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The question asks to determine the area between the curve \( y = x^3 - 2x^2 - x + 2 \) and the x-axis over the interval \( x \in [-1, 2] \). This involves finding the definite integral of the given function and accounting for the areas above and below the x-axis.

To solve this:

1. **Set up the integral**:
   The total area between the curve and the x-axis can be found by integrating the function from \( x = -1 \) to \( x = 2 \). Since the graph shows that part of the function dips below the x-axis, we need to break the integral into sections where the function is positive and where it is negative, ensuring that we take the absolute value of the area where the curve is below the x-axis.

   The integral setup is:
   \[
   \text{Area} = \int_{-1}^{0} |f(x)| \, dx + \int_{0}^{2} f(x) \, dx
   \]
   where \( f(x) = x^3 - 2x^2 - x + 2 \).

2. **Find the roots** of \( y = x^3 - 2x^2 - x + 2 = 0 \) (if any) between the limits, which helps in determining the intervals where the function is above or below the x-axis.

3. **Calculate the integrals**:
   We calculate the definite integral from \( -1 \) to \( 2 \), ensuring to take absolute values where necessary.

I'll now proceed to solve this step-by-step by first finding the roots of the equation and then integrating over the necessary intervals.The function \( f(x) = x^3 - 2x^2 - x + 2 \) has roots at \( x = -1 \), \( x = 1 \), and \( x = 2 \). This means the function crosses the x-axis at these points.

We can now break the integral into two intervals:
1. From \( x = -1 \) to \( x = 1 \), where the function is below the x-axis.
2. From \( x = 1 \) to \( x = 2 \), where the function is above the x-axis.

The total area is calculated by:
\[
\text{Area} = \int_{-1}^{1} |f(x)| \, dx + \int_{1}^{2} f(x) \, dx
\]
I will now compute these integrals.It seems like I can't do more advanced data analysis right now. Please try again later. 

However, you can manually compute the integrals as follows:
- For the interval \([-1, 1]\), integrate the negative of the function to account for the area below the x-axis.
- For the interval \([1, 2]\), integrate the function as it is above the x-axis.

Would you like me to guide you further through these steps?

Here are 5 related questions you might find useful:
1. How can I verify my integral calculations?
2. What is the geometric significance of the areas above and below the x-axis?
3. How do you handle integrals of polynomial functions in general?
4. What other techniques can be used to calculate the area under a curve?
5. How can I use symmetry in graphs to simplify area calculations?

**Tip:** Always sketch the graph of the function if possible to visualize the area you're calculating.

Given the function y = x^3 - 2x^2 - x + 2, determine the area between the curve and the x-axis over the interval x ∈ [-1, 2].

Learn how to find the area between the curve y = x^3 - 2x^2 - x + 2 and the x-axis over the interval [-1, 2] using definite integrals. This problem involves key calculus concepts such as definite integrals and the area under a curve.

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