Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x^3 and y = x.?

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  • Mike G
    Lv 7
    3 years ago

    At intersection x^3-x = 0

    x^2 = 1

    x = 1 or -1

    As the curve is symmetrical use double the area between 0 and 1

    n = 4

    Width of each rectangle = 0.25

    Height of each rectangle x-x^3

    h(x) = x-x^3

    Midpoints of the rectangles = 0.125, 0.375, 0.625, 0.875

    h(0.125) = 0.123

    h(0.375) = 0.322

    h(0.625) = 0.381

    h(0.875) = 0.205

    Area of one loop = 0.25*(0.123+0.322+0.381+0.205) = 0.25775

    Area of region = 2*0.25775 =

    0.5155

    See Graph

    https://www.desmos.com/calculator/o8qtwoq4qp

  • ?
    Lv 7
    3 years ago

    x = x³

    x³ - x = 0

    x(x+1)(x-1) = 0

    The two curves intersect at x = -1, 0, 1

    The area of the region bounded by the two curves is given by:

    A = ∫|x³-x|dx

    integration over the interval -1 ≤ x ≤ 1

    Performing the integration, A = 1/2

    Approximating with the midpoint rule with n = 4:

    Midpoints are:

    x = -3/4, -1/4, 1/4, 3/4

    Δx = 1/2

    At the midpoints:

    f(x) = |x³ - x|

    f(-3/4) = |(-3/4)³ - (-3/4)| = 3/4 - 27/64 = (48-27)/64 = 21/64

    f(-1/4) = |(-1/4)³ - (-1/4)| = 1/4 - 1/64 = 15/64

    f( 1/4) = |(1/4)³ - (1/4)| = 1/4 - 1/64 = 15/64

    f(3/4) = |(3/4)² - (3/4)| = 3/4 - 27/64 = 21/64

    Area = (1/2)(21/64 + 15/64 + 15/64 + 21/64)

    = (21+15)/64

    = 36/64

    = 9/16

    = 0.5625

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