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 where they intersect. We set the equations equal to each other: \[ 2x + 3 = x^2 \] Rearranging gives: \[ x^2 - 2x - 3 = 0 \] This can be factored as: \[ (x - 3)(x + 1) = 0 \] So the points of intersection are \( x = 3 \) and \( x = -1 \). Next, we integrate the difference between the two functions from \( x = -1 \) to \( x = 3 \): \[ \text{Area} = \int_{-1}^{3} ((2x + 3) - (x^2)) \, dx \] Simplifying the integrand: \[ 2x + 3 - x^2 = -x^2 + 2x + 3 \] Now we compute the integral: \[ \int_{-1}^{3} (-x^2 + 2x + 3) \, dx \] Calculating the integral: \[ = \left[-\frac{x^3}{3} + x^2 + 3x \right]_{-1}^{3} \] Evaluating at \( x = 3 \): \[ -\frac{3^3}{3} + 3^2 + 3(3) = -9 + 9 + 9 = 9 \] Now evaluating at \( x = -1 \): \[ -\frac{(-1)^3}{3} + (-1)^2 + 3(-1) = \frac{1}{3} + 1 - 3 = \frac{1}{3} + 1 - 3 = \frac{1}{3} - \frac{9}{3} = -\frac{8}{3} \] Now substituting these values into the expression: \[ \text{Area} = 9 - \left(-\frac{8}{3}\right) = 9 + \frac{8}{3} \] Converting \( 9 \) to thirds: \[ 9 = \frac{27}{3} \] So: \[ \text{Area} = \frac{27}{3} + \frac{8}{3} = \frac{35}{3} \] Thus, the area enclosed between the lines \( y = 2x + 3 \) and \( y = x^2 \) is \[ \frac{35}{3} \text{ square units.} \]