The diagram shows part of the curve $y = \dfrac{3}{9-2x}$. The point $P\ (3,\ 1)$ lies on the curve and the normal to the curve at $P$ meets the x-axis at $Q$.

1. Find the derivative of the curve equation to determine the gradient function.

$y = 3(9-2x)^{-1}$

$\frac{dy}{dx} = -3(9-2x)^{-2} \cdot (-2)$

$\frac{dy}{dx} = \frac{6}{(9-2x)^2}$

2. Calculate the gradient of the tangent at point $P(3, 1)$.

At $x = 3$,
$\frac{dy}{dx} = \frac{6}{(9-2(3))^2} = \frac{6}{3^2} = \frac{6}{9} = \frac{2}{3}$

3. Determine the gradient of the normal at $P$.

Gradient of normal, $m_n = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}$

4. Find the equation of the normal and solve for the x-intercept $Q$.

Equation of normal at $P(3, 1)$:
$y - 1 = -\frac{3}{2}(x - 3)$

Since $Q$ lies on the x-axis, substitute $y = 0$:
$0 - 1 = -\frac{3}{2}(x - 3)$
$1 = \frac{3}{2}(x - 3)$
$\frac{2}{3} = x - 3$
$x = 3\frac{2}{3} = \frac{11}{3}$

Coordinates of $Q$ are $\left(\frac{11}{3}, 0\right)$.

1. State the coordinates of $Q$ from part (a) and identify the two regions that make up the shaded area

From part (a), the gradient of the curve at $P(3, 1)$ is $\frac{2}{3}$, so the normal has gradient $-\frac{3}{2}$. The equation of the normal is $y - 1 = -\frac{3}{2}(x - 3)$, giving $Q\left(\frac{11}{3}, 0\right)$.

The shaded area consists of the area under the curve from $x = 0$ to $x = 3$, and the area of the triangle formed by the normal from $x = 3$ to $x = \frac{11}{3}$.

2. Integrate the equation of the curve to find the area under the curve

Area under curve $= \int_{0}^{3} \frac{3}{9-2x} \, dx$
$= \left[ \frac{3 \ln(9-2x)}{-2} \right]_{0}^{3}$

3. Evaluate the definite integral

$= -\frac{3}{2} \ln(9 - 2(3)) - \left( -\frac{3}{2} \ln(9 - 2(0)) \right)$
$= -\frac{3}{2} \ln 3 + \frac{3}{2} \ln 9$
$= -\frac{3}{2} \ln 3 + \frac{3}{2} (2 \ln 3)$
$= \frac{3}{2} \ln 3$

4. Calculate the area of the triangle and the total area

Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height}$
$= \frac{1}{2} \times \left(\frac{11}{3} - 3\right) \times 1$
$= \frac{1}{2} \times \frac{2}{3} \times 1 = \frac{1}{3}$

Total area $= \frac{3}{2} \ln 3 + \frac{1}{3} \approx 1.98$ (to 3 s.f.)