Mohon bantuannya kaka, beserta dengan caranya
1. Luas = internal [1 - 2] {(2 - x) - (x²)} dx
= {2x - (1/2)x² - (1/3)x³) [1 - 2]
= (4 - 2 - 8/3) - (2 - 1/2 - 1/3)
= (6 - 8/3) - (2 - 1/6)
= 4 - 15/6
= 9/6
= 3/2
3. -x² + 4 = -2x + 4
x² - 2x = 0
x(x - 2) = 0
x = 0 atau x = 2
Volume = π internal [0 - 2] {(-2x + 4)² - (-x² + 4)²) dx
2. Luas = internal [-3 - 1] (x² + 2x - 3) dx
= (1/3 x³ + x² - 3x) [-3 - 1]
= (1/3 + 1 - 3) - (-9 + 9 + 9)
= (4/3 - 3) - 9
= 4/3 - 12
= 4/3 - 36/3
= -32/3 ( - adalah di kuadran 3)
maaf jiks salah dan smga membantu
= {2x - (1/2)x² - (1/3)x³) [1 - 2]
= (4 - 2 - 8/3) - (2 - 1/2 - 1/3)
= (6 - 8/3) - (2 - 1/6)
= 4 - 15/6
= 9/6
= 3/2
3. -x² + 4 = -2x + 4
x² - 2x = 0
x(x - 2) = 0
x = 0 atau x = 2
Volume = π internal [0 - 2] {(-2x + 4)² - (-x² + 4)²) dx
2. Luas = internal [-3 - 1] (x² + 2x - 3) dx
= (1/3 x³ + x² - 3x) [-3 - 1]
= (1/3 + 1 - 3) - (-9 + 9 + 9)
= (4/3 - 3) - 9
= 4/3 - 12
= 4/3 - 36/3
= -32/3 ( - adalah di kuadran 3)
maaf jiks salah dan smga membantu
1 votes
Thanks 1
MathSolver74
jawaban sama persis
bohong dosa
More Questions From This User See All
cari batas2nya dulu
x² = 2 - x
x² + x - 2 = 0
(x + 2)(x - 1) = 0
x = -2 atau x = 1
y = 4 maka x = 2
karena di kuadran I maka
Luas = int [1 - 2] {(2 - x) - (x²)} dx
= {2x - (1/2)x² - (1/3)x³) [1 - 2]
= (4 - 2 - 8/3) - (2 - 1/2 - 1/3)
= (6 - 8/3) - (2 - 1/6)
= 4 - 15/6
= 9/6
= 3/2 satuan luas
2) langsung aja
Luas = int [-3 - 1] (x² + 2x - 3) dx
= (1/3 x³ + x² - 3x) [-3 - 1]
= (1/3 + 1 - 3) - (-9 + 9 + 9)
= (4/3 - 3) - 9
= 4/3 - 12
= 4/3 - 36/3
= -32/3
tanda (-) menunjukan di kuadran III
3) Cari batasnya dulu
-x² + 4 = -2x + 4
x² - 2x = 0
x(x - 2) = 0
x = 0 atau x = 2
Volume = π int [0 - 2] {(-2x + 4)² - (-x² + 4)²) dx
4) ubah jd x = ...