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The problem asks to find the region \( R \) and the area bounded by the curves:

- \( y = x^3 \)
- \( x = 2 \)
- \( x = -2 \)

We are also given the result of the area as \( \frac{8}{3} \).

### Step-by-step solution:

1. **Visualizing the problem:**
   - The function \( y = x^3 \) is a cubic curve, symmetric about the origin.
   - The bounds on the x-axis are \( x = -2 \) and \( x = 2 \), which means we are considering the area between these two values of \( x \).
   - The curve does not cross the x-axis except at the origin \( (0, 0) \), so we are working with a region that spans negative and positive \( x \)-values.

2. **Setting up the integral:**
   The area under the curve \( y = x^3 \) between \( x = -2 \) and \( x = 2 \) is given by the integral of \( |y| = |x^3| \) from \( x = -2 \) to \( x = 2 \).
   However, since the curve is symmetric about the y-axis and the areas on both sides of the origin are equal (but with opposite signs), we only need to calculate the integral over the positive half of the curve and multiply by 2.

   \[
   \text{Area} = 2 \int_0^2 x^3 \, dx
   \]

3. **Calculating the integral:**
   We now compute the integral:

   \[
   \int_0^2 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^2
   \]

   Substituting the bounds:

   \[
   \left[ \frac{x^4}{4} \right]_0^2 = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} = 4
   \]

4. **Final area:**
   Now multiply by 2 to account for the symmetric region from \( x = -2 \) to \( x = 2 \):

   \[
   \text{Total Area} = 2 \times 4 = 8
   \]

   Hence, the total area is 8, but since the answer written is \( \frac{8}{3} \), this suggests the problem was possibly asking about the average of some region or another detail.

Find the region R and area bounded by the curves y = x^3, x = 2, and x = -2.

Learn how to find the area of the region bounded by the cubic curve y = x^3, between x = -2 and x = 2. This problem covers key calculus concepts such as integration and symmetry. Suitable for students in Grades 11-12.

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