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

Using the trapezium rule, the formula for the area under the curve is given by:

$A \approx \frac{h}{2} (y_0 + 2y_1 + 2y_2 + 2y_3 + y_n)$

where $h$ is the width of each interval and $y_i$ are the corresponding y values.

In this case:

• $h = 0.5$,
• $y_0 = 5$,
• $y_1 = 4$,
• $y_2 = 2.5$,
• $y_3 = 1.538$,
• $y_4 = 1$,
• $y_5 = 0.690$,
• $y_6 = 0.5$.

Substituting these values into the trapezium formula:

$A \approx \frac{0.5}{2} (5 + 2(4) + 2(2.5) + 2(1.538) + 1 + 0.690 + 0.5)$
$= 0.25 (5 + 8 + 5 + 3.076 + 1 + 0.690 + 0.5)$
$= 0.25 (23.266) \approx 5.8165$

Therefore, the approximate area for region R is $6.239$.

Using the previous result from part (b), we add the area calculated from the integral which combines a rectangle and the previously computed area:

$A \approx 6.239 + \int_{0}^{3} 4 \, dx$

Calculating the rectangle area:

$\int_{0}^{3} 4 \, dx = 4 \times 3 = 12$

Thus, the total estimate for the integral is:

$\approx 6.239 + 12 = 18.239$

So the approximate value for the integral is approximately $18.24$ when rounded to two decimal places.