a) y = 5^x + ext{log}(x + 1), ext{ } 0 ext{ } ext{ } ext{ } < x < 2 Complete the table below, by giving the value of y when x = 1 - Edexcel - A-Level Maths Pure - Question 4 - 2017 - Paper 3

The trapezium rule states:

$T = \frac{b - a}{2n} [f(a) + 2 \sum_{i=1}^{n-1} f(x_i) + f(b)]$

Here, we have n = 4 (subintervals between x = 0 and x = 2) with y values: 1, 2.821, 5.301, 12.502, and 26.585.

Calculating:

$$= \frac{2}{8} (1 + 2(2.821 + 5.301 + 12.502) + 26.585) \ = 0.25 [1 + 2(20.624) + 26.585] \ = 0.25 [1 + 41.248 + 26.585] \ = 0.25 [68.833] \ = 17.20825. Thus, the approximate integral value is 17.21 (to 2 decimal places).$$

To solve \int_0^2 (5 + 5^x + \text{log}(x + 1)) dx:

We can break this down:

\int_0^2 5 , dx + \int_0^2 5^x , dx + \int_0^2 \text{log}(x + 1) , dx.

1. For \int_0^2 5 , dx: $= 5(x) |_{0}^{2} = 5(2 - 0) = 10.$

2. For \int_0^2 5^x , dx: $= \frac{5^x}{\ln(5)} |_{0}^{2} = \frac{5^2 - 5^0}{\ln(5)} = \frac{25 - 1}{\ln(5)} = \frac{24}{\ln(5)}.$ Approximating \ln(5) \approx 1.609: $\approx \frac{24}{1.609} \approx 14.9.$

3. The last integral \int_0^2 \text{log}(x + 1) , dx is approximated as 17.21.

Combining:

$10 + 14.9 + 17.21 \approx 42.11.$

Thus, our final result is approximately 42.11 (to 2 decimal places).