Matematika

Pertanyaan

hitunglah luas daerah yang dibatasi oleh kurva y = x2 - 2x - 3 dan sumbu x

2 Jawaban

  • y=x²-2x-3
    y=(x+1)(x-3)

    x1=-1
    x2=3

    [tex] \int\limits^3_{-1} { x^{2}-2x-3 } \, dx[/tex]
    [tex]= \frac{1}{3} x^3 - x^2 - 3x \left \{ {{3} \atop {-1}} \right. [/tex]
    [tex]= \frac{1}{3}(3)^3 - (3)^2 - 3(3) - ( \frac{1}{3}(-1)^3 - (-1)^2 - 3(-1)[/tex]
    [tex]=13(27)-9-9-(-1/3-1-(-3)[/tex]
    [tex]=-9-(-1-3+9/3) [/tex]
    [tex]=-9-(5/3) = -27-5/3 = -32/3[/tex]


  • Ada 3 cara

    1) titik potong : x^2 - 2x - 3 = 0
    => (x - 3)(x + 1) = 0
    => x = 3 atau x = -1
    Kurva berada di bawah sumbu x
    Maka
    Luas = Integral -(x^2 - 2x - 3) dx
    = Integral (-x^2 + 2x + 3) dx
    = -1/3 x^3 + x^2 + 3x | batas -1 sampai 3
    = -1/3 . 3^3 + 3^2 + 3(3) - (-1/3 (-1)^3 + (-1)^2 + 3(-1))
    = -9 + 9 + 9 - (1/3 + 1 - 3)
    = 9 - 1/3 + 2
    = 11 - 1/3
    = 32/3
    = 10 2/3

    2) y = x^2 - 2x - 3
    D = b^2 - 4ac = (-2)^2 - 4(1)(-3) = 4 + 12 = 16
    Luas = D√D/6a^2 = 16√16 / 6 (1)^2 = 8 . 4 / 3 = 32/3 = 10 2/3

    3) y = x^2 - 2x - 3 = (x - 3)(x + 1) => x = 3 atau x = -1
    Titik tengah : x = (3 + (-1))/2 => x = 1
    Panjang = |x1 - x2| = |3 - (-1)| = 4
    Lebar (x = 1) = y = |1^2 - 2(1) - 3| = |1 - 2 - 3| = 4
    Luas = 2/3 x luas persegi panjang
    = 2/3 x 4 . 4
    = 32/3
    = 10 2/3

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