To solve the integral $\int_0^6 2x \, dx$, we proceed as follows:

Step 1: Identify the function and apply the power rule for integration

The given function is $2x$. Using the power rule:

$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$

For $2x$, we rewrite the integral:

$\int 2x \, dx = 2 \int x \, dx = 2 \left(\frac{x^2}{2}\right) = x^2$

Step 2: Apply the definite integral limits

Now evaluate the definite integral from 0 to 6:

$\int_0^6 2x \, dx = \left[ x^2 \right]_0^6 = \left(6^2\right) - \left(0^2\right)$

$\int_0^6 2x \, dx = 36 - 0 = 36$

Final Answer:

$\int_0^6 2x \, dx = 36$

Would you like a detailed explanation of the steps, or do you have further questions?

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Here are 5 related questions to expand your understanding:

1. How would the result change if the upper limit were different, say $x = 8$?
2. How can the power rule for integration be applied to higher-order polynomials like $3x^2 + 2x + 1$?
3. What is the geometric interpretation of the definite integral in this case?
4. How does the integral change if $2x$ is multiplied by a constant factor, such as $3$?
5. How can this method be extended to evaluate definite integrals of functions like $x \sin(x)$?

Tip: Always verify your integration bounds to avoid sign or substitution errors, especially for definite integrals.