Tolong jawab soal tersebut !

InTegraL
• Luas

y1 = 6 - x
y2 = x²

Batas IntegraL :
y1 = y2
x² + x - 6 = 0
(x + 3)(x - 2) = 0
x = -3 ; x = 2

Luas
= ∫(y1 - y2) dx [2_-3]
= ∫(6 - x - x²) dx
= 6x - 1/2 x² - 1/3 x³
= 6(2 - (-3)) - 1/2 (2² - (-3)²) - 1/3 (2³ - (-3)³)
= 30 + 5/2 - 35/3
= 30 + (15 - 70)/6
= 30 - 55/6
= 30 - 9 1/6
= 20 5/6 satuan luas

y = x² 
y = 6-x
-Cari limit integrasi dl
y=y
x²=6-x
x²+x-6=0
(x-2)(x+3)=0
x=2 dan -3 
- Selesaikan integralnya.
[tex] \int\ {x^{2} - (6-x)} \, dx = \frac{1}{3} x^{3}-6x+ \frac{1}{2} x^{2} [/tex]
-Masukkan limit, terus dikurangkan
[tex][ \frac{1}{3}(2)^3 + \frac{1}{2} (2)^2 -6(2)] - [\frac{1}{3}(-3)^3 + \frac{1}{2} (-3)^2 -6(-3)] = \frac{-125}{6} [/tex]

Tapi nilai utk luas ngk bs negatif jd anggap jawabannya positif aj, yaitu 125/6 atau [tex] 20\frac{5}{6} [/tex]