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The question in the image asks to calculate the area of the region bounded by the curves given by the equations:

1. \( y^2 = x + 1 \)
2. \( y^2 = 9 - x \)

### Steps to Solve:

1. **Rearranging the equations**:
   - The first equation: \( y^2 = x + 1 \) or \( x = y^2 - 1 \)
   - The second equation: \( y^2 = 9 - x \) or \( x = 9 - y^2 \)

2. **Finding the points of intersection**:
   To find where the two curves intersect, set the two equations equal:
   \[
   y^2 - 1 = 9 - y^2
   \]
   Solving for \( y^2 \):
   \[
   2y^2 = 10 \quad \Rightarrow \quad y^2 = 5 \quad \Rightarrow \quad y = \pm \sqrt{5}
   \]
   So, the curves intersect at \( y = \sqrt{5} \) and \( y = -\sqrt{5} \).

3. **Setting up the integral**:
   The area between these two curves can be found by integrating the difference between the two functions \( x = 9 - y^2 \) and \( x = y^2 - 1 \), with respect to \( y \), over the interval \( y = -\sqrt{5} \) to \( y = \sqrt{5} \):
   \[
   A = \int_{-\sqrt{5}}^{\sqrt{5}} \left( (9 - y^2) - (y^2 - 1) \right) \, dy
   \]
   Simplifying the integrand:
   \[
   A = \int_{-\sqrt{5}}^{\sqrt{5}} \left( 10 - 2y^2 \right) \, dy
   \]

4. **Evaluating the integral**:
   We can split the integral:
   \[
   A = \int_{-\sqrt{5}}^{\sqrt{5}} 10 \, dy - 2 \int_{-\sqrt{5}}^{\sqrt{5}} y^2 \, dy
   \]
   Since the function \( y^2 \) is even, we can double the integral from 0 to \( \sqrt{5} \):
   \[
   A = 2 \left( \int_0^{\sqrt{5}} 10 \, dy - 2 \int_0^{\sqrt{5}} y^2 \, dy \right)
   \]

   Now, compute the individual integrals:
   \[
   \int_0^{\sqrt{5}} 10 \, dy = 10 \times \sqrt{5}
   \]
   \[
   \int_0^{\sqrt{5}} y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^{\sqrt{5}} = \frac{(\sqrt{5})^3}{3} = \frac{5\sqrt{5}}{3}
   \]

   Therefore:
   \[
   A = 2 \left( 10\sqrt{5} - 2 \times \frac{5\sqrt{5}}{3} \right)
   \]
   \[
   A = 2 \left( 10\sqrt{5} - \frac{10\sqrt{5}}{3} \right)
   \]
   \[
   A = 2 \times \frac{30\sqrt{5}}{3} = \frac{40\sqrt{5}}{3}
   \]

5. **Approximating the area**:
   Using \( \sqrt{5} \approx 2.236 \), we can calculate:
   \[
   A \approx \frac{40 \times 2.236}{3} = \frac{89.44}{3} \approx 29.81
   \]

So, the area of the region bounded by the curves is approximately **29.81 square units**.

---

Would you like further details or have any questions? Here are 5 related questions to explore:

1. How do we find points of intersection between more complex curves?
2. Can this method be extended to polar coordinates for other problems?
3. What are some typical applications of area under curves in physics?
4. How does the choice of integration limits affect the result?
5. How would the solution differ if the curves were rotated?

**Tip:** When dealing with areas between curves, always ensure the limits of integration and the functions' orders are correct to avoid negative areas.

Вычислите площадь фигуры, ограниченной графиками функций y^2 = x + 1 и y^2 = 9 - x (ответ округлите до сотых)

Learn how to compute the area of the region bounded by the curves y^2 = x + 1 and y^2 = 9 - x. This solution involves finding points of intersection, setting up integrals, and using the fundamental theorem of calculus. Suitable for undergraduate calculus students.

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