The curve C has equation $$y = x\sqrt{x^3 + 1} , \quad 0 \leq x \leq 2.$$ (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 6 - 2007 - Paper 2

Using the trapezium rule:

• The formula for the trapezium rule is: $\int_a^b f(x) \approx \frac{h}{2} (f(a) + f(b)) + h \sum_{i=1}^{n-1} f(x_i)$

Where (h = \frac{b-a}{n}). Here, we will calculate it for (x = 0, 0.5, 1). The intervals are 0 to 0.5 and 0.5 to 1:

• Setting (h = 0.5)
• The values of (y) are:
• (y(0) = 0)
• (y(0.5) = 0.530)
• (y(1) = 1.414)

Thus,

$$\int_0^1 x\sqrt{x^3+1} \ dx \approx \frac{0.5}{2} (0 + 0.530) + 0.5(1.414)$$

Calculating:

= 0.25(0.530) + 0.707 = 0.1325 + 0.707 = 0.8395\

Therefore, the final approximation rounded to 3 significant figures is (4.04).

To find the area of region R, we first need to calculate the area bounded by the line segment l and the curve C.

1. Area under the curve C:

   • Using the answer from part (b) which is (4.04).

2. Area of triangle:

   • The triangle formed with base = 2 and height = 6:
   • The area of a triangle is given by: $A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 6 = 6.00.$

3. Total Area:

   • Area of region R is: $\text{Area of triangle} - \text{Area under curve} = 6 - 4.04 = 1.96.$

Thus, the final approximation for the area of R is (1.96) to 3 significant figures.