Calculate Area of the region bnurided by \( F(x)=-x^{2}-2 x \) and \( x \)-axis

To calculate the area of the region bounded by the curve \( F(x) = -x^2 - 2x \) and the \( x \)-axis, follow these steps: ### 1. **Find the Points of Intersection with the \( x \)-Axis** Set \( F(x) \) equal to zero to find the \( x \)-intercepts: \[ -x^2 - 2x = 0 \\ -x(x + 2) = 0 \\ x = 0 \quad \text{or} \quad x = -2 \] So, the curve intersects the \( x \)-axis at \( x = -2 \) and \( x = 0 \). ### 2. **Determine the Sign of \( F(x) \) Between the Intercepts** Let's analyze \( F(x) \): \[ F(x) = -x^2 - 2x = -(x^2 + 2x) = -(x + 1)^2 + 1 \] This is a downward-opening parabola with its vertex at \( x = -1 \) and \( F(-1) = 1 \). Evaluating \( F(x) \) at points between \( x = -2 \) and \( x = 0 \): - At \( x = -2 \): \( F(-2) = 0 \) - At \( x = -1 \): \( F(-1) = 1 \) - At \( x = 0 \): \( F(0) = 0 \) Since \( F(x) \) is non-negative between \( x = -2 \) and \( x = 0 \), the area can be calculated without taking the absolute value. ### 3. **Set Up the Integral for the Area** The area \( A \) is given by the definite integral of \( F(x) \) from \( x = -2 \) to \( x = 0 \): \[ A = \int_{-2}^{0} (-x^2 - 2x) \, dx \] ### 4. **Compute the Integral** \[ \int (-x^2 - 2x) \, dx = -\frac{x^3}{3} - x^2 + C \] Evaluate from \( x = -2 \) to \( x = 0 \): \[ A = \left[ -\frac{0^3}{3} - 0^2 \right] - \left[ -\frac{(-2)^3}{3} - (-2)^2 \right] \\ A = (0) - \left( \frac{8}{3} - 4 \right) \\ A = 0 - \left( \frac{8}{3} - \frac{12}{3} \right) \\ A = 0 - \left( -\frac{4}{3} \right) \\ A = \frac{4}{3} \] ### **Final Answer** The area bounded by the curve \( F(x) = -x^2 - 2x \) and the \( x \)-axis is **\(\frac{4}{3}\) square units**. **Answer:** The bounded area is four-thirds square units.