y = √(5x + 2) (a) Complete the table below, giving the values of y to 3 decimal places - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 2

To use the trapezium rule for the approximation of

$\int_0^2 \sqrt{5x + 2} \, dx$, we first note the calculated y values:

• y(0) = 1.732
• y(0.5) = 1.581
• y(1) = 2.646
• y(1.5) = 3.082
• y(2) = 3.464

Next, applying the trapezium rule:

$\int_a^b f(x) \, dx \approx \frac{h}{2} \left( f(a) + 2 \sum_{i=1}^{n-1} f(x_i) + f(b) \right)$

where:

• h = 0.5 (the width of each sub-interval),
• n = total number of intervals = 4.

Applying these values:

$\int_0^2 \sqrt{5x + 2} \, dx \approx \frac{0.5}{2} \left( 1.732 + 2(1.581 + 2.646 + 3.082) + 3.464 \right) \approx \frac{0.5}{2} \left( 1.732 + 2(7.309) + 3.464 \right)$

Calculating further:

$= \frac{0.5}{2} \left( 1.732 + 14.618 + 3.464 \right)$
$= \frac{0.5}{2} \times 19.814 \approx \frac{0.5 \times 19.814}{2} = 4.954$

Thus, the approximate value of the integral is 4.954.