" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
∫ ∫ ∫ z dz dy dx
yang pertama di integralin z dz:
1/2 z², masukan batasan:
1/2 (1 - x - y)², maka:
∫ ∫ 1/2 (x² + y² + 2xy - 2x - 2y + 1) dy dx
trus integral y dy:
= ∫ ∫ (x²y)/2 + xy² - 2xy + y³/3 - y² + y, masukkan batasan:
= ∫ (x² (1 - x))/2 + x (1 - x)² - 2x (1 - x) + ((1 - x)³)/3 - (1 - x)² + 1 - x
= ∫ (x² - x³)/2 + x - x³ - 2x + 2x² + (1 - 3x + 3x² - x³)/3 - (1 - 2x + x²) + 1 - x
= ∫ (x² - x³)/2 + (1 - 3x + 3x² - x³)/3 + x - x³ - 2x + 2x² - (1 - 2x + x²) + 1 - x
= ∫ (x² - x³)/2 + (1 - 3x + 3x² - x³)/3 + x - x³ - 2x + 2x² - 1 + 2x - x² + 1 - x
= ∫ (x² - x³)/2 + (1 - 3x + 3x² - x³)/3 - x³ + x² dx
∫ (x² - x³)/2 + (1 - 3x + 3x² - x³)/3 - x³ + x² dx
integral x dx:
= 1/6 ((-11x⁴)/4 + x³ - 3x² + 2x), masukkan batasan:
= 1/6 ((-11(1)⁴)/4 + (1)³ - 3(1)² + 2(1))
= 1/6 ((-11)/4 + 1 - 3 + 2)
= 1/6 (-11/4)
= -11/24