The finite region R, as shown in Figure 2, is bounded by the x-axis and the curve with equation y = 27 - 2x - 9rac{16}{x^2}, \, x > 0 The curve crosses the x-axis at the points (1, 0) and (4, 0) - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 4

To find the area using the trapezium rule, we use the following formula:

$A \approx \frac{1}{2}h(y_0 + 2(y_1 + y_2 + y_3 + y_4) + y_n)$

Where:

• h = width of the intervals = (x_n - x_0)/n = (4 - 1)/5 = 0.6,
• y_0 = 5.866, y_1 = 5.210, y_2 = 5.000, y_3 = 4.600, y_4 = 9.200, y_n = 0.

Now, substituting these values:

$A \approx \frac{1}{2} \times 0.6 \times (5.866 + 2(5.210 + 5.000 + 4.600 + 9.200) + 0)$

Calculating:

• Total = 5.866 + 2(24.02) = 5.866 + 48.04 = 53.906.
• Now, area = \frac{1}{2} \times 0.6 \times 53.906 \approx 16.1718.

Thus, the approximate value for the area of R is:

Area ≈ 16.17.

To find the exact area of the region R, we integrate the function:

$\int_1^4 \left(27 - 2x - 9\sqrt{\frac{16}{x^2}}\right) dx.$

Breaking down the integral:

1. Start with: $\int(27 - 2x - 9 \cdot \frac{4}{x}) dx = \int 27 dx - \int 2x dx - \int 36x^{-1} dx.$

2. Calculating each integral:

   • $\int 27 dx = 27x$,
   • $\int 2x dx = x^2$,
   • $\int 36x^{-1} dx = 36 \ln|x|.$

3. Thus, the full integral becomes: $27x - x^2 - 36\ln|x| + C.$

4. Evaluating from 1 to 4: $\left[27(4) - (4)^2 - 36\ln(4)\right] - \left[27(1) - (1)^2 - 36\ln(1)\right].$

   • Simplifying: $\left[108 - 16 - 36\ln(4)\right] - \left[27 - 1 - 0\right].$
   • This results in: $92 - 26 - 36\ln(4) = 66 - 36\ln(4).$

Hence, the exact value for the area of R is:

Area = 66 - 36\ln(4).