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 \( y = x^2 \), follow these steps: 1. **Find the Points of Intersection:** Set the two equations equal to each other to find the \( x \)-values where they intersect: \[ 2x + 3 = x^2 \] Rearrange the equation: \[ x^2 - 2x - 3 = 0 \] Solve the quadratic equation: \[ x = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2} \] \[ x = 3 \quad \text{and} \quad x = -1 \] So, the points of intersection are at \( x = -1 \) and \( x = 3 \). 2. **Determine Which Function is on Top:** Choose a test point between \( x = -1 \) and \( x = 3 \), say \( x = 0 \): \[ y_1 = 2(0) + 3 = 3 \] \[ y_2 = (0)^2 = 0 \] Since \( y_1 > y_2 \) at \( x = 0 \), \( y = 2x + 3 \) is the upper function between \( x = -1 \) and \( x = 3 \). 3. **Set Up the Integral for the Area:** The area \( A \) between the curves from \( x = -1 \) to \( x = 3 \) is: \[ A = \int_{-1}^{3} \left( (2x + 3) - x^2 \right) \, dx \] Simplify the integrand: \[ A = \int_{-1}^{3} (-x^2 + 2x + 3) \, dx \] 4. **Integrate and Evaluate:** Find the antiderivative: \[ \int (-x^2 + 2x + 3) \, dx = -\frac{x^3}{3} + x^2 + 3x \] Evaluate from \( x = -1 \) to \( x = 3 \): - At \( x = 3 \): \[ -\frac{3^3}{3} + 3^2 + 3(3) = -9 + 9 + 9 = 9 \] - At \( x = -1 \): \[ -\frac{(-1)^3}{3} + (-1)^2 + 3(-1) = \frac{1}{3} + 1 - 3 = -\frac{5}{3} \] Subtract the lower limit from the upper limit: \[ A = 9 - \left( -\frac{5}{3} \right) = 9 + \frac{5}{3} = \frac{32}{3} \] 5. **Conclusion:** The area enclosed between the lines \( y = 2x + 3 \) and \( y = x^2 \) is: \[ \boxed{\dfrac{32}{3}} \]