Find the area enclosed by the given curves. \( y=2 x-x^{2}, y=2 x-4 \)

To find the area enclosed by the curves \( y = 2x - x^2 \) and \( y = 2x - 4 \), we first need to determine their points of intersection. Setting the equations equal to each other: \[ 2x - x^2 = 2x - 4 \] Simplifying, we find: \[ -x^2 + 4 = 0 \] \[ x^2 = 4 \] \[ x = 2 \quad \text{or} \quad x = -2 \] Next, we'll find the area between these curves from \( x = -2 \) to \( x = 2 \). The area \( A \) is given by the integral: \[ A = \int_{-2}^{2} ((2x - x^2) - (2x - 4)) \, dx \] \[ = \int_{-2}^{2} (4 - x^2) \, dx \] Calculating the integral: \[ A = \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2} \] Evaluating at the bounds: For \( x = 2 \): \[ 4(2) - \frac{(2)^3}{3} = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3} \] For \( x = -2 \): \[ 4(-2) - \frac{(-2)^3}{3} = -8 + \frac{8}{3} = -\frac{24}{3} + \frac{8}{3} = -\frac{16}{3} \] Now, subtracting the two evaluations gives: \[ A = \frac{16}{3} - \left(-\frac{16}{3}\right) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3} \] Thus, the area enclosed by the curves is \( \frac{32}{3} \) square units.