Indeed, when integrating with respect to $x$, vertical slices parallel to the $y$-axis are used. If the upper or lower boundary of the region is defined by different functions over different intervals, you need to split the region into subregions and calculate the areas separately. Then, you sum these areas to get the total area. The formula you mentioned is indeed used to find the area between two curves $y = f(x)$ and $y = g(x)$ over an interval $[a, b]$ where $f(x) \ge g(x)$:

$A = \int_a^b \left( f(x) - g(x) \right) \, dx$

Let's apply this method step by step to our problem:

Step-by-Step Solution

1. Find the points of intersection between the curves $y = x$ and $y = x^2 - 2$: $x = x^2 - 2$ $x^2 - x - 2 = 0$ $(x - 2)(x + 1) = 0$ So, the intersection points are $x = 2$ and $x = -1$.

2. Determine the region of interest: The problem states the shaded region is below the x-axis, above the parabola, and above the line $y = x$ in the third quadrant. This corresponds to the interval $x \in [-2, -1]$.

3. Set up the integral: The area $A$ of the shaded region is given by: $A = \int_{-2}^{-1} \left[ (x^2 - 2) - x \right] \, dx$ Simplify the integrand: $A = \int_{-2}^{-1} (x^2 - x - 2) \, dx$

4. Evaluate the integral: Find the antiderivative of $x^2 - x - 2$: $\int (x^2 - x - 2) \, dx = \frac{x^3}{3} - \frac{x^2}{2} - 2x$ 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?

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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.