Tolong dengan caranya kaka

#F

Luas daerah y = x² dan y = 2 – x

y1 – y2 = 0

x² -(2 – x ) = 0

x² + x – 2 = 0

(x + 2)(x – 1) = 0

x= – 2 atau x = 1

luas = a b∫ garis – kurva dx

Luas = ₋₂¹ ∫ 2 – x – (x²) dx

Luas = ₋₂¹∫( -x² – x – 2) dx

L = ₋₂¹ ∫ – (x² + x – 2)dx

L = – [ 1/3 x³ + 1/2 x² – 2x]¹₋₂

L = – [1/3 (1³ – (-2)³) + 1/2 (1² -(-2)²) – 2 (1 – (-2))]

L = – [ 1/3 (1 +8) + 1/2 (1 – 4) – 2( 1 + 2)]

L = – [9/3 – 3/2 – 6] = – [3 – 3/2 0 6 ] = – [-4 ¹/₂]

L = 4 ¹/₂ satuan