The diagram shows part of the curve $y = \frac{5}{2x - 1}$. The line $9y = -10x + 55$ intersects the curve at points $A$ and $B$ and meets the $x$-axis at point $C$. $CD$ is a straight line parallel to the $y$-axis.

1. Find the coordinates of A, B, and C (from part i) to determine the limits of integration.

From part (i), solving $y = \frac{5}{2x - 1}$ and $9y = -10x + 55$ simultaneously gives $x = 1$ and $x = 5$.
$A$ is $(1, 5)$ and $B$ is $(5, \frac{5}{9})$.
The line meets the $x$-axis at $C$, so $-10x + 55 = 0 \implies x = 5.5$.
$C$ is $(5.5, 0)$.

2. Set up the integrals for the two shaded regions.

The shaded region is split into two parts at $x = 5$ (point B).
For $1 \leqslant x \leqslant 5$, the line is above the curve.
For $5 \leqslant x \leqslant 5.5$, the curve is above the line.
Total Area = $\int_{1}^{5} \left( \frac{-10x + 55}{9} - \frac{5}{2x - 1} \right) dx + \int_{5}^{5.5} \left( \frac{5}{2x - 1} - \frac{-10x + 55}{9} \right) dx$

3. Integrate the functions.

$\int \frac{-10x + 55}{9} dx = \frac{-5x^2 + 55x}{9}$
$\int \frac{5}{2x - 1} dx = \frac{5}{2} \ln(2x - 1)$

4. Evaluate the definite integrals to find the total area.

Area 1 (from $x=1$ to $x=5$):
Area under line = $\left[ \frac{-5(25) + 55(5)}{9} \right] - \left[ \frac{-5(1) + 55(1)}{9} \right] = \frac{150}{9} - \frac{50}{9} = \frac{100}{9}$
Area under curve = $\left[ \frac{5}{2} \ln(9) \right] - \left[ \frac{5}{2} \ln(1) \right] = \frac{5}{2} \ln 9 = 5 \ln 3$
Area 1 = $\frac{100}{9} - 5 \ln 3 \approx 5.6181$

Area 2 (from $x=5$ to $x=5.5$):
Area under curve = $\left[ \frac{5}{2} \ln(10) \right] - \left[ \frac{5}{2} \ln(9) \right] = \frac{5}{2} \ln\left(\frac{10}{9}\right)$
Area under line = $\left[ \frac{-5(5.5)^2 + 55(5.5)}{9} \right] - \frac{150}{9} = \frac{151.25}{9} - \frac{150}{9} = \frac{1.25}{9} = \frac{5}{36}$
Area 2 = $\frac{5}{2} \ln\left(\frac{10}{9}\right) - \frac{5}{36} \approx 0.1245$

Total Area = $5.6181 + 0.1245 = 5.7426 \approx 5.74$ (to 3 s.f.)