Complete the table below, giving the missing value of y to 3 decimal places - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 4

Using the trapezium rule:

$A \approx \frac{h}{2}(f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + 2f(x_4) + f(x_5))$ where $h = 0.5$, the width of each segment.

Calculating:

$A \approx \frac{0.5}{2} \left(5 + 2(4) + 2(2.5) + 2(1.538) + 2(0.690) + 0.5 \right)$

Simplifying:

1. Total the values: $5 + 2(4) + 2(2.5) + 2(1.538) + 2(0.690) + 0.5 = 6.239$
2. Multiply and find: $A \approx 0.5 \times 6.239 = 3.1195 \approx 6.239$

For the integral:

The required integral can be rewritten as: $\int_{0}^{3} \frac{4 + 5}{(x^2+1)} dx = \int_{0}^{3} \left(\frac{4}{(x^2+1)} + \frac{5}{(x^2+1)}\right) dx$

Using the trapezium rule from part (b): $\approx 6.239 + 12 = 18.239$

Thus, the approximate value of the integral is:

$\int_{0}^{3} \frac{4+5}{(x^2+1)} dx \approx 18.24$