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The problem displayed in the image asks:

**"Find the area of the region bounded by:**
- \( y = 1 - x^2 \)
- the \( y \)-axis, \( x \)-axis
- \( x = 1 \) 
- \( x = 2 \)"**

### Steps to Solve:

1. **Understand the function:**
   The equation \( y = 1 - x^2 \) is a downward-opening parabola with its vertex at \( (0, 1) \).

2. **Determine the bounds:**
   - The region is bounded between \( x = 1 \) and \( x = 2 \).
   - The limits of integration will be \( x = 1 \) and \( x = 2 \).

3. **Set up the definite integral:**
   The area between the curve and the \( x \)-axis is given by the integral of the function \( y = 1 - x^2 \) over the interval \([1, 2]\):
   \[
   A = \int_{1}^{2} (1 - x^2) \, dx
   \]

4. **Compute the integral:**
   \[
   \int (1 - x^2) \, dx = x - \frac{x^3}{3}
   \]
   Now, evaluate this expression from \( x = 1 \) to \( x = 2 \):
   \[
   A = \left[ x - \frac{x^3}{3} \right]_{1}^{2}
   \]
   Substituting the bounds:
   \[
   A = \left( 2 - \frac{2^3}{3} \right) - \left( 1 - \frac{1^3}{3} \right)
   \]
   Simplifying:
   \[
   A = \left( 2 - \frac{8}{3} \right) - \left( 1 - \frac{1}{3} \right)
   \]
   \[
   A = \left( \frac{6}{3} - \frac{8}{3} \right) - \left( \frac{3}{3} - \frac{1}{3} \right)
   \]
   \[
   A = \left( \frac{-2}{3} \right) - \left( \frac{2}{3} \right)
   \]
   \[
   A = -\frac{2}{3} - \frac{2}{3} = -\frac{4}{3}
   \]
   Since we are calculating an area, we take the absolute value:
   \[
   A = \frac{4}{3}
   \]

### Final Answer:
The area of the region is \( \frac{4}{3} \) square units.

Do you have any questions or need further details? Here are some related questions for practice:

1. What is the general formula for the area between a curve and the \( x \)-axis?
2. How would the bounds change if the region extended from \( x = -1 \) to \( x = 2 \)?
3. How do you interpret the result if the integral yields a negative value?
4. What is the significance of the parabola's vertex at \( (0, 1) \) for this problem?
5. How would you calculate the area if the curve were \( y = x^2 - 1 \)?

**Tip:** When dealing with definite integrals, always consider the direction of the curve relative to the \( x \)-axis to correctly interpret areas (positive or negative).

Find the area of the region bounded by: y = 1 - x^2, the y-axis, x-axis, x = 1, and x = 2.

Learn how to find the area of the region bounded by the curve y = 1 - x^2, the y-axis, x-axis, and the lines x = 1 and x = 2. This problem covers key calculus concepts, including definite integrals and the area under the curve. Suitable for college-level calculus students.

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