hitunglah luas daerah yang dibatasi oleh kurva y = x2 - 2x - 3 dan sumbu x
Matematika
herherhyera
Pertanyaan
hitunglah luas daerah yang dibatasi oleh kurva y = x2 - 2x - 3 dan sumbu x
2 Jawaban
-
1. Jawaban PermadiPutraAmin
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] -
2. Jawaban arsetpopeye
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