y = 5x + log_e(x + 1), 0 ≤ x < 2 Complete the table below, by giving the value of y when x = 1 | x | 0 | 0.5 | 1 | 1.5 | 2 | |-----|-----|-------|---------|-------|--------| | y | 1 | 2.821 | 12.502 | 26.585| | - Edexcel - A-Level Maths Pure - Question 4 - 2017 - Paper 3

Using the trapezium rule:

$ext{Area} = \frac{h}{2} (y_0 + 2y_1 + 2y_2 + y_3)$

where h = 0.5, y_0 = 1, y_1 = 2.821, y_2 = 5.693, y_3 = 12.502:

Calculating the area:

$= \frac{0.5}{2} (1 + 2(2.821) + 2(5.693) + 12.502)$

$= 0.25 (1 + 5.642 + 11.386 + 12.502)$

$= 0.25 (30.530) \approx 7.6325$

Thus, the approximate value for the integral is:

7.63 (to 2 decimal places).

From part (b), we have:

$\int_0^2 (5 + log_e(x + 1)) \, dx \approx 7.63$

To find:

$\int_0^2 (5 + 5x + log_e(x + 1)) \, dx$,

we can decompose it:

$= \int_0^2 5 \, dx + \int_0^2 5x \, dx + \int_0^2 log_e(x + 1) \, dx$

We know that:

1. $\int_0^2 5 \, dx = 5x \Big|_0^2 = 10$.
2. $\int_0^2 5x \, dx = \frac{5x^2}{2} \Big|_0^2 = 10$.

Thus, the whole integral becomes:

$10 + 10 + 7.63 = 27.63$

So, the approximate value is:

27.63 (to 2 decimal places).