The finite region R, as shown in Figure 2, is bounded by the x-axis and the curve with equation y = 27 - 2x - 9 \sqrt{\frac{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 4 - 2013 - Paper 6

To use the trapezium rule:

• The formula is given by:

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

• Where:

  • $h$ is the width of each interval (0.5 in this case)
  • $y_0 = 5.866, y_1 = 5.210, y_2 = 6.272, y_3 = 6.634, y_4 = 1.856, y_5 = 0$

Now substituting these values:

$ext{Area} \approx \frac{0.5}{2}(5.866 + 2(5.210) + 2(6.272) + 2(6.634) + 2(1.856) + 0)$

Calculating:

$= 0.25(5.866 + 10.420 + 12.544 + 13.268 + 3.712) = 0.25 \times 55.810 = 13.9525$

So, the approximate area is about $\text{Area} \approx 11.42$.

To find the exact area under the curve:

• We calculate the integral:

  $\int_{1}^{4} \left( 27 - 2x - 9 \sqrt{\frac{16}{x^2}} \right) dx$

• This can be simplified to:

  $= \int_{1}^{4} \left( 27 - 2x - \frac{36}{x} \right) dx$

  Evaluating this:

  $= \left[ 27x - x^2 - 36 \ln|x| \right]_{1}^{4}$

• Now, substituting the limits:

  $= \left( 108 - 16 - 36 \ln(4) \right) - \left( 27 - 1 - 36 \ln(1) \right)$

• Since $\ln(1) = 0$:

  $= 92 - 26 + 36 \ln(4) = 66 - 36 \ln(4)$

Thus, the exact value for the area of region R is $66 - 36 \ln(4)$.