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To find the area under the curve \( y = 3x^2 - 5x - 3 \) between \( x = 0 \) and \( x = 2 \), we need to compute the definite integral of the function over this interval. Here's the step-by-step process:

1. **Set up the integral**:
   The area under the curve from \( x = 0 \) to \( x = 2 \) is given by:

   \[
   \text{Area} = \int_{0}^{2} (3x^2 - 5x - 3) \, dx
   \]

2. **Find the antiderivative**:
   We integrate term by term:
   - The integral of \( 3x^2 \) is \( x^3 \),
   - The integral of \( -5x \) is \( -\frac{5}{2}x^2 \),
   - The integral of \( -3 \) is \( -3x \).

   So, the antiderivative of \( 3x^2 - 5x - 3 \) is:

   \[
   F(x) = x^3 - \frac{5}{2}x^2 - 3x
   \]

3. **Evaluate the definite integral**:
   Now, evaluate \( F(x) \) at the limits \( x = 2 \) and \( x = 0 \):

   \[
   F(2) = 2^3 - \frac{5}{2}(2^2) - 3(2) = 8 - \frac{5}{2}(4) - 6 = 8 - 10 - 6 = -8
   \]
   \[
   F(0) = 0^3 - \frac{5}{2}(0^2) - 3(0) = 0
   \]

   Therefore, the definite integral is:

   \[
   \text{Area} = F(2) - F(0) = -8 - 0 = -8
   \]

4. **Interpretation**:
   Since the area cannot be negative, we take the absolute value:

   \[
   \text{Area} = 8
   \]

Thus, the area under the curve between \( x = 0 \) and \( x = 2 \) is \( 8 \) square units.

Would you like further clarification or more details on any part of the solution?

Here are some related questions:
1. What is the geometric significance of a negative integral result?
2. How do you find the area between two curves?
3. What is the fundamental theorem of calculus and how does it apply to this problem?
4. How does the integral change if the limits of integration are reversed?
5. How would you approach this problem if the curve were shifted vertically?

**Tip**: Always consider whether the curve dips below the x-axis when interpreting definite integrals—negative values may indicate area below the axis.

Find the area under the curve y = 3x^2 - 5x - 3 between the lines x = 0 and x = 2.

Learn how to calculate the area under the curve y = 3x^2 - 5x - 3 using definite integrals, evaluated between x = 0 and x = 2. This step-by-step solution covers integral calculus concepts suitable for Grade 11-12 students.

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