The curve C has equation $$y = x \sqrt{x^3 + 1}, \quad 0 \leq x \leq 2.\n a) Complete the table below, giving the values of y to 3 decimal places at x = 1 and x = 1.5 - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 2

The trapezium rule formula is given by:

$\int_a^b f(x) \, dx \approx \frac{b - a}{2} (f(a) + f(b))$

For this problem, we will break the integral into two intervals:

1. From 0 to 0.5: $f(0) = 0, \, f(0.5) = 0.530$ Area1 = \frac{0.5 - 0}{2} (0 + 0.530) = \frac{0.5}{2} \cdot 0.530 = 0.1325$$

2. From 0.5 to 1: $f(0.5) = 0.530, \, f(1) = 1.414$ Area2 = \frac{1 - 0.5}{2} (0.530 + 1.414) = \frac{0.5}{2} \cdot (0.530 + 1.414) = 0.5 \cdot 0.967 = 0.4835$$

3. From 1 to 1.5: $f(1) = 1.414, \, f(1.5) = 3.137$ Area3 = \frac{1.5 - 1}{2} (1.414 + 3.137) = \frac{0.5}{2} \cdot (1.414 + 3.137) = 0.5 \cdot 2.2755 = 0.568875$$

4. From 1.5 to 2: $f(1.5) = 3.137, \, f(2) = 6$ Area4 = \frac{2 - 1.5}{2} (3.137 + 6) = \frac{0.5}{2} \cdot (3.137 + 6) = 0.5 \cdot 4.5685 = 2.28425$$

The total area is:

$Total Area \approx 0.1325 + 0.4835 + 0.568875 + 2.28425 \approx 3.469125$

Rounding to 3 significant figures gives us approximately 3.47.

The area of the region R, which is bounded by the curve C and the line segment l, can be calculated using:

$\text{Area of triangle} - \text{Approximate area under the curve}$

1. The area of triangle formed by the points (0,0), (2,6), and (0,6) is: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 6 = 6$

2. Using the previously calculated area of the region under the curve (3.469125), the area of R is: $\text{Area of R} = 6 - 3.469125 \approx 2.530875$

Thus, rounding to 3 significant figures, the approximate area of R is 2.53.