3. Figure 1 shows the finite region R bounded by the x-axis, the y-axis, the line x = \frac{\pi}{2} and the curve with equation $y = \sec\left(\frac{1}{2} x\right), \quad 0 \leq x \leq \frac{\pi}{2}.$ The table shows corresponding values of x and y for y = \sec\left(\frac{1}{2} x\right) - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 9

To use the trapezium rule, we approximate the area as:

$A \approx \frac{h}{2} \left( y_0 + 2 \sum_{i=1}^{n-1} y_i + y_n \right)$

where h is the width of each interval. Here, we have:

• Interval widths: (h = \frac{\pi/2 - 0}{3} = \frac{\pi}{6})
• Values: (y_0 = 1, y_1 = 1.035276, y_2 = 2.000000, y_3 = 1.414214)

Applying the trapezium rule:

$A \approx \frac{\frac{\pi}{6}}{2} \left( 1 + 2 \cdot (1.035276 + 2.000000) + 1.414214 \right)$

Calculating:

$A \approx \frac{\frac{\pi}{6}}{2} \left( 1 + 4.070552 + 1.414214 \right) = \frac{\frac{\pi}{6}}{2} \cdot 6.484766$

After evaluating, the area is approximately 1.7787.

To calculate the volume of the solid formed by rotating region R around the x-axis, we use the formula:

$V = \pi \int_{0}^{\frac{\pi}{2}} \left[ y \right]^2 dx$

Substituting for y:

$V = \pi \int_{0}^{\frac{\pi}{2}} \left[ \sec\left( \frac{1}{2} x \right) \right]^2 dx$

This simplifies to:

$V = \pi \int_{0}^{\frac{\pi}{2}} \sec^2\left( \frac{1}{2} x \right) dx$

Using the substitution (u = \frac{1}{2} x), we have:

• du = (\frac{1}{2} dx) implies dx = 2 du,
• and changing the limits gives (u = 0) to (u = \frac{\pi}{4}).

Thus:

$V = 2\pi \int_{0}^{\frac{\pi}{4}} \sec^2(u) du = 2\pi \left[ \tan(u) \right]_{0}^{\frac{\pi}{4}} = 2\pi(1 - 0) = 2\pi.$

The exact volume of the solid is therefore (2\pi).