Answer: \(\frac{25\pi^2}{4}\).

Use differentiation to find the gradient information needed. Stationary points occur when the derivative is zero, and tangents or normals are found from the gradient at the given point.

Differentiate \(y=2x\cos x\) using the product rule:

\(\frac{\mathrm dy}{\mathrm dx}=2\cos x-2x\sin x.\)

At \(x=\pi\),

\(\frac{\mathrm dy}{\mathrm dx}=2\cos\pi-2\pi\sin\pi=-2.\)

So the gradient of the normal is \(\frac12\).

The normal passes through \((\pi,-2\pi)\), so

\(y+2\pi=\frac12(x-\pi).\)

For the \(x\)-intercept, put \(y=0\):

\(2\pi=\frac12(x-\pi)\), giving \(x=5\pi\).

For the \(y\)-intercept, put \(x=0\):

\(y+2\pi=-\frac{\pi}{2}\), so \(y=-\frac{5\pi}{2}\).

Therefore triangle \(POQ\) has intercept lengths \(5\pi\) and \(\frac{5\pi}{2}\). Its area is

\(\frac12\cdot5\pi\cdot\frac{5\pi}{2}=\frac{25\pi^2}{4}.\)

This gives the required answer: \(\frac{25\pi^2}{4}\).