1. The problem asks for the area of the region bounded by the curve $y=-x^2 - x - 2$, the x-axis ($y=0$), and the vertical lines $x=-2$ and $x=2$. 2. First, find where the parabola intersects the x-axis by solving $-x^2 - x - 2 = 0$: $$-x^2 - x - 2 = 0 \implies x^2 + x + 2 = 0$$ The discriminant is $\Delta = 1^2 - 4 \cdot 1 \cdot 2 = 1 - 8 = -7 < 0$, so no real roots. 3. Since the curve is below the x-axis (leading coefficient of $-x^2$ is negative and no real roots), the bounded region is below the x-axis and above the parabola. 4. We calculate the area between the curve and the x-axis by integrating the absolute value of the function from $x=-2$ to $x=2$: $$\text{Area} = \int_{-2}^{2} |y| \, dx = \int_{-2}^{2} -(-x^2 - x - 2) \, dx = \int_{-2}^{2} (x^2 + x + 2) \, dx$$ 5. Evaluate the integral: $$\int_{-2}^{2} (x^2 + x + 2) \, dx = \int_{-2}^{2} x^2 \, dx + \int_{-2}^{2} x \, dx + \int_{-2}^{2} 2 \, dx$$ 6. Calculate each term: $$\int_{-2}^{2} x^2 \, dx = \left[\frac{x^3}{3}\right]_{-2}^{2} = \frac{2^3}{3} - \frac{(-2)^3}{3} = \frac{8}{3} - \left(-\frac{8}{3}\right) = \frac{16}{3}$$ $$\int_{-2}^{2} x \, dx = \left[\frac{x^2}{2}\right]_{-2}^{2} = \frac{2^2}{2} - \frac{(-2)^2}{2} = 2 - 2 = 0$$ $$\int_{-2}^{2} 2 \, dx = 2[x]_{-2}^{2} = 2(2 - (-2)) = 2 \times 4 = 8$$ 7. Summing up: $$\text{Area} = \frac{16}{3} + 0 + 8 = \frac{16}{3} + \frac{24}{3} = \frac{40}{3}$$ 8. Therefore, the total area bounded is $\frac{40}{3}$, but this is not among the given options. Rechecking the prompt options and problem: The problem asks for the area bounded by $y = -x^2 - x - 2$, the x-axis, and $x = -2$ and $x = 2$. Since the parabola is always below the x-axis in the interval, the area bounded is exactly the integral of $-(y)$ from $-2$ to $2$, which we computed. Hence, the answer is $\frac{40}{3}$, which is not among the provided options a, b, c, d. Double-checking the polynomial: $y = -x^2 - x - 2 = -(x^2 + x + 2)$ has discriminant negative, so no roots. Therefore, the area bounded is $\frac{40}{3}$. Since none of the given options match this, possibly the problem expects the answer without multiplying by 2 or with different limits. If instead we consider the curves between intersection points of the parabola and x-axis, no real intersection means area is calculated as above. Hence, the closest step is to trust the calculation: **Final answer:** The area is $\frac{40}{3}$. (If options are restricted, none fit exactly.)