The given expression represents the expected value $E(x^2)$ for a probability distribution. Specifically, it seems like the expression is related to the Gaussian (normal) distribution. The integral is:

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

Let's simplify this and solve the integral.

1. The integral can be rewritten as: $E(x^2) = \int_0^\infty x^4 \cdot e^{-\frac{x^2}{2}} \, dx$

2. To evaluate this integral, we use the substitution: $u = \frac{x^2}{2}, \quad du = x dx$ Thus, $x^2 = 2u$, and $dx = \frac{du}{x} = \frac{du}{\sqrt{2u}}$.

3. The new limits change from $0$ to $\infty$, and the integral becomes: $E(x^2) = \int_0^\infty (2u)^2 \cdot e^{-u} \cdot \frac{du}{\sqrt{2u}}$

I will now proceed to solve this integral step by step.The solution to the given integral is:

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

This represents the expected value $E(x^2)$, computed for the given expression involving a Gaussian-like probability distribution.

Would you like further explanation or details on the steps involved? Here are some related questions you might find useful:

1. How is this integral related to the Gaussian distribution?
2. What are the properties of expected value and variance in this context?
3. How can this result be generalized for higher moments of $x$?
4. Can this integral be solved using different techniques like the Gamma function?
5. What is the role of substitution in solving integrals like these?

Tip: When solving integrals involving $e^{-x^2}$, it's often helpful to consider substitutions related to Gaussian integrals, as they simplify the process.