(i) To find the \(x\)-coordinate of \(M\), we need to find the derivative of \(y = x^2 e^{-\frac{1}{2}x}\) and set it to zero. Using the product rule, the derivative is:

\(\frac{d}{dx}(x^2 e^{-\frac{1}{2}x}) = 2x e^{-\frac{1}{2}x} - \frac{1}{2}x^2 e^{-\frac{1}{2}x}\).

Setting the derivative to zero:

\(2x e^{-\frac{1}{2}x} - \frac{1}{2}x^2 e^{-\frac{1}{2}x} = 0\).

Factor out \(e^{-\frac{1}{2}x}\):

\(e^{-\frac{1}{2}x} (2x - \frac{1}{2}x^2) = 0\).

Since \(e^{-\frac{1}{2}x} \neq 0\), solve \(2x - \frac{1}{2}x^2 = 0\):

\(x(2 - \frac{1}{2}x) = 0\).

\(x = 0\) or \(x = 4\).

Since \(x = 0\) is not a maximum, \(x = 4\) is the maximum point.

(ii) To find the area of the shaded region, integrate \(y = x^2 e^{-\frac{1}{2}x}\) from \(x = 0\) to \(x = 1\):

Use integration by parts: \(\int x^2 e^{-\frac{1}{2}x} \, dx\).

Let \(u = x^2\), \(dv = e^{-\frac{1}{2}x} \, dx\).

Then \(du = 2x \, dx\), \(v = -2e^{-\frac{1}{2}x}\).

\(\int x^2 e^{-\frac{1}{2}x} \, dx = -2x^2 e^{-\frac{1}{2}x} + 4 \int x e^{-\frac{1}{2}x} \, dx\).

Integrate \(\int x e^{-\frac{1}{2}x} \, dx\) by parts again.

Let \(u = x\), \(dv = e^{-\frac{1}{2}x} \, dx\).

Then \(du = dx\), \(v = -2e^{-\frac{1}{2}x}\).

\(\int x e^{-\frac{1}{2}x} \, dx = -2x e^{-\frac{1}{2}x} + 4 \int e^{-\frac{1}{2}x} \, dx\).

\(\int e^{-\frac{1}{2}x} \, dx = -2e^{-\frac{1}{2}x}\).

Complete the integration:

\(-2(x^2 + 4x + 8)e^{-\frac{1}{2}x}\).

Evaluate from \(x = 0\) to \(x = 1\):

\([-2(1^2 + 4 \cdot 1 + 8)e^{-\frac{1}{2}}] - [-2(0^2 + 4 \cdot 0 + 8)e^{0}]\).

\(= -26e^{-\frac{1}{2}} + 16\).

Final area: \(16 - 26e^{-\frac{1}{2}}\).