https://file.aiquickdraw.com/mathbot/file/17236410572217h3qcate.jpg

To understand how \( k^2 \) is taken as a common factor in the given equations, let's break down the process.

The equation shown in the image is:

\[
\text{Var}(kx) = k^2 \text{Var}(x)
\]

Further expanded, it appears as:

\[
\text{Var}(kx) = \int_{-\infty}^{\infty} (kx)^2 f(x) \, dx - \left[\int_{-\infty}^{\infty} kx f(x) \, dx\right]^2
\]

This simplifies to:

\[
k^2 \int_{-\infty}^{\infty} x^2 f(x) \, dx - k^2 \left[\int_{-\infty}^{\infty} x f(x) \, dx\right]^2
\]

### Step-by-Step Explanation:

1. **Distribute the Square:** 
   - \( (kx)^2 \) becomes \( k^2 x^2 \) when you square \( kx \).
   - This affects the integral term: \( \int (kx)^2 f(x) \, dx \) becomes \( k^2 \int x^2 f(x) \, dx \).

2. **Taking \( k^2 \) as a Common Factor:**
   - In the first term, \( k^2 \) is a constant and can be factored out of the integral: \( k^2 \int x^2 f(x) \, dx \).
   - In the second term, the square applies to the entire expression, so when \( k \) is squared and taken out, it becomes \( k^2 \left[\int x f(x) \, dx\right]^2 \).

3. **Final Expression:**
   - Both terms have a common factor of \( k^2 \), so the expression can be rewritten as:
     \[
     k^2 \left[\int_{-\infty}^{\infty} x^2 f(x) \, dx - \left[\int_{-\infty}^{\infty} x f(x) \, dx\right]^2 \right]
     \]
   - This confirms that \( k^2 \) is factored out correctly.

This process shows that the common factor \( k^2 \) comes from both terms in the variance expression after distributing and simplifying.

Do you want further details or have any other questions?

Here are some related questions you might consider:
1. What is the significance of the variance in probability distributions?
2. How does the scaling factor \( k \) affect the variance?
3. Can you show how \( k \) affects the mean and variance separately?
4. What are the properties of variance under linear transformations?
5. How does one derive the variance formula for continuous distributions?
6. What is the physical interpretation of \( \text{Var}(kx) = k^2 \text{Var}(x) \)?
7. What happens to variance if \( k \) is negative?
8. Can this process be generalized for other moments of a distribution?

**Tip:** Remember that variance scales quadratically with the scaling factor \( k \). This is because variance is a second-order moment, which involves squaring the deviation from the mean.

Explore how k^2 is taken as a common factor in variance equations through detailed mathematical derivation and integration principles. This explanation covers integral calculus and variance properties, suitable for advanced undergraduate students and researchers.

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