(i) To find the stationary point, we first find the derivative of the curve: \(\frac{dy}{dx} = 4x^3 + 4\). Setting this to zero gives \(4x^3 + 4 = 0\), which simplifies to \(x^3 = -1\), so \(x = -1\). Substituting \(x = -1\) back into the original equation gives \(y = (-1)^4 + 4(-1) + 9 = 1 - 4 + 9 = 6\). Thus, the stationary point is \((-1, 6)\).

To determine the nature, we find the second derivative: \(\frac{d^2y}{dx^2} = 12x^2\). Substituting \(x = -1\) gives \(12(-1)^2 = 12\), which is positive, indicating a minimum point.

(ii) To find the area, we integrate the curve from \(x = 0\) to \(x = 1\). The integral of \(y = x^4 + 4x + 9\) is \(\int (x^4 + 4x + 9) \, dx = \frac{x^5}{5} + 2x^2 + 9x\). Evaluating from 0 to 1 gives:

\(\left[ \frac{x^5}{5} + 2x^2 + 9x \right]_0^1 = \left( \frac{1^5}{5} + 2(1)^2 + 9(1) \right) - \left( \frac{0^5}{5} + 2(0)^2 + 9(0) \right) = \frac{1}{5} + 2 + 9 = 11.2\).