(i) To find the \(x\)-coordinate of the minimum point \(M\), we first find the derivative of \(y = x^{\frac{1}{2}} \ln x\) using the product rule:
\(\frac{d}{dx}(x^{\frac{1}{2}} \ln x) = \frac{1}{2}x^{-\frac{1}{2}} \ln x + x^{\frac{1}{2}} \cdot \frac{1}{x} = \frac{1}{2}x^{-\frac{1}{2}} \ln x + x^{-\frac{1}{2}}\).
Set the derivative to zero to find critical points:
\(\frac{1}{2}x^{-\frac{1}{2}} \ln x + x^{-\frac{1}{2}} = 0\).
\(\ln x = -2 \Rightarrow x = e^{-2}\).
(ii) To find the area of the shaded region, use integration by parts on \(\int x^{\frac{1}{2}} \ln x \, dx\) from \(x = 1\) to \(x = 4\).
Let \(u = \ln x\) and \(dv = x^{\frac{1}{2}} \, dx\), then \(du = \frac{1}{x} \, dx\) and \(v = \frac{2}{3}x^{\frac{3}{2}}\).
Integration by parts gives:
\(\int x^{\frac{1}{2}} \ln x \, dx = \frac{2}{3}x^{\frac{3}{2}} \ln x - \int \frac{2}{3}x^{\frac{3}{2}} \cdot \frac{1}{x} \, dx\).
\(= \frac{2}{3}x^{\frac{3}{2}} \ln x - \frac{2}{3} \int x^{\frac{1}{2}} \, dx\).
\(= \frac{2}{3}x^{\frac{3}{2}} \ln x - \frac{2}{3} \cdot \frac{2}{3}x^{\frac{3}{2}} + C\).
Evaluate from \(x = 1\) to \(x = 4\):
\(\left[ \frac{2}{3}x^{\frac{3}{2}} \ln x - \frac{4}{9}x^{\frac{3}{2}} \right]_1^4\).
Calculate the definite integral:
\(\left( \frac{2}{3} \cdot 8 \cdot \ln 4 - \frac{4}{9} \cdot 8 \right) - \left( 0 - \frac{4}{9} \right) = 4.28\).
