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 points where they intersect: \[ 2x + 3 = x^2 \] Rearrange the equation: \[ x^2 - 2x - 3 = 0 \] Solve for \( x \): \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2} \] Thus, the solutions are: \[ x = 3 \quad \text{and} \quad x = -1 \] 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 \( x = 3 \) to determine which function is above the other. Let \( x = 0 \): \[ y_{\text{line}} = 2(0) + 3 = 3 \] \[ y_{\text{parabola}} = (0)^2 = 0 \] Since \( y_{\text{line}} > y_{\text{parabola}} \), the line \( y = 2x + 3 \) is above the parabola \( y = x^2 \) in this interval. ### 3. **Set Up the Integral for the Area** The area \( A \) between the two curves from \( x = -1 \) to \( x = 3 \) is: \[ A = \int_{-1}^{3} \left[(2x + 3) - x^2\right] \, dx \] ### 4. **Compute the Integral** Integrate term by term: \[ \int (2x + 3 - x^2) \, dx = \int 2x \, dx + \int 3 \, dx - \int x^2 \, dx = x^2 + 3x - \frac{x^3}{3} \] Evaluate from \( x = -1 \) to \( x = 3 \): \[ A = \left[3^2 + 3(3) - \frac{3^3}{3}\right] - \left[(-1)^2 + 3(-1) - \frac{(-1)^3}{3}\right] \] \[ A = \left[9 + 9 - 9\right] - \left[1 - 3 + \frac{1}{3}\right] = 9 - \left(-\frac{5}{3}\right) = \frac{32}{3} \] ### **Final Answer** \[ \boxed{\dfrac{32}{3}} \]