1. **State the problem:** Find the shaded area between the curve $y = x^2$ and the line $y = 2x + 3$ bounded by their points of intersection. 2. **Find points of intersection:** Set $x^2 = 2x + 3$ to find where the parabola and line meet. $$x^2 = 2x + 3$$ $$x^2 - 2x - 3 = 0$$ 3. **Solve the quadratic equation:** $$x^2 - 2x - 3 = 0$$ Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=-2$, $c=-3$: $$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2}$$ So, $$x = \frac{2 + 4}{2} = 3 \quad \text{or} \quad x = \frac{2 - 4}{2} = -1$$ 4. **Determine which function is on top between $x=-1$ and $x=3$:** At $x=0$, $y_{line} = 2(0) + 3 = 3$ and $y_{parabola} = 0^2 = 0$, so the line is above the parabola in this interval. 5. **Set up the integral for the shaded area:** $$\text{Area} = \int_{-1}^{3} \big[(2x + 3) - x^2\big] \, dx$$ 6. **Integrate:** $$\int_{-1}^{3} (2x + 3 - x^2) \, dx = \int_{-1}^{3} ( -x^2 + 2x + 3) \, dx$$ Calculate the antiderivative: $$\left[-\frac{x^3}{3} + x^2 + 3x \right]_{-1}^{3}$$ 7. **Evaluate at the bounds:** At $x=3$: $$-\frac{3^3}{3} + 3^2 + 3(3) = -\frac{27}{3} + 9 + 9 = -9 + 9 + 9 = 9$$ At $x=-1$: $$-\frac{(-1)^3}{3} + (-1)^2 + 3(-1) = -\frac{-1}{3} + 1 - 3 = \frac{1}{3} + 1 - 3 = \frac{1}{3} - 2 = -\frac{5}{3}$$ 8. **Subtract to find the area:** $$9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{27}{3} + \frac{5}{3} = \frac{32}{3}$$ **Final answer:** $$\boxed{\frac{32}{3}}$$ This is the shaded area between the parabola and the line from $x=-1$ to $x=3$.