The diagram shows part of the curve $y = \sqrt{3x-2}$. The normal to the curve at $B$ meets the $x$-axis at $C$. Given that the $x$-coordinate of $B$ is 9, find

1. State the coordinates of $A$, $B$, and $C$ found in part (a)

From part (a), we have $A = (\frac{2}{3}, 0)$, $B = (9, 5)$, and $C = (10.5, 0)$.

2. Set up the integral for the area under the curve from $A$ to the $x$-coordinate of $B$

Area under curve $= \int_{\frac{2}{3}}^{9} (3x-2)^{\frac{1}{2}} \, dx$

3. Evaluate the integral

$\int_{\frac{2}{3}}^{9} (3x-2)^{\frac{1}{2}} \, dx = \left[ \frac{(3x-2)^{\frac{3}{2}}}{\frac{3}{2} \times 3} \right]_{\frac{2}{3}}^{9} = \left[ \frac{2}{9}(3x-2)^{\frac{3}{2}} \right]_{\frac{2}{3}}^{9}$
$= \frac{2}{9}(25)^{\frac{3}{2}} - \frac{2}{9}(0)^{\frac{3}{2}} = \frac{2}{9}(125) = \frac{250}{9}$

4. Calculate the area of the triangle formed by the normal and the $x$-axis

The region under the normal from $x=9$ to $x=10.5$ is a triangle with base $10.5 - 9 = 1.5$ and height $5$.
Area of triangle $= \frac{1}{2} \times 1.5 \times 5 = \frac{15}{4}$

5. Find the total area of the shaded region

Total area $= \frac{250}{9} + \frac{15}{4} = \frac{1000 + 135}{36} = \frac{1135}{36} = 31\frac{19}{36}$