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

To calculate the area enclosed between the line \( y = 2x + 3 \) and the parabola \( y = x^2 \), follow these steps: ### 1. **Find the Points of Intersection** Set the two equations equal to each other to find their points of intersection: \[ 2x + 3 = x^2 \] Rearrange the equation: \[ x^2 - 2x - 3 = 0 \] Solve the quadratic equation using the quadratic formula: \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2} \] So, the solutions are: \[ x = 3 \quad \text{and} \quad x = -1 \] Thus, 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 \( x = 0 \): \[ y_{\text{line}} = 2(0) + 3 = 3 \] \[ y_{\text{parabola}} = 0^2 = 0 \] Since \( y_{\text{line}} > y_{\text{parabola}} \) at \( x = 0 \), 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** First, find the antiderivative: \[ \int (2x + 3 - x^2) \, dx = x^2 + 3x - \frac{1}{3}x^3 \] Evaluate from \( x = -1 \) to \( x = 3 \): \[ \left[ x^2 + 3x - \frac{1}{3}x^3 \right]_{-1}^{3} \] **At \( x = 3 \):** \[ 3^2 + 3(3) - \frac{1}{3}(3)^3 = 9 + 9 - 9 = 9 \] **At \( x = -1 \):** \[ (-1)^2 + 3(-1) - \frac{1}{3}(-1)^3 = 1 - 3 + \frac{1}{3} = -2 + \frac{1}{3} = -\frac{5}{3} \] **Subtract the two results:** \[ 9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{27}{3} + \frac{5}{3} = \frac{32}{3} \] ### **Final Answer** The area enclosed between the two curves is \( \frac{32}{3} \) square units. \[ \boxed{\dfrac{32}{3}} \]