1. The function $y=x e^{-a x^{2}}$ is given, where $a$ is a positive parameter.

a) Examine the behavior of the graph of this function for any $a$ and sketch the graph for $a=2$.

b) The extreme points of the given function change their position in the $x O y$ plane as $a$ varies. Determine the curve $y=f(x)$ on which these extreme points lie.

c) Determine the area $P(b)$ of the region bounded by the arc of the graph of the given function, the $x$-axis, and the line $x=b(b>0)$. What is the limit of the area $P(b)$ as $b \rightarrow+\infty$?
Solution. a) The given function is defined for every $x \in \mathbb{R}$ and is odd. The zero and the sign of the function coincide with the zero and the sign of the argument. From $y' = (1 - 2ax^2)e^{-ax^2}$, it follows that the function has stationary points at $x_1 = -\frac{1}{\sqrt{2a}}$ and $x_2 = \frac{1}{\sqrt{2a}}$. At these points, the function has a minimum $y_1 = y(x_1) = -\frac{1}{\sqrt{2ae}}$ and a maximum $y_2 = y(x_2) = \frac{1}{\sqrt{2ae}}$. The intervals of monotonicity and convexity are shown in the given graph. Furthermore, from $y'' = 2ax(2ax^2 - 3)e^{-ax^2}$, we conclude that the function has three inflection points, namely $\left(-\sqrt{\frac{3}{2a}}, -\sqrt{\frac{3}{2ae^3}}\right)$, $\left(\sqrt{\frac{3}{2a}}, \sqrt{\frac{3}{2ae^3}}\right)$, and $(0,0)$.

![](https://cdn.mathpix.com/cropped/2024_06_05_e4cff208225cb6887089g-4.jpg?height=303&width=733&top_left_y=245&top_left_x=496)

b) From $y(x_1) = -\frac{1}{\sqrt{2ae}}$ and $y(x_2) = \frac{1}{\sqrt{2ae}}$, it follows that all extremal points lie on the line $y = \frac{x}{\sqrt{e}}$.

c) We have:

$$
P(b) = \int_{0}^{b} x e^{-ax^2} dx = \frac{1}{2a}\left(1 - e^{-ab^2}\right), \lim_{b \to \infty} P(b) = \frac{1}{2a}
$$