The curve C has equation $y = x\sqrt{(x^3 + 1)}$, $0 \leq x \leq 2$ - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 2

To apply the trapezium rule, we use the values from the completed table:

• $y_0 = 0$
• $y_1 = 0.530$
• $y_2 = 1.414$
• $y_3 = 3.137$

The formula for the trapezium rule is: $ext{Area} \approx \frac{h}{2} (y_0 + 2y_1 + 2y_2 + y_3)$ where $h$ is the width of each segment. Here, $h = 0.5$.

Calculating: $\text{Area} \approx \frac{0.5}{2}(0 + 2(0.530) + 2(1.414) + 3.137)$ $= 0.25(0 + 1.060 + 2.828 + 3.137)$ $= 0.25(7.025) = 1.75625$

Thus, the answer for part (b) is approximately $1.756$, rounded to 3 significant figures it is $1.76$.

To find the area of region $R$, we first recall from part (b) that we calculated an approximate area under the curve:

• Area under curve $C$ = $1.756$.

The area of the triangle formed by the line segment $l$ from $(0,0)$ to $(2,6)$ is calculated as: $\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 6 = 6.$

Finally, the area of region $R$ is given by: $\text{Area of } R = \text{Area of triangle} - \text{Area under curve}$ $= 6 - 1.756 = 4.244.$

Rounding to 3 significant figures gives us the final area for region $R$ as approximately $4.24$.