Determine the area of the shaded region bounded by \( y=-x^{2}+6 x \) and \( y=x^{2}-4 x \). The area of the region is \( \square \). (Type an exact answer.)

To find the area of the shaded region between the curves \( y = -x^2 + 6x \) and \( y = x^2 - 4x \), we first need to find the points of intersection by setting the equations equal to each other: \[ -x^2 + 6x = x^2 - 4x \] Rearranging gives: \[ 0 = 2x^2 - 10x \] Factoring out \( 2x \): \[ 2x(x - 5) = 0 \] This gives us the solutions \( x = 0 \) and \( x = 5 \). Next, we calculate the area between the curves from \( x = 0 \) to \( x = 5 \). The top curve is \( y = -x^2 + 6x \), and the bottom curve is \( y = x^2 - 4x \). The area \( A \) is given by: \[ A = \int_{0}^{5} \left[(-x^2 + 6x) - (x^2 - 4x)\right] dx \] Simplifying the integrand: \[ A = \int_{0}^{5} (-2x^2 + 10x) dx \] Now, we compute the integral: \[ A = \left[-\frac{2}{3}x^3 + 5x^2\right]_{0}^{5} \] Evaluating at the limits: \[ A = \left[-\frac{2}{3}(5^3) + 5(5^2)\right] - \left[-\frac{2}{3}(0^3) + 5(0^2)\right] \] Calculating: \[ A = \left[-\frac{2}{3}(125) + 5(25)\right] \] \[ A = \left[-\frac{250}{3} + 125\right] \] Converting \( 125 \) to thirds: \[ 125 = \frac{375}{3} \] So: \[ A = -\frac{250}{3} + \frac{375}{3} = \frac{125}{3} \] Thus, the area of the shaded region is \[ \boxed{\frac{125}{3}}. \]