7 Multiple Choice 1 point Determine the area under the function \( f(x)=6 x-x^{3} \) and the \( x \)-axis on the interval \( [-2,2] \).

To find the area under the curve \( f(x) = 6x - x^3 \) on the interval \( [-2, 2] \), we will first identify where the function intersects the \( x \)-axis by solving \( 6x - x^3 = 0 \). Factoring gives us \( x(6 - x^2) = 0 \), leading to roots at \( x = 0 \) and \( x = \pm \sqrt{6} \). Between -2 and 2, the relevant roots for this range are \( x = -\sqrt{6} \approx -2.45 \) (out of the interval) and \( x = 0 \). Now we can compute the definite integral over \( [-2, 0] \) and \( [0, 2] \): \[ \text{Area} = \int_{-2}^{2} |f(x)| \, dx = \int_{-2}^{0} (6x - x^3) \, dx + \int_{0}^{2} (6x - x^3) \, dx \] Calculating each integral: 1. **From -2 to 0**: \[ \int (6x - x^3) \, dx = 3x^2 - \frac{x^4}{4} \] Evaluating: \[ \left[ 3(0)^2 - \frac{(0)^4}{4} \right] - \left[ 3(-2)^2 - \frac{(-2)^4}{4} \right] = 0 - \left[ 3(4) - \frac{16}{4} \right] = - (12 - 4) = -8 \] 2. **From 0 to 2**: \[ \int (6x - x^3) \, dx = 3x^2 - \frac{x^4}{4} \] Evaluating: \[ \left[ 3(2)^2 - \frac{(2)^4}{4} \right] - \left[ 3(0)^2 - \frac{(0)^4}{4} \right] = \left[ 12 - 4 \right] - 0 = 8 \] Now, combine the absolute values of the areas: \[ \text{Total Area} = |-8| + |8| = 8 + 8 = 16 \] Thus, the area under the function \( f(x) = 6x - x^3 \) and the \( x \)-axis on the interval \( [-2, 2] \) is **16 square units**.