(i) To find the coordinates of \(M\), we first find the derivative of \(y = x^3 \ln x\) using the product rule:
\(\frac{dy}{dx} = 3x^2 \ln x + x^2\).
Set the derivative to zero to find the critical points:
\(3x^2 \ln x + x^2 = 0\).
Factor out \(x^2\):
\(x^2 (3 \ln x + 1) = 0\).
Since \(x \neq 0\), solve \(3 \ln x + 1 = 0\):
\(\ln x = -\frac{1}{3}\).
\(x = e^{-1/3}\).
Substitute \(x = e^{-1/3}\) back into the original equation to find \(y\):
\(y = (e^{-1/3})^3 \ln(e^{-1/3}) = -\frac{1}{3e}\).
Thus, the coordinates of \(M\) are \(\left( e^{-1/3}, -\frac{1}{3e} \right)\).
(ii) To find the area of the shaded region, integrate \(y = x^3 \ln x\) from \(x = 1\) to \(x = 2\):
\(\int_1^2 x^3 \ln x \, dx\).
Using integration by parts, let \(u = \ln x\) and \(dv = x^3 \, dx\):
\(du = \frac{1}{x} \, dx\), \(v = \frac{x^4}{4}\).
\(\int x^3 \ln x \, dx = \frac{x^4}{4} \ln x - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx\).
\(= \frac{x^4}{4} \ln x - \frac{1}{4} \int x^3 \, dx\).
\(= \frac{x^4}{4} \ln x - \frac{1}{16} x^4 + C\).
Evaluate from \(x = 1\) to \(x = 2\):
\(\left[ \frac{x^4}{4} \ln x - \frac{1}{16} x^4 \right]_1^2\).
\(= \left( \frac{16}{4} \ln 2 - \frac{1}{16} \cdot 16 \right) - \left( \frac{1}{4} \ln 1 - \frac{1}{16} \right)\).
\(= 4 \ln 2 - 1 - 0 + \frac{1}{16}\).
\(= 4 \ln 2 - \frac{15}{16}\).
