Determine the area of the shaded region in the figure.

xy

y equals xy=x

y equals x squared minus 2y=x2−2

A coordinate system has an unlabeled horizontal x-axis and an unlabeled vertical y-axis. A curve labeled y equals x squared minus 2 is a parabola that opens upward from its vertex on the negative y-axis. A line labeled y equals x rises from left to right and passes through the origin. The curve and line intersect at two points, one in the third quadrant and one in the first quadrant. The region below the x-axis, above the parabola, and above the line in the third quadrant is shaded.

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Part 1

The area of the shaded region is 

enter your response here.

To determine the area of the shaded region bounded by the curves \( y = x^2 - 2 \) and \( y = x \) below the x-axis, we need to follow these steps:

1. **Find the points of intersection between the curves \( y = x \) and \( y = x^2 - 2 \):**
   
   Set \( x = x^2 - 2 \):
   \[
   x = x^2 - 2
   \]
   \[
   x^2 - x - 2 = 0
   \]
   Factor the quadratic equation:
   \[
   (x - 2)(x + 1) = 0
   \]
   Thus, the solutions are:
   \[
   x = 2 \quad \text{and} \quad x = -1
   \]

2. **Determine the bounds of integration:**
   
   The curves intersect at \( x = -1 \) and \( x = 2 \). However, we are interested in the shaded region below the x-axis and above the parabola, which lies between \( x = -2 \) and \( x = -1 \) in the third quadrant.

3. **Set up the integral to find the area:**
   
   The integral representing the area between the curves \( y = x^2 - 2 \) and \( y = x \) from \( x = -2 \) to \( x = -1 \) is:
   \[
   \text{Area} = \int_{-2}^{-1} [(x^2 - 2) - x] \, dx
   \]
   Simplify the integrand:
   \[
   \text{Area} = \int_{-2}^{-1} (x^2 - x - 2) \, dx
   \]

4. **Evaluate the integral:**
   \[
   \int (x^2 - x - 2) \, dx = \frac{x^3}{3} - \frac{x^2}{2} - 2x
   \]
   Evaluate this antiderivative from \( x = -2 \) to \( x = -1 \):
   \[
   \left[ \frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{-2}^{-1}
   \]
   Calculate the value at the bounds:
   \[
   \left( \frac{(-1)^3}{3} - \frac{(-1)^2}{2} - 2(-1) \right) - \left( \frac{(-2)^3}{3} - \frac{(-2)^2}{2} - 2(-2) \right)
   \]
   \[
   = \left( -\frac{1}{3} - \frac{1}{2} + 2 \right) - \left( -\frac{8}{3} - 2 + 4 \right)
   \]
   Simplify each part:
   \[
   -\frac{1}{3} - \frac{1}{2} + 2 = -\frac{2}{6} - \frac{3}{6} + \frac{12}{6} = \frac{7}{6}
   \]
   \[
   -\frac{8}{3} - 2 + 4 = -\frac{8}{3} - \frac{6}{3} + \frac{12}{3} = -\frac{8}{3} - \frac{6}{3} + \frac{12}{3} = -\frac{2}{3}
   \]
   So, we have:
   \[
   \frac{7}{6} - \left( -\frac{2}{3} \right) = \frac{7}{6} + \frac{2}{3} = \frac{7}{6} + \frac{4}{6} = \frac{11}{6}
   \]

Thus, the area of the shaded region is:
\[
\boxed{\frac{11}{6}}
\]

Would you like more details or have any questions about this solution?

---

1. How do you determine the intersection points of two curves?
2. What is the importance of the bounds of integration in definite integrals?
3. How do you factor a quadratic equation?
4. What steps are involved in setting up an integral for finding area between curves?
5. How do you simplify integrands before integrating?
6. Why is it necessary to evaluate the antiderivative at specific bounds?
7. What techniques are used to simplify fractions in integrals?
8. How can you verify the correctness of your calculated area?

**Tip:** Always verify the bounds of integration carefully to ensure they match the region you are interested in.

Learn how to calculate the area of the shaded region between the curves y = x^2 - 2 and y = x below the x-axis. This problem involves finding intersection points, setting up definite integrals, and evaluating antiderivatives. Suitable for advanced high school students in Grades 11-12.

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