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Calculus

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Find the area of the region enclosed by
y−2x^2 ≥ 0 and |x|+|y| ≤ 1

  • Calculus -

    the graph of |x|+|y| ≤ 1 is a square with vertices
    (1,0), (0,1), (-1,0) and (0,-1)
    y ≥ 2x^2 is the region above y = 2x^2
    so we need their intersection:
    The line of the square in the first quadrant is
    y = -x + 1
    then 2x^2 = -x + 1
    2x^2 + x - 1 = 0
    (2x-1)(x+1) = 0
    x = 1/2 or x = -1

    from the diagram we want x=1/2, then y = 1/2)

    From symmetry we can take the region form x = 0 to x = 1/2, then double that answer.

    Area = ∫(-x + 1 - 2x^2) dx from 0 to 1/2
    = -x^2/2 + x - 2x^3/3 | from 0 to 1/2
    = -1/8 + 1/2 - 1/12 - 0
    = 7/24

  • continuation - Calculus -

    oops, forgot to double our answer ...

    since this was only the area in the first quadrant, the area in II would be the same, so the
    total area = 14/24 or

    7/12 square units

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