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Limits When Determining Area between two Graphs

  1. Jun 13, 2005 #1
    Hi all having a little problem with finding the limits on the area between 2 graphs.

    i can find the easy one such as:

    Find the area between y=x^2 and y = 2x
    which is:
    x^2 = 2x
    x^2 - 2x = 0
    x(x-2) = 0

    x = 0 & 2

    but when i have a question like:
    Find the area between y=2-x^2 & y =x

    i cant work it out i got to x(1+x)= 2

    but im sooo lost
    any help appreciated
  2. jcsd
  3. Jun 13, 2005 #2
    Nvermind, I understand what your saying. To find the points of intersection between those two graphs, set them equal to each other.

    [tex] 2-x^2 = x [/tex]

    [tex] x^2 + x = 2 [/tex]

    An obvious one is x=1.

    Try quadratic formula.
    Last edited: Jun 13, 2005
  4. Jun 13, 2005 #3
    sorry whozum i dont think i explained the question well, i need to work out the points of intersection i have no problems working out the area.

    yeh ive already got 1. so using the quadratic formula i should be able to find the points out?
  5. Jun 13, 2005 #4
    so the intersecting points are -2 & 1?
  6. Jun 13, 2005 #5
    There you go. Graph it to make sure.
  7. Jun 13, 2005 #6
    hello there

    well first of all you need to find where both functions actually intersect this is done by making 2-x^2=x then using the quadratic formulae to find where they intersect, and so you will find that they will intersect at 1 and at -2 now if you want to find the area between these functions its best that you graph it and then split up the area which should correspond to the addition to a couple of integrals
    [tex]\int_0^1 2-x-x^2 dx+\int_{-\sqrt{2}}^0 2-x^2+x dx-\int_{-2}^{-\sqrt 2} x -2+x^2 dx[/tex]
    by integrating you will be able to find the area between those two functions?
    by the way y=2-x^2 has roots at +/-sqrt{2}
    the area is 2.5 units hopefully with out any small errors
    Last edited: Jun 13, 2005
  8. Jun 13, 2005 #7
    thanks guys!
  9. Jun 13, 2005 #8


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    Science Advisor

    Why in the world should one do such a thing? For all x between -2 and 1, 2- x2 is larger than x so 2-x2- x is positive and is the "height" of a thin rectangle between the two. The area is
    [tex]\int_{-2}^1 2- x- x^2 dx= \frac{9}{2}= 4.5[/tex].
  10. Jun 13, 2005 #9
    yeh thats tha answer i got 9/2
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