Consider the region D bounded by
x = 0, y = x5 + 1,
and
y = 3 − x2
shown in the figure. Find the area
A(D)
of the region D.

To find the area of the region \( D \) bounded by the curves \( y = x^5 + 1 \), \( y = 3 - x^2 \), and \( x = 0 \), we can proceed with the following steps:

### Step-by-Step Solution:
1. **Identify the Bounds of Integration**:
   - The curves intersect where \( y = x^5 + 1 \) and \( y = 3 - x^2 \). Set the equations equal to each other:
     \[
     x^5 + 1 = 3 - x^2
     \]
     Simplify the equation:
     \[
     x^5 + x^2 - 2 = 0
     \]
   - This is a nonlinear equation and may require numerical methods to solve exactly, but we can analyze approximate solutions by testing values within feasible bounds.
     - Plugging in \( x = 1 \):
       \[
       1^5 + 1^2 - 2 = 1 + 1 - 2 = 0
       \]
       Thus, \( x = 1 \) is a solution.
     - For \( x = -1 \):
       \[
       (-1)^5 + (-1)^2 - 2 = -1 + 1 - 2 = -2 \quad (\text{not a solution})
       \]
   - With \( x = 1 \), it becomes a solution. We now need to verify other potential bounds using approximations or advanced methods. 

2. **Set Up the Integral**:
   - The region is bounded between \( x = 0 \) and \( x = 1 \).
   - Determine which curve is on top within the region. For \( x \in [0, 1] \), substitute \( x = 0.5 \):
     - \( y = x^5 + 1 = (0.5)^5 + 1 \approx 1.03125 \)
     - \( y = 3 - x^2 = 3 - (0.5)^2 = 3 - 0.25 = 2.75 \)
   - Thus, \( y = 3 - x^2 \) is on top, and \( y = x^5 + 1 \) is on the bottom.

3. **Compute the Integral**:
   - The area \( A(D) \) is given by:
     \[
     A(D) = \int_{0}^{1} \left( (3 - x^2) - (x^5 + 1) \right) \, dx
     \]
     Simplify the integrand:
     \[
     A(D) = \int_{0}^{1} \left( 2 - x^2 - x^5 \right) \, dx
     \]

4. **Evaluate the Integral**:
   - Compute each term separately:
     \[
     \int 2 \, dx = 2x \Big|_0^1 = 2
     \]
     \[
     \int x^2 \, dx = \frac{x^3}{3} \Big|_0^1 = \frac{1}{3}
     \]
     \[
     \int x^5 \, dx = \frac{x^6}{6} \Big|_0^1 = \frac{1}{6}
     \]
   - Now, plug these back into the expression for the area:
     \[
     A(D) = 2 - \frac{1}{3} - \frac{1}{6} = 2 - \frac{1}{2} = \frac{3}{2} \approx 1.5
     \]

### Conclusion:
The area \( A(D) \) of the region \( D \) is approximately \( \boxed{1.5} \).

Would you like more details on any step? Here are a few related questions you might find interesting:

1. How do you determine which curve is above or below the other within a region?
2. What are some numerical methods to solve higher-degree polynomial equations?
3. How can we check for additional solutions to the boundary intersection?
4. How does the area change if the boundaries extend to other points?
5. What happens to the integral setup if the region was rotated around the y-axis?

**Tip**: When calculating areas between curves, always confirm which function is on top within the limits to avoid sign errors in the integral.

Consider the region D bounded by x = 0, y = x^5 + 1, and y = 3 − x^2 shown in the figure. Find the area A(D) of the region D.

Explore how to calculate the area of the region bounded by the curves y = x^5 + 1 and y = 3 - x^2, using definite integrals and the area between curves. This advanced calculus problem is suitable for students in Grades 11-12.

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