Luas daerah yang diarsir pada gambar
adalah6⅔
Pembahasan:
y₁ = x² – 4x + 3
y₂ = –x² + 6x – 5
Titik potong :
y₁ = y₂
⇒ x² – 4x + 3 = –x² + 6x – 5
⇒ 2x² – 10x + 8 = 0
⇒ x² – 5x + 4 = 0
⇒ (x – 4)(x – 1) = 0
x = 4 atau x = 1
Luas daerah arsiran:
₁∫³y₂–y₁ dx = ₁∫³(–x²+6 –5) – (x²– 4x+3) dx
= ₁∫³ –2x² + 10x – 8 dx
= |–⅔x³ + 5x² – 8x|³₁
= (–⅔(3)³ + 5(3)² – 8.3) – (–⅔(1)³ + 5(1)² – 8.1)
= (–18 + 45 – 24) – (–⅔ – 3)
= 3 – (-⅔ – 9/3)
= 3 + 11/3
= 6⅔