx[/itex]
    if a< 1.

    Try both possiblilites and see if you can solve for a. Then of course use b=-1-a and c= a.
    Last edited: Mar 27, 2007
  6. Mar 24, 2007 #5
    you devided the answers into 2 possiblities
    depending if the parbula is positive or negative
    if a>1
    or a<1
    however the direction of the parabule is always towards the positive
    part of the Y axes

    because the coefficient of X^2 is 1
    it cannot flip to the opposite side

  7. Mar 26, 2007 #6
    what is that sigh
    or int??

    regarding the question why i have an integral from 0 to 1
    it is one of the areas

    the second area is between 1 and the second point

    what is that sigh
    or int??
    Last edited: Mar 26, 2007
  8. Mar 26, 2007 #7
    The int and sup are tex code to make it easy to read. maybe this will be clearer now
    Last edited: Mar 26, 2007
  9. Mar 27, 2007 #8
    on the second possibility when i check the area between 1 and a
    this area is under the X axes line
    aren"t we suppose to do [ 0-f(x) ] dx

    if i do mesure the are by this [ 0-f(x) ] dx method
    i get that the area equals to zero

    why it is wronge?
    Last edited: Mar 27, 2007
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