The given distribution function is:

$f(x) = x e^{-x/\lambda}, \quad 0 < x < \infty$

To find the mean value of $x$, denoted as $\langle x \rangle$, we need to calculate the expected value of $x$:

$\langle x \rangle = \int_0^\infty x f(x) \, dx = \int_0^\infty x \cdot x e^{-x/\lambda} \, dx$

This simplifies to:

$\langle x \rangle = \int_0^\infty x^2 e^{-x/\lambda} \, dx$

We can solve this using integration by parts or by using a known integral formula for gamma functions. The integral is of the form:

$\int_0^\infty x^n e^{-ax} \, dx = \frac{n!}{a^{n+1}} \quad \text{for } a > 0$

In our case, $n = 2$ and $a = \frac{1}{\lambda}$, so:

$\langle x \rangle = 2! \cdot \lambda^3 = 2 \lambda^3 / \lambda^2 = 2 \lambda$

Thus, the mean value of $x$ is $\langle x \rangle = 2 \lambda$.

The correct answer is:

(b) $2 \lambda$

Would you like any further explanation or details on this process?

Here are 5 related questions you might explore next:

1. How do you compute the variance of $x$ for this distribution?
2. What is the normalization constant for this distribution function?
3. Can you derive the cumulative distribution function (CDF) from the given PDF?
4. What is the skewness of this distribution?
5. How does the expected value change if $f(x)$ is defined over a different interval?

Tip: When dealing with integrals involving exponential functions, the gamma function can often simplify calculations.