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 determine the points at which these two curves intersect. Setting the two equations equal to each other: \[ 2x + 3 = x^2 \] Rearranging gives us: \[ x^2 - 2x - 3 = 0 \] Factoring the quadratic: \[ (x - 3)(x + 1) = 0 \] Thus, the intersection points are \( x = 3 \) and \( x = -1 \). Next, we set up the integral to find the area between the curves from \( x = -1 \) to \( x = 3 \). The area \( A \) is given by: \[ A = \int_{-1}^{3} [(2x + 3) - (x^2)] \, dx \] Calculating this: 1. First, simplify the integrand: \[ 2x + 3 - x^2 = -x^2 + 2x + 3 \] 2. Now integrate: \[ A = \int_{-1}^{3} (-x^2 + 2x + 3) \, dx \] Calculating the integral term-by-term: \[ \int (-x^2) \, dx = -\frac{x^3}{3}, \quad \int 2x \, dx = x^2, \quad \int 3 \, dx = 3x \] Putting it all together, we have: \[ A = \left[ -\frac{x^3}{3} + x^2 + 3x \right]_{-1}^{3} \] Now we evaluate this from \( x = -1 \) to \( x = 3 \): First, calculate at \( x = 3 \): \[ -\frac{3^3}{3} + 3^2 + 3 \cdot 3 = -\frac{27}{3} + 9 + 9 = -9 + 9 + 9 = 9 \] Next, calculate at \( x = -1 \): \[ -\frac{(-1)^3}{3} + (-1)^2 + 3 \cdot (-1) = -\frac{-1}{3} + 1 - 3 = \frac{1}{3} + 1 - 3 = \frac{1}{3} - 2 = \frac{1 - 6}{3} = -\frac{5}{3} \] Now, subtract the two results: \[ A = 9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{27}{3} + \frac{5}{3} = \frac{32}{3} \] Thus, the area enclosed between the lines \( y = 2x + 3 \) and \( y = x^2 \) is \[ \boxed{\frac{32}{3}}. \]