Complete the table below, giving the missing value of y to 3 decimal places - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 4

Using the trapezium rule, the area A can be approximated using:

$A \approx \frac{1}{2} \cdot (b_0 + b_n) \cdot h + \sum_{i=1}^{n-1} b_i \cdot h$

Where:

• ( b_0 = y(0) = 5 ),
• ( b_1 = y(0.5) = 4 ),
• ( b_2 = y(1) = 2.5 ),
• ( b_3 = y(1.5) \approx 1.538 ),
• ( b_4 = y(2.5) = 0.690 ),
• ( b_5 = y(3) = 0.5 )
• and ( h = 0.5 ) (the width between x points).

Calculating, we get:

$A \approx \frac{1}{2} (5 + 0.5) \cdot 0.5 + (4 + 2.5 + 1.538 + 0.690) \cdot 0.5$

Calculating the sums:

$A \approx 0.5(5.5) + (4 + 2.5 + 1.538 + 0.690) \cdot 0.5 \approx 2.75 + 3.339 \approx 6.089$

To find the approximate value of the integral, we can adjust the area calculated in part (b) using the fact that:

$\int_0^3 \frac{4 + 5}{(x^2 + 1)} \, dx = \int_0^3 4 \, dx + \int_0^3 \frac{5}{(x^2 + 1)} \, dx$

1. The first integral, ( \int_0^3 4 , dx ) evaluates to:

   • ( 4 \cdot (3 - 0) = 12 )

2. The second integral corresponds to the area calculated earlier:

   • Approximated by adding the two areas discovered:
   • Approximate area from part (b): ( 6.089 )

Thus, the total approximate value is:

$12 + 6.089 = 18.089 \approx 18.24$

Therefore, the approximate value for ( \int_0^3 \frac{4 + 5}{(x^2 + 1)} , dx ) is approximately 18.24.