(a)
Trapezoidal rule: \( \text{Area} \approx \frac{h}{2} [y_0 + y_4 + 2(y_1 + y_2 + y_3)] \)
Heights: \( y_0 = 0 \), \( y_1 = 3 \), \( y_2 = 8 \), \( y_3 = 9 \), \( y_4 = 0 \), \( h = 10 \)
\( \frac{10}{2} \times (0 + 0 + 2(3 + 8 + 9)) = 5 \times (0 + 2 \times 20) = 5 \times 40 = 200 \, \text{cm}^2 \)

Result: 200 cm² [2]

(b)(i)
Area under curve: \( \int_0^{40} (0.04x^2 – 0.001x^3) \, dx \)

Result: \( \int_0^{40} (0.04x^2 – 0.001x^3) \, dx \) [2]

(b)(ii)
Antiderivative: \( F(x) = \frac{0.04}{3}x^3 – \frac{0.001}{4}x^4 = \frac{1}{75}x^3 – \frac{1}{4000}x^4 \)
Evaluate: \( F(40) = \frac{1}{75} \times 64000 – \frac{1}{4000} \times 2560000 = 853.333 – 640 = 213.333 \)
\( F(0) = 0 \), so area = 213.333 cm² \(\approx\) 213.33 cm²

Result: 213.33 cm² [2]

(c)
Percentage error: \( \left| \frac{213.33 – 200}{213.33} \right| \times 100 \approx 0.0625 \times 100 = 6.25\% \)

Result: 6.25% [2]