To determine the total area of the two regions enclosed by the functions $f(x) = x \cdot e^{-x^2}$ and $g(x) = \frac{1}{e} x$, follow these steps:

Step 1: Find the Points of Intersection

First, we need to find the points where $f(x)$ intersects $g(x)$.

Set $f(x) = g(x)$: $x \cdot e^{-x^2} = \frac{1}{e} x$

Divide both sides by $x$ (assuming $x \neq 0$): $e^{-x^2} = \frac{1}{e}$

Take the natural logarithm of both sides: $-x^2 = -1$ $x^2 = 1$ $x = \pm 1$

So, the points of intersection are $x = 1$ and $x = -1$.

Step 2: Set Up the Integrals for the Enclosed Areas

The total area enclosed by the two curves is the sum of the areas of the individual regions between these points. The two regions are symmetric about the y-axis.

The area of the region from $x = 0$ to $x = 1$: $A_1 = \int_0^1 \left( \frac{1}{e} x - x e^{-x^2} \right) \, dx$

The area of the region from $x = -1$ to $x = 0$: $A_2 = \int_{-1}^0 \left( \frac{1}{e} x - x e^{-x^2} \right) \, dx$

Since the areas are symmetric: $A_1 = A_2$

Step 3: Calculate the Area of One Region

Calculate $A_1$: $A_1 = \int_0^1 \left( \frac{1}{e} x - x e^{-x^2} \right) \, dx$

Integral Calculation

1. $\int_0^1 \frac{1}{e} x \, dx$: $\frac{1}{e} \int_0^1 x \, dx = \frac{1}{e} \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{e} \cdot \frac{1}{2} = \frac{1}{2e}$

2. $\int_0^1 x e^{-x^2} \, dx$: Let $u = x^2$. Then $du = 2x \, dx$ and $x \, dx = \frac{1}{2} \, du$. $\int_0^1 x e^{-x^2} \, dx = \int_0^1 e^{-x^2} x \, dx = \int_0^1 e^{-u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int_0^1 e^{-u} \, du = \frac{1}{2} \left[ -e^{-u} \right]_0^1 = \frac{1}{2} \left( -\frac{1}{e} + 1 \right) = \frac{1}{2} \left( 1 - \frac{1}{e} \right) = \frac{1}{2} - \frac{1}{2e}$

So, $A_1 = \frac{1}{2e} - \left( \frac{1}{2} - \frac{1}{2e} \right) = \frac{1}{2e} - \frac{1}{2} + \frac{1}{2e} = \frac{1}{e} - \frac{1}{2}$

Since the areas are symmetric: $A = 2A_1 = 2 \left( \frac{1}{e} - \frac{1}{2} \right) = \frac{2}{e} - 1$

Final Answer:

The total area of the two regions enclosed by $f(x)$ and $g(x)$ is: $\boxed{\frac{2}{e} - 1}$

If you have any further questions or need more details, feel free to ask! Here are some questions you might be interested in:

1. How can I verify the intersection points of the curves?
2. What is the significance of using definite integrals in this problem?
3. Can you provide a graph for better visualization of the areas?
4. How would the answer change if the functions were different?
5. Are there alternative methods to solve this problem?

Tip: Understanding the properties of exponential functions and their intersections with linear functions can help simplify complex integral calculations.