(a) To find the coordinates of \(M\), we need to find the maximum point of the function \(y = xe^{-\frac{1}{4}x^2}\). Differentiate \(y\) with respect to \(x\):
\(\frac{dy}{dx} = e^{-\frac{1}{4}x^2} - \frac{x^2}{2}e^{-\frac{1}{4}x^2}\).
Set \(\frac{dy}{dx} = 0\) to find critical points:
\(e^{-\frac{1}{4}x^2} (1 - \frac{x^2}{2}) = 0\).
Since \(e^{-\frac{1}{4}x^2} \neq 0\), solve \(1 - \frac{x^2}{2} = 0\):
\(x^2 = 2\) \(\Rightarrow x = \sqrt{2}\).
Substitute \(x = \sqrt{2}\) back into the original equation to find \(y\):
\(y = \sqrt{2}e^{-\frac{1}{4}(\sqrt{2})^2} = \sqrt{2}e^{-\frac{1}{2}}\).
Thus, the coordinates of \(M\) are \(\left( \sqrt{2}, \sqrt{2}e^{-\frac{1}{2}} \right)\).
(b) To find the area under the curve from \(x = 0\) to \(x = 3\), use the substitution \(x = \sqrt{u}\), so \(dx = \frac{1}{2\sqrt{u}} du\).
The integral becomes:
\(\int xe^{-\frac{1}{4}x^2} \, dx = \int \frac{1}{2} e^{-\frac{1}{4}u} \, du\).
Change the limits: when \(x = 0, u = 0\) and when \(x = 3, u = 9\).
Integrate:
\(\frac{1}{2} \int e^{-\frac{1}{4}u} \, du = -2e^{-\frac{1}{4}u} \bigg|_0^9\).
Calculate the definite integral:
\(-2e^{-\frac{9}{4}} + 2e^0 = 2 - 2e^{-\frac{9}{4}}\).
Thus, the exact area of the shaded region is \(2 - 2e^{-\frac{9}{4}}\).
