Jawab:
Integral Tertentu
Luas daerah antara kurva dan garis
y kurva = y garis
L = D√D/ (6 a²)
Penjelasan dengan langkah-langkah:
y= x² + x dan y = 2x + 6
x² + x = 2x + 6
x² +x - 2x - 6 = 0
x² - x - 6= 0
a = 1 , b = - 1 , c = - 6
D = b² - 4ac = (-1)² - 4(1)(-6) =1 + 24 = 25
Luas = D√D/( 6. a²)
Luas = (25 √ 25 ) / (6. 1²)
Luas = 25 (5) / 6
Luas = 125/6 = 20⁵/₆ satuan
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Jawab:
Integral Tertentu
Luas daerah antara kurva dan garis
y kurva = y garis
L = D√D/ (6 a²)
Penjelasan dengan langkah-langkah:
y= x² + x dan y = 2x + 6
x² + x = 2x + 6
x² +x - 2x - 6 = 0
x² - x - 6= 0
a = 1 , b = - 1 , c = - 6
D = b² - 4ac = (-1)² - 4(1)(-6) =1 + 24 = 25
Luas = D√D/( 6. a²)
Luas = (25 √ 25 ) / (6. 1²)
Luas = 25 (5) / 6
Luas = 125/6 = 20⁵/₆ satuan