1. **State the problem:** We need to sketch the curves $y = x^2 + 3$ and $y = 7 - 3x$ and find the area enclosed between them using integration. 2. **Find the points of intersection:** Set the two functions equal to find the limits of integration: $$x^2 + 3 = 7 - 3x$$ Rearranging: $$x^2 + 3x - 4 = 0$$ Factor: $$(x + 4)(x - 1) = 0$$ So, $x = -4$ and $x = 1$ are the points of intersection. 3. **Determine which curve is on top between the intersection points:** Check at $x=0$: $y_1 = 0^2 + 3 = 3$ $y_2 = 7 - 3(0) = 7$ Since $7 > 3$, $y = 7 - 3x$ is above $y = x^2 + 3$ between $x = -4$ and $x = 1$. 4. **Set up the integral for the enclosed area:** $$\text{Area} = \int_{-4}^{1} \left[(7 - 3x) - (x^2 + 3)\right] dx = \int_{-4}^{1} (4 - 3x - x^2) dx$$ 5. **Integrate:** $$\int (4 - 3x - x^2) dx = 4x - \frac{3x^2}{2} - \frac{x^3}{3} + C$$ 6. **Evaluate the definite integral:** $$\left[4x - \frac{3x^2}{2} - \frac{x^3}{3}\right]_{-4}^{1} = \left(4(1) - \frac{3(1)^2}{2} - \frac{1^3}{3}\right) - \left(4(-4) - \frac{3(-4)^2}{2} - \frac{(-4)^3}{3}\right)$$ Calculate each part: At $x=1$: $$4 - \frac{3}{2} - \frac{1}{3} = 4 - 1.5 - 0.3333 = 2.1667$$ At $x=-4$: $$-16 - \frac{3(16)}{2} - \frac{-64}{3} = -16 - 24 + 21.3333 = -18.6667$$ 7. **Subtract:** $$2.1667 - (-18.6667) = 20.8334$$ **Final answer:** The area enclosed by the curves is approximately $20.83$ square units.