y = 5x + log_2(x + 1), 0 ≤ x ≤ 2 Complete the table below, by giving the value of y when x = 1: x | 0 | 0.5 | 1.5 | 2 y | 1 | 2.821 | 12.502| 26.585 (b) Use the trapezium rule, with all the values of y from the completed table, to find an approximate value for ∫_0^2 (5x + log_2(x + 1)) dx giving your answer to 2 decimal places - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 2

Using the trapezium rule:

The formula for the trapezium rule is: ext{Area} = rac{b - a}{2} imes (f(a) + f(b)) where: a = 0, b = 2, f(0) = 1, f(0.5) = 2.821, f(1) = 6, f(1.5) = 12.502, f(2) = 26.585.

Thus we calculate:

Area = (rac{2 - 0}{2} \times (1 + 2.821 + 6 + 12.502 + 26.585)) (= 1 \times (49.908) = 49.908.)

For further refining, we apply the trapezium rule with intervals: (\frac{0.5}{2} [1 + 2.821] + \frac{0.5}{2} [2.821 + 6] + \frac{0.5}{2} [6 + 12.502] + \frac{0.5}{2} [12.502 + 26.585]). Calculating that results in an approximate value around 17.56 (to 2 decimal places).

We can approximate:

a = 5, b = 5\int_0^2 x dx + (17.56)

The integrals can be calculated: igg[5x\bigg]_0^2 = 5(2) - 5(0) = 10.

So, combining: 10 + 17.56 = 27.56. Thus, the approximate value is: 27.56 (to 2 decimal places).