Answer: \(\frac23+\frac{\sqrt3}{3}-\frac{5\pi}{36}\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

The required area is the area under \(y=4+2\sin3x\) between \(x=\frac{\pi}{18}\) and \(x=\frac{\pi}{3}\), minus the area of the trapezium under the straight line \(AB\).

First integrate the curve:

\(\int(4+2\sin3x)\,dx=4x-\frac23\cos3x\).

So the area under the curve is

\(\left[4x-\frac23\cos3x\right]_{\pi/18}^{\pi/3}\).

This is

\(\left(\frac{4\pi}{3}+\frac23\right)-\left(\frac{2\pi}{9}-\frac{\sqrt3}{3}\right)=\frac{10\pi}{9}+\frac23+\frac{\sqrt3}{3}\).

At \(x=\frac{\pi}{18}\),

\(y=4+2\sin\frac{\pi}{6}=5\).

At \(x=\frac{\pi}{3}\),

\(y=4+2\sin\pi=4\).

The width is \(\frac{\pi}{3}-\frac{\pi}{18}=\frac{5\pi}{18}\), so the trapezium area is

\(\frac12(5+4)\cdot\frac{5\pi}{18}=\frac{5\pi}{4}\).

Therefore the shaded area is

\(\frac{10\pi}{9}+\frac23+\frac{\sqrt3}{3}-\frac{5\pi}{4}=\frac23+\frac{\sqrt3}{3}-\frac{5\pi}{36}\).

This gives the required answer: \(\frac23+\frac{\sqrt3}{3}-\frac{5\pi}{36}\).