Calculate the area enclosed between the lines \( y = 2x + 3 \) and \( y = x^2 \).

To find the area enclosed between the lines \( y = 2x + 3 \) and \( y = x^2 \), we first need to identify the points where these two curves intersect by setting them equal to each other: \[ 2x + 3 = x^2 \] Rearranging this gives us: \[ x^2 - 2x - 3 = 0 \] Factoring the quadratic: \[ (x - 3)(x + 1) = 0 \] Thus, the solutions are: \[ x = 3 \quad \text{and} \quad x = -1 \] Next, we determine the area between the curves from \( x = -1 \) to \( x = 3 \). The area \( A \) can be calculated using the integral: \[ A = \int_{-1}^{3} ((2x + 3) - x^2) \, dx \] Evaluating the integral: 1. Set up the integral: \[ A = \int_{-1}^{3} (2x + 3 - x^2) \, dx \] 2. Integrate the function: \[ \int (2x + 3 - x^2) \, dx = x^2 + 3x - \frac{x^3}{3} + C \] 3. Evaluate from \( -1 \) to \( 3 \): \[ A = \left[ x^2 + 3x - \frac{x^3}{3} \right]_{-1}^{3} \] Calculating at the upper limit: \[ A(3) = 3^2 + 3(3) - \frac{3^3}{3} = 9 + 9 - 9 = 9 \] Calculating at the lower limit: \[ A(-1) = (-1)^2 + 3(-1) - \frac{(-1)^3}{3} = 1 - 3 + \frac{1}{3} = -2 + \frac{1}{3} = -\frac{6}{3} + \frac{1}{3} = -\frac{5}{3} \] Thus, we find: \[ A = 9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{27}{3} + \frac{5}{3} = \frac{32}{3} \] The area enclosed between the lines \( y = 2x + 3 \) and \( y = x^2 \) is \[ \frac{32}{3} \text{ square units.} \]