https://file.aiquickdraw.com/mathbot/file/1727057473283furcb1kk.jpg

The image shows the calculation of the expectation value \( E(x^2) \), which is typically done for a normal distribution, where the probability density function (PDF) is given by \( f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2 / 2} \). Let's go through each step:

1. **First Line**: \( E(x^2) = \int_{-\infty}^{\infty} x^2 f(x) \, dx \)
   - This is the definition of the expected value of \( x^2 \), where \( f(x) \) is the probability density function (presumably for a normal distribution).

2. **Second Line**: \( E(x^2) = \int_{-\infty}^{\infty} x^2 e^{-x^2 / 2} \, dx \)
   - The PDF \( f(x) \) is replaced with \( e^{-x^2 / 2} \), which implies a standard normal distribution (ignoring constants).

3. **Third Line**: A substitution is made: \( u = \frac{x^2}{2} \), which gives \( x^2 = 2u \) and \( du = x dx \).
   - The integral is transformed using this substitution, and the bounds change accordingly. This leads to:
     \[
     \int_0^\infty (2u) e^{-u} du
     \]

4. **Fourth Line**: Simplifies to:
   \[
   4 \int_0^\infty u e^{-u} du
   \]
   This is a standard Gamma function integral \( \Gamma(n) = \int_0^\infty u^{n-1} e^{-u} du \), with \( n = 2 \).

5. **Final Line**: The result of the Gamma function \( \Gamma(2) = 1! = 1 \), so the integral becomes:
   \[
   4 \times 1 = 4
   \]

### Conclusion:
The steps are correct, and the result \( E(x^2) = 4 \) is true, assuming the constants associated with the normal distribution (like \( \frac{1}{\sqrt{2\pi}} \)) are omitted in the setup.

Let me know if you'd like more details on any of these steps!

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Here are some related questions to expand your understanding:
1. What is the role of the Gamma function in solving integrals?
2. How would the result change if the normalization constant \( \frac{1}{\sqrt{2\pi}} \) was included?
3. Can you explain the method of substitution used in this problem?
4. What is the expected value of \( x^2 \) for a non-standard normal distribution (with mean and variance)?
5. How does this approach differ when calculating the variance \( \text{Var}(x) = E(x^2) - (E(x))^2 \)?

**Tip**: When performing substitutions in integrals, always remember to change the limits of integration to match the new variable of substitution.

Explore the calculation of the expectation value E(x²) for a normal distribution, including the use of integration by substitution and the Gamma function. Suitable for university students studying advanced calculus or probability theory.

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