4. Figure 1 shows a horizontal line \( y=3 \) intersecting the curve \( y=4 x-x^{2} \) at the points \( A \) and \( B \). Find the area of shaded reg

To find the area of the shaded region between the line \( y = 3 \) and the curve \( y = 4x - x^2 \), we first need to determine the points of intersection. Setting \( 4x - x^2 = 3 \) gives us the equation \( x^2 - 4x + 3 = 0 \). Factoring, we find \( (x - 1)(x - 3) = 0 \), so the points of intersection are \( x = 1 \) and \( x = 3 \). The area between the curve and the line can be calculated by integrating the difference of the functions from \( x = 1 \) to \( x = 3 \): \[ \text{Area} = \int_{1}^{3} \left( (4x - x^2) - 3 \right) \, dx \] This simplifies to: \[ \text{Area} = \int_{1}^{3} (4x - x^2 - 3) \, dx = \int_{1}^{3} (-x^2 + 4x - 3) \, dx \] Calculating the integral: 1. The antiderivative of \( -x^2 + 4x - 3 \) is \( -\frac{x^3}{3} + 2x^2 - 3x \). 2. Evaluating from \( 1 \) to \( 3 \): \[ \left[ -\frac{3^3}{3} + 2(3^2) - 3(3) \right] - \left[ -\frac{1^3}{3} + 2(1^2) - 3(1) \right] \] Calculating at \( x = 3 \): \[ -\frac{27}{3} + 18 - 9 = -9 + 18 - 9 = 0 \] Calculating at \( x = 1 \): \[ -\frac{1}{3} + 2 - 3 = -\frac{1}{3} - 1 = -\frac{4}{3} \] Now subtract: \[ 0 - (-\frac{4}{3}) = \frac{4}{3} \] So, the area of the shaded region is \( \frac{4}{3} \) square units.