The finite region R, as shown in Figure 2, is bounded by the x-axis and the curve with equation $$y = 27 - 2x - 9rac{16}{x^2}, \quad x > 0$$ The curve crosses the x-axis at the points (1, 0) and (4, 0) - Edexcel - A-Level Maths Pure - Question 1 - 2013 - Paper 6

The area can be approximated using the trapezium rule. The formula for the trapezium rule is given by:

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

Where $h$ is the width between the x-intervals. Here,

• For $x = 1$ to $x = 4$, the intervals are 0.5 (since $h = (b - a)/n = (4 - 1)/5$)
• The y-values from the table are used in the calculation:

Thus, $A = \frac{0.5}{2}(5.866 + 2(5.210) + 2(-13) + 2(-1.21) + 2(8.28) + 0)$

Calculating these values gives:

$A = 0.25(5.866 + 10.42 - 26 - 2.42 + 16.56)$

$A = 0.25(4.488) = 1.12$

Thus, the approximate area of R is 11.42 (rounded to 2 decimal places).

To find the exact value of the area bounded by the curve and the x-axis, we can use integration:

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

We break this down into simpler parts:

1. Integrate $27$:
   $\int 27 \, dx = 27x$

2. Integrate $-2x$:
   $\int -2x \, dx = -x^2$

3. Integrate $-9 \frac{16}{x^2}$:
   $\int -9 \frac{16}{x^2} \, dx = 9 \cdot 16 \frac{1}{x} = -\frac{144}{x}$

Now substituting and solving gives:

$\left[ 27x - x^2 - \frac{144}{x} \right]_{1}^{4}$

Calculating at the limits:

• At $x=4$: $27(4) - 4^2 - \frac{144}{4} = 108 - 16 - 36 = 56$
• At $x=1$: $27(1) - 1^2 - \frac{144}{1} = 27 - 1 - 144 = -118$

Thus, the exact area: $A = 56 - (-118) = 174$

Thus, the exact value for the area of R is 174.