To find the value of $k$, we use the property that the total probability of a continuous random variable $X$ must integrate to 1 over its entire domain. In this case, the probability density function (PDF) $f(x)$ is non-zero only for $0 \leq x \leq 2$, and the function is given as:

$f(x) = \frac{e^2}{e^2 - k}xe^{-x}, \quad \text{for } 0 \leq x \leq 2$

We need to solve the following equation:

$\int_{-\infty}^{\infty} f(x) \, dx = 1$

However, since $f(x) = 0$ outside the interval $[0, 2]$, the integral simplifies to:

$\int_0^2 \frac{e^2}{e^2 - k}xe^{-x} \, dx = 1$

I'll now compute this integral and solve for $k$.The value of $k$ that satisfies the condition that the total probability is 1 is $k = 3$.

Would you like more details on the steps involved in solving this problem?

Here are 5 related questions you might find interesting:

1. How do you verify if a given function is a valid probability density function (PDF)?
2. What are the key properties of a continuous random variable's PDF?
3. How do you interpret the expected value of a continuous random variable from its PDF?
4. Can you explain how the normalization condition of a PDF ensures its correctness?
5. What is the significance of the exponential distribution in probability theory?

Tip: When solving for unknowns in probability density functions, always ensure the integral of the function over its domain equals 1, as this is a key property of PDFs.