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The problem asks to find the total area of the shaded regions between two curves: 

- \( y = \frac{x^3}{3} - x \) (the cubic curve)
- \( y = \frac{x}{3} \) (the straight line)

The points of intersection given on the graph are:  
- \( (-2, -\frac{2}{3}) \)
- \( (0, 0) \)
- \( (3, 6) \)

### Steps to solve:

1. **Find the points of intersection:**
   The intersection points are already given as \( x = -2 \), \( x = 0 \), and \( x = 3 \).

2. **Set up the integral for the total area:**
   We need to compute the areas in two regions:
   - From \( x = -2 \) to \( x = 0 \), where the cubic curve is above the straight line.
   - From \( x = 0 \) to \( x = 3 \), where the straight line is above the cubic curve.

3. **Area between curves formula:**
   The area between two curves is given by:
   \[
   A = \int_{a}^{b} \left[ f(x) - g(x) \right] dx
   \]
   where \( f(x) \) is the upper curve and \( g(x) \) is the lower curve.

4. **Set up the integrals:**
   For \( x \in [-2, 0] \), the cubic function \( y = \frac{x^3}{3} - x \) is above the line \( y = \frac{x}{3} \). So:
   \[
   A_1 = \int_{-2}^{0} \left( \left[ \frac{x^3}{3} - x \right] - \frac{x}{3} \right) dx
   \]

   For \( x \in [0, 3] \), the line \( y = \frac{x}{3} \) is above the cubic function \( y = \frac{x^3}{3} - x \). So:
   \[
   A_2 = \int_{0}^{3} \left( \frac{x}{3} - \left[ \frac{x^3}{3} - x \right] \right) dx
   \]

5. **Compute the integrals:**

   - First integral \( A_1 \):
     \[
     A_1 = \int_{-2}^{0} \left( \frac{x^3}{3} - x - \frac{x}{3} \right) dx = \int_{-2}^{0} \left( \frac{x^3}{3} - \frac{4x}{3} \right) dx
     \]
   
   - Second integral \( A_2 \):
     \[
     A_2 = \int_{0}^{3} \left( \frac{x}{3} - \left( \frac{x^3}{3} - x \right) \right) dx = \int_{0}^{3} \left( \frac{4x}{3} - \frac{x^3}{3} \right) dx
     \]

6. **Calculate each integral:**

   Let's compute the total area by integrating these two expressions step-by-step.

I will now calculate the results for each integral.The total area of the shaded regions is:

- The area from \( x = -2 \) to \( x = 0 \) (Region 1) is \( \frac{4}{3} \).
- The area from \( x = 0 \) to \( x = 3 \) (Region 2) is \( \frac{-3}{4} \), but we take the absolute value, so it's \( \frac{3}{4} \).

Thus, the total area of the shaded regions is:
\[
\frac{25}{12} \approx 2.08 \text{ square units}.
\]

Would you like more details on any step or clarification? Here are some related questions:

1. How is the area between curves formula derived?
2. What does taking the absolute value of the integral represent in this context?
3. How does changing the order of functions affect the integral result?
4. Can we visualize the integrals without using calculus?
5. What happens if we shift the curves vertically or horizontally?

**Tip:** Always check the order of the curves when setting up integrals for area, as it affects whether the integrand is positive or negative.

Find the total areas of the shaded region between the curves y = x^3 / 3 - x and y = x / 3, as shown in the graph.

This problem explores how to find the total area between the cubic curve y = x^3 / 3 - x and the linear function y = x / 3 using definite integrals. The solution involves computing the area in two regions based on given intersection points.

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