To find the area bounded by the line \( y = 3 - x \), the parabola \( y = x^2 - 9 \), and the conditions \( x \geq -4 \) and \( y \geq 0 \), we need to: 1. Find the points of intersection of the line and the parabola by solving the system of equations: \[ 3 - x = x^2 - 9 \] Rearranging the equation: \[ x^2 + x - 12 = 0 \] Solving this quadratic equation: \[ (x + 4)(x - 3) = 0 \] So, the points of intersection are \( x = -4 \) and \( x = 3 \). 2. Set up the integral to find the area between the curves. We calculate the area between the line and the parabola from \( x = -4 \) to \( x = 3 \): \[ \text{Area} = \int_{-4}^{3} [(3 - x) - (x^2 - 9)] \, dx \] Simplifying the integrand: \[ \text{Area} = \int_{-4}^{3} (12 - x - x^2) \, dx \] 3. Evaluate the integral: \[ \int (12 - x - x^2) \, dx = 12x - \frac{x^2}{2} - \frac{x^3}{3} \] Now, evaluate from \( x = -4 \) to \( x = 3 \): \[ \left[ 12x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-4}^{3} \] Substituting \( x = 3 \): \[ 12(3) - \frac{(3)^2}{2} - \frac{(3)^3}{3} = 36 - \frac{9}{2} - 9 = 36 - 4.5 - 9 = 22.5 \] Substituting \( x = -4 \): \[ 12(-4) - \frac{(-4)^2}{2} - \frac{(-4)^3}{3} = -48 - \frac{16}{2} + \frac{64}{3} = -48 - 8 + 21.33 = -34.67 \] Therefore, the area is: \[ \text{Area} = 22.5 - (-34.67) = 22.5 + 34.67 = 57.17 \]