Set up a definite integral for each area of the enclosed regions. Then calculate the area. a. \[ \begin{array}{c} y_{1}=x^{2}-4 x+3 \\ y_{2}=-x^{2}+2 x+3 \end{array} \] 1 a. \qquad (2 pts) 1a. Area \( = \) \qquad (1 pt)

To find the area of the enclosed region between the curves \( y_1 = x^2 - 4x + 3 \) and \( y_2 = -x^2 + 2x + 3 \), we will follow these steps: 1. **Find the points of intersection** of the two curves by setting \( y_1 = y_2 \). 2. **Set up the definite integral** for the area between the curves. 3. **Calculate the area** using the definite integral. ### Step 1: Find the Points of Intersection Set the equations equal to each other: \[ x^2 - 4x + 3 = -x^2 + 2x + 3 \] Combine like terms: \[ x^2 + x^2 - 4x - 2x + 3 - 3 = 0 \] This simplifies to: \[ 2x^2 - 6x = 0 \] Factoring out \( 2x \): \[ 2x(x - 3) = 0 \] Setting each factor to zero gives: \[ x = 0 \quad \text{or} \quad x = 3 \] ### Step 2: Set Up the Definite Integral The area \( A \) between the curves from \( x = 0 \) to \( x = 3 \) is given by the integral of the upper curve minus the lower curve. Here, \( y_2 \) is above \( y_1 \) in this interval. Thus, the area can be expressed as: \[ A = \int_{0}^{3} (y_2 - y_1) \, dx = \int_{0}^{3} \left( (-x^2 + 2x + 3) - (x^2 - 4x + 3) \right) \, dx \] This simplifies to: \[ A = \int_{0}^{3} \left( -x^2 + 2x + 3 - x^2 + 4x - 3 \right) \, dx \] Combining like terms: \[ A = \int_{0}^{3} \left( -2x^2 + 6x \right) \, dx \] ### Step 3: Calculate the Area Now we will calculate the definite integral: \[ A = \int_{0}^{3} (-2x^2 + 6x) \, dx \] Calculating the integral: \[ A = \left[ -\frac{2}{3}x^3 + 3x^2 \right]_{0}^{3} \] Evaluating at the bounds: \[ A = \left( -\frac{2}{3}(3)^3 + 3(3)^2 \right) - \left( -\frac{2}{3}(0)^3 + 3(0)^2 \right) \] Calculating the first term: \[ = -\frac{2}{3}(27) + 3(9) = -18 + 27 = 9 \] Thus, the area \( A \) is: \[ A = 9 \] ### Final Answer 1a. Area \( = 9 \)