(i) The curve is given by \(y = 2\sqrt{x}\). To find the equation of the normal, we first find the derivative \(\frac{dy}{dx}\). For \(y = 2\sqrt{x}\), \(\frac{dy}{dx} = x^{-\frac{1}{2}}\).

At \(x = 4\), \(\frac{dy}{dx} = \frac{1}{2}\). The slope of the normal is the negative reciprocal, so \(m = -2\).

The equation of the normal line at point B (4, 4) is \(y - 4 = -2(x - 4)\), which simplifies to \(y = -2x + 12\).

(ii) To find the area of the shaded region, we integrate the curve from \(x = 1\) to \(x = 4\). The integral is \(\int 2\sqrt{x} \, dx = \int 2x^{0.5} \, dx = \frac{2}{1.5}x^{1.5}\).

Evaluating from 1 to 4 gives \(\left[ \frac{2}{1.5}x^{1.5} \right]_1^4 = \frac{2}{1.5}(4^{1.5} - 1^{1.5}) = \frac{2}{1.5}(8 - 1) = \frac{2}{1.5} \times 7 = \frac{28}{3}\).