We need to evaluate the integral:

\[
I = \int_0^{\infty} x^2 e^{-2x} \, dx
\]

### Step 1: Use a known result for Laplace transforms

This is a standard form of an integral that can be solved using the Gamma function or directly using a Laplace transform. The general form of the Gamma function is:

\[
\Gamma(n) = \int_0^{\infty} x^{n-1} e^{-ax} \, dx
\]

For your integral, we have \( x^2 = x^{3-1} \), so we are dealing with the case where \( n = 3 \) and \( a = 2 \).

### Step 2: Apply the formula for the Gamma function

We know that:

\[
\int_0^{\infty} x^{n-1} e^{-ax} \, dx = \frac{\Gamma(n)}{a^n}
\]

In your case:
- \( n = 3 \)
- \( a = 2 \)

So, we can write the integral as:

\[
I = \frac{\Gamma(3)}{2^3}
\]

### Step 3: Simplify

The Gamma function \( \Gamma(3) = 2! = 2 \). Therefore, the integral becomes:

\[
I = \frac{2}{8} = \frac{1}{4}
\]

### Final Answer:

\[
\int_0^{\infty} x^2 e^{-2x} \, dx = \frac{1}{4}
\]

Would you like more details on how this method works, or have any other questions?

### Related Questions:
1. How can the Gamma function be used to solve integrals with different powers of \( x \)?
2. What is the general formula for the Laplace transform of \( x^n \)?
3. How does the Gamma function generalize the factorial function for non-integer values of \( n \)?
4. Can this method be applied to other exponential integrals with different exponents?
5. How would the integral change if the exponent on \( e^{-2x} \) were different, such as \( e^{-3x} \)?

**Tip:** Understanding how the Gamma function generalizes factorials and powers can help simplify many integrals in mathematical analysis and physics.

Evaluate the integral \( \int_0^{\infty} x^2 e^{-2x} \, dx \).

Learn how to evaluate the integral \( \int_0^{\infty} x^2 e^{-2x} \, dx \) using the Gamma function and Laplace transforms. This step-by-step guide covers key concepts in integral calculus and provides a clear explanation of the solution.

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