1. **Stating the problem:** Find the area of the region bounded by the curve $y = x^2 - 3x$ and the line $y = 3 - x$. 2. **Find the points of intersection:** Set the two equations equal to find $x$ values where they intersect: $$x^2 - 3x = 3 - x$$ Rearrange: $$x^2 - 3x + x - 3 = 0$$ $$x^2 - 2x - 3 = 0$$ 3. **Solve the quadratic equation:** $$x^2 - 2x - 3 = 0$$ Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=-2$, $c=-3$: $$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2}$$ $$x = \frac{2 \pm 4}{2}$$ So, $$x = 3 \quad \text{or} \quad x = -1$$ 4. **Determine which function is on top between $x=-1$ and $x=3$:** Test at $x=0$: $$y_{curve} = 0^2 - 3(0) = 0$$ $$y_{line} = 3 - 0 = 3$$ Since $3 > 0$, the line is above the curve in this interval. 5. **Set up the integral for the area:** $$\text{Area} = \int_{-1}^{3} \left[(3 - x) - (x^2 - 3x)\right] dx = \int_{-1}^{3} (3 - x - x^2 + 3x) dx = \int_{-1}^{3} (3 + 2x - x^2) dx$$ 6. **Integrate:** $$\int (3 + 2x - x^2) dx = 3x + x^2 - \frac{x^3}{3} + C$$ 7. **Evaluate definite integral:** $$\left[3x + x^2 - \frac{x^3}{3}\right]_{-1}^{3} = \left(3(3) + 3^2 - \frac{3^3}{3}\right) - \left(3(-1) + (-1)^2 - \frac{(-1)^3}{3}\right)$$ Calculate each part: $$= (9 + 9 - 9) - (-3 + 1 + \frac{1}{3}) = 9 - (-1.666\ldots) = 9 + 1.666\ldots = 10.666\ldots$$ 8. **Final answer:** $$\text{Area} = \frac{32}{3}$$ This is the area of the region bounded by the curve and the line.