(i) To find the minimum point \(M\), we first find the derivative of \(y = x^2 \ln x\) using the product rule:
\(\frac{dy}{dx} = 2x \ln x + x\).
Set the derivative to zero to find critical points:
\(2x \ln x + x = 0\).
\(x(2 \ln x + 1) = 0\).
Since \(x = 0\) is not in the domain, solve \(2 \ln x + 1 = 0\):
\(\ln x = -\frac{1}{2}\).
\(x = e^{-\frac{1}{2}}\).
Substitute back to find \(y\):
\(y = (e^{-\frac{1}{2}})^2 \ln(e^{-\frac{1}{2}}) = e^{-1}(-\frac{1}{2}) = -\frac{1}{2}e^{-1}\).
Thus, \(M \left( e^{-\frac{1}{2}}, -\frac{1}{2}e^{-1} \right)\).
(ii) To find the area of the shaded region, integrate \(y = x^2 \ln x\) from \(x = 1\) to \(x = e\):
Use integration by parts:
Let \(u = \ln x\), \(dv = x^2 dx\).
Then \(du = \frac{1}{x} dx\), \(v = \frac{x^3}{3}\).
\(\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx\).
\(= \frac{x^3}{3} \ln x - \frac{1}{3} \int x^2 \, dx\).
\(= \frac{x^3}{3} \ln x - \frac{1}{9} x^3\).
Evaluate from \(x = 1\) to \(x = e\):
\(\left[ \frac{x^3}{3} \ln x - \frac{1}{9} x^3 \right]_1^e\).
\(= \left( \frac{e^3}{3} \cdot 1 - \frac{1}{9} e^3 \right) - \left( \frac{1}{3} \cdot 0 - \frac{1}{9} \right)\).
\(= \frac{e^3}{3} - \frac{1}{9} e^3 + \frac{1}{9}\).
\(= \frac{1}{9}(2e^3 + 1)\).
