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 $x$ values where they intersect: $$x^2 + 3 = 7 - 3x$$ Rearrange: $$x^2 + 3x - 4 = 0$$ Factor: $$(x + 4)(x - 1) = 0$$ So, $x = -4$ or $x = 1$. 3. **Determine which curve is on top between the intersection points:** Test a point between $-4$ and $1$, say $x=0$: $$y_1 = 0^2 + 3 = 3$$ $$y_2 = 7 - 3(0) = 7$$ Since $7 > 3$, $y = 7 - 3x$ is above $y = x^2 + 3$ in this interval. 4. **Set up the integral for the enclosed area:** The area $A$ is given by: $$A = \int_{-4}^{1} \left[(7 - 3x) - (x^2 + 3)\right] dx = \int_{-4}^{1} (4 - 3x - x^2) dx$$ 5. **Integrate:** $$\int (4 - 3x - x^2) dx = 4x - \frac{3x^2}{2} - \frac{x^3}{3} + C$$ 6. **Evaluate the definite integral:** $$A = \left[4x - \frac{3x^2}{2} - \frac{x^3}{3}\right]_{-4}^{1}$$ Calculate at $x=1$: $$4(1) - \frac{3(1)^2}{2} - \frac{(1)^3}{3} = 4 - \frac{3}{2} - \frac{1}{3} = 4 - 1.5 - 0.3333 = 2.1667$$ Calculate at $x=-4$: $$4(-4) - \frac{3(-4)^2}{2} - \frac{(-4)^3}{3} = -16 - \frac{3(16)}{2} - \frac{-64}{3} = -16 - 24 + 21.3333 = -18.6667$$ Subtract: $$2.1667 - (-18.6667) = 20.8334$$ 7. **Final answer:** The area enclosed by the curves is approximately $20.83$ square units.