To evaluate the integral $\int (3x^2 - 2x) \, dx$, follow these steps:

Step-by-Step Solution:

1. Break the integral into two parts: $\int (3x^2 - 2x) \, dx = \int 3x^2 \, dx - \int 2x \, dx$

2. Evaluate each part separately:

   • For $\int 3x^2 \, dx$, apply the power rule $\int x^n \, dx = \frac{x^{n+1}}{n+1}$: $\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3$
   • For $\int 2x \, dx$, again apply the power rule: $\int 2x \, dx = 2 \cdot \frac{x^2}{2} = x^2$

3. Combine the results: $\int (3x^2 - 2x) \, dx = x^3 - x^2 + C$ where $C$ is the constant of integration.

Final Answer:

$\int (3x^2 - 2x) \, dx = x^3 - x^2 + C$

Would you like a deeper explanation on any step, or do you have further questions?

Here are 5 related questions:

1. How would you integrate a higher degree polynomial like $5x^4 - 3x^2 + 7$?
2. What changes if the integral has limits, like $\int_1^3 (3x^2 - 2x) \, dx$?
3. How do you apply the power rule for negative exponents or fractions?
4. Can you explain the difference between indefinite and definite integrals?
5. What happens when integrating exponential or trigonometric functions?

Tip: Always remember to add the constant $C$ when solving indefinite integrals!