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The given problem involves finding \( E[X^2] \) and \( \text{Var}(X) \) for a continuous random variable \( X \) with the probability density function (pdf):

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

### Step 1: Compute \( E[X^2] \)
The expected value \( E[X^2] \) is given by:

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

Since the pdf is non-zero only for \( x > 0 \), we only integrate from 0 to infinity:

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

To solve this, we can use integration by parts or recognize it as a gamma function integral. The result of this standard integral is:

\[
E[X^2] = \frac{3}{2}
\]

### Step 2: Compute \( E[X] \)
To find the variance, we also need \( E[X] \), the expected value of \( X \):

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

Again, this is a standard integral, and the result is:

\[
E[X] = \frac{1}{2}
\]

### Step 3: Compute \( \text{Var}(X) \)
Variance is defined as:

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

Substituting the values:

\[
\text{Var}(X) = \frac{3}{2} - \left( \frac{1}{2} \right)^2 = \frac{3}{2} - \frac{1}{4} = \frac{5}{4}
\]

### Final Answers:
- \( E[X^2] = \frac{3}{2} \)
- \( \text{Var}(X) = \frac{5}{4} \)

Would you like further explanation on any step or additional details?

Here are some related questions for practice:
1. How would you find \( E[X^3] \) for this distribution?
2. What is the cumulative distribution function (CDF) of this random variable?
3. How would you verify that \( f(x) \) is a valid probability density function?
4. Can you calculate the skewness of this distribution?
5. How does the expected value change if the pdf is modified to \( f(x) = \lambda e^{-\lambda x} \)?

**Tip**: Always check that the total integral of a pdf equals 1 to ensure it's properly normalized.

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 E[X^2] and the variance of a continuous random variable X with a given pdf f(x) = 2e^(-2x) for x > 0. This detailed solution includes step-by-step calculations using integration and properties of exponential distributions. Suitable for undergraduate students.

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