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The problem states that \( X \) is a continuous random variable with the probability density function (pdf) given by:

\[
f(x) = \begin{cases} 
2e^{-2x}, & x > 0 \\
0, & \text{otherwise}
\end{cases}
\]

We are asked to find \( E[X^2] \) and \( \text{Var}(X) \).

### Step 1: Finding \( E[X^2] \)

The expectation \( E[X^2] \) is defined as:

\[
E[X^2] = \int_{-\infty}^{\infty} x^2 f(x) \, dx
\]

Given the pdf, we only need to integrate from 0 to \( \infty \) because \( f(x) = 0 \) for \( x \leq 0 \). Thus:

\[
E[X^2] = \int_0^{\infty} x^2 \cdot 2e^{-2x} \, dx
\]

We can use integration by parts to evaluate this integral. Alternatively, the integral of the form \( \int_0^{\infty} x^n e^{-ax} \, dx = \frac{n!}{a^{n+1}} \) can be applied here. For \( n = 2 \) and \( a = 2 \), we get:

\[
E[X^2] = \frac{2!}{2^3} = \frac{2}{8} = \frac{1}{4}
\]

### Step 2: Finding \( E[X] \)

To calculate \( \text{Var}(X) \), we also need \( E[X] \), which is defined as:

\[
E[X] = \int_{0}^{\infty} x \cdot 2e^{-2x} \, dx
\]

Using the same rule for integrals \( \int_0^{\infty} x^n e^{-ax} \, dx = \frac{n!}{a^{n+1}} \), with \( n = 1 \) and \( a = 2 \), we get:

\[
E[X] = \frac{1!}{2^2} = \frac{1}{4}
\]

### Step 3: Finding \( \text{Var}(X) \)

The variance of \( X \), \( \text{Var}(X) \), is given by:

\[
\text{Var}(X) = E[X^2] - (E[X])^2
\]

Substituting the values we found:

\[
\text{Var}(X) = \frac{1}{4} - \left( \frac{1}{4} \right)^2 = \frac{1}{4} - \frac{1}{16} = \frac{4}{16} - \frac{1}{16} = \frac{3}{16}
\]

### Final Answers:

- \( E[X^2] = \frac{1}{4} \)
- \( \text{Var}(X) = \frac{3}{16} \)

Let me know if you'd like any further details or clarification!

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Here are 5 follow-up questions related to this topic:

1. What are the conditions for a function to be a valid probability density function (pdf)?
2. How is the expected value of a continuous random variable related to the concept of center of mass in physics?
3. Can you compute \( E[X^3] \) for the same pdf? What does this represent?
4. How would the variance change if the pdf was scaled by a factor?
5. What is the physical interpretation of variance in terms of probability distributions?

**Tip:** When dealing with exponential functions in probability, knowing key integrals of the form \( \int_0^\infty x^n e^{-ax} \, dx \) can save a lot of time!

Let X be a continuous random variable with pdf given by f(x) = 2e^(-2x), if x > 0, 0 otherwise. Find E[X^2] and Var(X).

Learn how to calculate the expected value and variance for a continuous random variable with PDF f(x) = 2e^(-2x) for x > 0. This problem involves key concepts in probability, such as integration and exponential distribution properties.

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