The given image depicts the expected value of $x^2$, which is the second moment of a probability distribution. The integral is formulated as follows:

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

In the second line of the image, we see that the probability density function (PDF), $f(x)$, is substituted by a normal distribution function of the form:

$E(x^2) = \int_0^{\infty} x^2 \cdot x^2 \cdot e^{-\frac{x^2}{2}} \, dx$

This form suggests we're working with a Gaussian (normal) distribution. Now, let's solve this integral step by step.

Step 1: Simplify the integral

We can rewrite the integral as:

$\int_0^{\infty} x^4 e^{-\frac{x^2}{2}} \, dx$

Step 2: Substitution

Let's use the substitution $u = \frac{x^2}{2}$, which means $du = x \, dx$ and $x^2 = 2u$. This transforms the integral:

$x^4 = 4u^2$ $dx = \frac{du}{x} = \frac{du}{\sqrt{2u}}$

Thus, the integral becomes:

$\int_0^{\infty} 4u^2 e^{-u} \frac{du}{\sqrt{2u}} = 4 \int_0^{\infty} u^{3/2} e^{-u} \, du$

Step 3: Solving the gamma function

This is a standard gamma function integral:

$\int_0^{\infty} u^{n-1} e^{-u} \, du = \Gamma(n)$

For this case, we have $n = \frac{5}{2}$, so:

$\Gamma\left(\frac{5}{2}\right) = \frac{3}{4} \sqrt{\pi}$

Step 4: Final calculation

Thus, the value of the original integral is:

$E(x^2) = 4 \times \frac{3}{4} \sqrt{\pi} = 3 \sqrt{\pi}$

This gives the expected value of $x^2$.

Would you like further details or explanations? Here are some related questions:

1. What is the general form of the Gaussian distribution?
2. How do we derive the gamma function from scratch?
3. What are the properties of the second moment in probability theory?
4. How does this expected value change for other distributions?
5. Can you explain the relationship between moments and variance?

Tip: The gamma function generalizes the factorial function to non-integer values, which is highly useful in integral calculus!