1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Generalisations of area between two curves

  1. May 28, 2015 #1
    1. The problem statement, all variables and given/known data
    The problem consists of investigating the area between two functions of the forms (Parabolic segment):
    : y = mx + c and y = ax^2 + bx + c

    The investigation involves finding a combination that has one of each of the above functions and finding an area of one. The area between the functions has to be in the first quarter (positive x and y).

    2. The attempt at a solution
    My investigation up to now has found the following:
    Function 1: 6x^2 - 2x + 8
    Function 2: 4x + 8

    These give points of intersection of: (0,8) and (1,12). These are the lower and upper bounds when integrating.

    Integration produces:

    Curve: 3x^2 - x^2 + 8x [Boundaries of 0 and 1]: Area under curve = 9 (between 0 and 1).

    Line: 2x^2 + 8x [Boundaries of 0 and 1]; Area under line = 10 (Between 0 and 1).

    => Area between line and curve is: 10 - 9 = 1

    3. Further investigations
    I have already found that the coefficient of A (Parabola) has to be a multiple of 3, otherwise there will be a recurring decimal when integrated.

    My next steps involve:
    : Producing any generalisations that affect finding this specific area with these two functions.
    : Proving/generalising using Algebra.
  2. jcsd
  3. May 28, 2015 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There are a lot of things you haven't specified about this problem. May the parabola open either upwards or downwards? Must the intersections of the parabola and line be in the first quadrant? Is the area you want to calculate the area bounded by the two functions between the intersection points? Must the complete region "between" the functions lie in the first quadrant?

    Also please show us how you got A must be a multiple of 3.

    [Edit, added]: I would suggest starting with a simple subcase. For example, consider the two functions ##(x-a)^2## and ##mx##. See if you can show that for any ##a\ge 0## there is a corresponding ##m## that works. Get the relationship between ##m## and ##a##. Then start on more complicated cases.
    Last edited: May 28, 2015
  4. May 28, 2015 #3
    The Parabola can open either upwards or downwards. The intersections of the parabola and the line must be in the first quarter. The area is bounded by the functions between the points and the complete region between the functions must lie in the first quadrant.

    I got that A must be a multiple of 3 because when the quadratic is integrated, the A value is divided by 3. If the value of A isn't a multiple of 3 then the result will always give a recurring decimal and will never produce an area of one.

    (x-a)^2 = x^2 − 2xa + a^2 is not a parabola though is it?
    Last edited: May 28, 2015
  5. May 28, 2015 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That is a parabolic function. The graph of ##y= (x-a)^2## is a parabola with vertex ##(a,0)## opening upwards. ##A## does not have to equal ##3##. Like I suggested, start with a real simple case. Take ##y=(x-1)^2## and ##y = mx##. It is obvious geometrically that there is some positive ##m## so that the area between the line and parabola is ##1##.
  6. May 29, 2015 #5
    Hello LCKurtz,

    I not hundred percent sure I understand the process here. I have tried different m values when y = (x-1)^2 and it produces an m of approximately 0.704 (3 dp). What is the aim of the investigation?
  7. May 29, 2015 #6
    May you can use the absolute value of $$ \int_0^1f_1(x)dx-\int_0^1f_2(x) $$
  8. May 29, 2015 #7
    The first function being the line and the second function being the parabola?
  9. May 29, 2015 #8
    If you need just area, take absolute value. If you need sign then check for this on initial problem.
  10. May 29, 2015 #9
    Just to check I understand what you are saying:
    I only need the area: specifically an area of 1 and if there are any generalisations that can be made.

    What is the absolute value?
  11. May 29, 2015 #10
    Integral ##\int f\,dx ## comes from area calculation between function graph and x axis.
  12. May 29, 2015 #11
    So, using that information.
    The area of the function that is furthest away from the x axis can't be less than 1?
  13. May 29, 2015 #12
    The integral have sign from f(x). The geometrical area is positive defined, like as geometrical distance. For area take the positive answer.
  14. May 29, 2015 #13
    This is because the area between the curves is above the x axis and the fact there is no such thing as a negative geometrical distance?
  15. May 29, 2015 #14
    Area is positive, integral may not. Distance is positive, position coordinate value may not.
  16. May 29, 2015 #15
    I can't have negative integrals though? because the area is bounded in the first quadrant?
  17. May 29, 2015 #16
    Your parabola is positive defined so every ##\int_a^bf_1(x)\,dx## must be positive. The other function may give negative integral on some limits.
  18. May 29, 2015 #17
    Why must the parabola be positive? is that because we are subtracting the other function away from it, thus a negative subtracting a negative will always be a negative. Therefore will never give an area of 1?

    Does the parabola have to be the first function? or can the line be function 1?
  19. May 29, 2015 #18
    Make a simple google graph to see the shapes.
  20. May 29, 2015 #19
    Both of these sets of functions are allowed in the constraints of this question.

    Attached Files:

  21. May 29, 2015 #20
    Function 1: 6x^2 - 2x + 8
    Function 2: 4x + 8
    This problem represented by the second graph.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted