(a)
Area = \( \int_{-1}^{2} (6 – x) \, dx \)
Antiderivative: \( 6x – \frac{x^2}{2} \)
Evaluate: \( \left[ 6x – \frac{x^2}{2} \right]_{-1}^{2} = \left( 12 – \frac{4}{2} \right) – \left( -6 – \frac{1}{2} \right) = 10 + 6.5 = 16.5 \)

Result: 16.5 square units [2]

(b)(i)
Area = \( \int_{-1}^{2} (1.5x^2 – 2.5x + 3) \, dx \)

Result: \( \int_{-1}^{2} (1.5x^2 – 2.5x + 3) \, dx \) [2]

(b)(ii)
Antiderivative: \( 0.5x^3 – 1.25x^2 + 3x \)
Evaluate: \( \left[ 0.5x^3 – 1.25x^2 + 3x \right]_{-1}^{2} = \left( 4 – 5 + 6 \right) – \left( -0.5 – 1.25 – 3 \right) = 5 + 4.75 = 9.75 \)

Result: 9.75 square units [1]

(c)
Area between curves = Area under \( y = 6 – x \) – Area under \( y = 1.5x^2 – 2.5x + 3 \)
\( = 16.5 – 9.75 = 6.75 \)
Alternative: \( \int_{-1}^{2} \left[ (6 – x) – (1.5x^2 – 2.5x + 3) \right] \, dx = \int_{-1}^{2} (-1.5x^2 + 1.5x + 3) \, dx \)
Antiderivative: \( -0.5x^3 + 0.75x^2 + 3x \)
Evaluate: \( \left[ -0.5x^3 + 0.75x^2 + 3x \right]_{-1}^{2} = 5 – (-1.75) = 6.75 \)

Result: 6.75 square units [2]