Figure 1 shows the graph of the curve with equation y = \frac{16}{x^2} - \frac{x}{2} + 1, \quad x > 0 The finite region R, bounded by the lines x = 1, the x-axis and the curve, is shown shaded in Figure 1 - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 4

Using the trapezium rule, we calculate the area as follows:

• The formula for the trapezium rule is given by:

  $A \approx \frac{1}{2}h \left( y_0 + 2y_1 + y_2 \right)\text{, where } h \text{ is the width of the intervals.}$

• For our values, we have:

  $h = 0.5\text{, with } y_0 = 0, y_1 = 4, y_2 = 2.31\text{.}$

• Plugging into the formula:

  $A \approx \frac{1}{2} \cdot 0.5 \left( 0 + 2 \times 4 + 2.31 \right) = \frac{1}{2} \cdot 0.5 \left( 0 + 8 + 2.31 \right) = \frac{1}{2} \cdot 0.5 \cdot 10.31 = \frac{10.31}{4} = 2.5775\text{.}$

• Rounding to 2 decimal places, the area is approximately:

  $A \approx 2.58.\text{ }$

To find the exact area under the curve y = \frac{16}{x^2} - \frac{x}{2} + 1 from x = 1 to x = 4, we integrate the function:

$\int_{1}^{4} \left( \frac{16}{x^2} - \frac{x}{2} + 1 \right) \mathrm{d}x\text{.}$

Calculating the integral:

1. Break it into parts:

   $= \int_{1}^{4} \frac{16}{x^2} \mathrm{d}x - \int_{1}^{4} \frac{x}{2} \mathrm{d}x + \int_{1}^{4} 1 \mathrm{d}x$

2. Integrate each term:

• $\int \frac{16}{x^2} \mathrm{d}x = -\frac{16}{x}$
• $\int \frac{x}{2} \mathrm{d}x = \frac{x^2}{4}$
• $\int 1 \mathrm{d}x = x$

3. Combine results:

$= \left[-\frac{16}{x} \right]_{1}^{4} - \left[\frac{x^2}{4} \right]_{1}^{4} + \left[x \right]_{1}^{4}$

4. Evaluate the boundaries:

• For $\left[-\frac{16}{x} \right]_{1}^{4} = -\frac{16}{4} - (-\frac{16}{1}) = -4 + 16 = 12$

• For $\left[\frac{x^2}{4} \right]_{1}^{4} = \frac{16}{4} - \frac{1}{4} = 4 - 0.25 = 3.75$

• For $\left[x \right]_{1}^{4} = 4 - 1 = 3$

5. Adding the results together gives:

$\text{Area } = 12 - 3.75 + 3 = 11.25\text{.}$

Thus, the exact area of region R is 11.25.