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}$ - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 9

To approximate the area using the trapezium rule, we divide the interval ([0, \frac{\pi}{2}] into intervals based on the x values:

1. We have:

   • Height 1 (for x = 0) = 1
   • Height 2 (for x = \frac{\pi}{6}) = 1.035276
   • Height 3 (for x = \frac{\pi}{3}) = 1.154701
   • Height 4 (for x = \frac{\pi}{2}) = 1.414214

2. The formula for the trapezium rule with n = 3 intervals is: $Area \approx \frac{h}{2} \left(y_0 + 2y_1 + 2y_2 + y_3\right)$ where (h = \frac{b - a}{n} = \frac{\frac{\pi}{2} - 0}{3} = \frac{\pi}{6}$$

3. Thus, $Area \approx \frac{\frac{\pi}{6}}{2} \left(1 + 2 \cdot 1.035276 + 2 \cdot 1.154701 + 1.414214\right)$

4. Evaluating this: $\approx \frac{\frac{\pi}{6}}{2} \left(1 + 2.070552 + 2.309402 + 1.414214\right)\approx \frac{\frac{\pi}{6}}{2} \cdot 6.794168\approx 1.778709023$

5. Therefore, the area of R is approximately (1.7787) to 4 decimal places.

To find the volume of the solid formed by rotating the region R around the x-axis, we use the formula: $V = \int_{a}^{b} \pi [f(x)]^2 dx$

1. Here, (f(x) = sec\left(\frac{1}{2}x\right)), (a = 0), and (b = \frac{\pi}{2}).

2. The volume can be calculated as: $V = \int_{0}^{\frac{\pi}{2}} \pi \left(sec\left(\frac{1}{2}x\right)\right)^2 dx$

3. Since (sec^2(t) = 1 + tan^2(t)), substituting with (t = \frac{1}{2}x), gives: $V = \int_{0}^{\frac{\pi}{2}} \pi \left(1 + tan^2\left(\frac{1}{2}x\right)\right) dx$

4. After integration and evaluation: $V = 2\pi\int_0^{\frac{\pi}{2}} sec^2\left(\frac{1}{2}x\right) dx$

5. The final volume can be evaluated as: $V = 2\pi (value)\approx 2\pi$ (exact value during calculation).