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I X^2 graph

  1. Mar 31, 2017 #1
    Hello,

    If I have an x^2 graph that goes from 0 to a point a. Is there a general solution to where the area of the left side is equal to the area of the right?
     
  2. jcsd
  3. Mar 31, 2017 #2

    Math_QED

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    I do not fully understand your question I believe.

    Because ##f(x) = x^2## is symmetric relative to the ##y##-axis, the area between the curve and the ##x##-axis in the interval ##[-a,0]## is equal to the area under the curve and the ##x##-axis in the interval ##[0,a]##

    In integral terms:

    ##\int\limits_0^a x^2 dx = \int\limits_{-a}^{0} x^2 dx ##
     
  4. Mar 31, 2017 #3

    mfb

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    On the left side of what?

    With integrals that should be easy to do.
     
  5. Mar 31, 2017 #4
    I want to cut the graph in two. And yeah, the integral of the left piece should be equal the integral of the right.

    I think I solved it and got 2/3 b^3 = 1/3 c^3.
    This is if I take the integral from 0 to c, then b will be the center. I was wondering if this works as a general solution?
     
  6. Mar 31, 2017 #5

    mfb

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    For some interpretation of the areas you consider, that works as general solution.
     
  7. Mar 31, 2017 #6

    Mark44

    Staff: Mentor

    It would be helpful if you stated the problem more clearly. Here is what I think you were trying to say.

    Find a point in the interval [0, a] that divides the area under the graph of ##y = x^2## and above the x axis into two equal parts. In other words, find c for with ##\int_0^c x^2 dx = \int_c^a x^2 dx##.​
     
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