1. **State the problem:** Find the area under the curve of the function $f(x) = x^2 - 3x - 4$ where the graph is below the x-axis. 2. **Find the roots of the function:** Solve $x^2 - 3x - 4 = 0$ to find the points where the parabola intersects the x-axis. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-3$, and $c=-4$. Calculate the discriminant: $$b^2 - 4ac = (-3)^2 - 4(1)(-4) = 9 + 16 = 25$$ Calculate the roots: $$x = \frac{-(-3) \pm \sqrt{25}}{2(1)} = \frac{3 \pm 5}{2}$$ So, $$x_1 = \frac{3 - 5}{2} = \frac{-2}{2} = -1$$ $$x_2 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$ 3. **Determine where the function is below the x-axis:** Since the parabola opens upward and roots are at $x=-1$ and $x=4$, the function is below the x-axis between $x=-1$ and $x=4$. 4. **Set up the integral for the area under the x-axis:** The area under the x-axis is the integral of the negative of the function (to get positive area) from $x=-1$ to $x=4$: $$\text{Area} = -\int_{-1}^{4} (x^2 - 3x - 4) \, dx$$ 5. **Calculate the integral:** Find the antiderivative: $$\int (x^2 - 3x - 4) \, dx = \frac{x^3}{3} - \frac{3x^2}{2} - 4x + C$$ Evaluate from $-1$ to $4$: $$\left[ \frac{x^3}{3} - \frac{3x^2}{2} - 4x \right]_{-1}^{4} = \left( \frac{4^3}{3} - \frac{3 \cdot 4^2}{2} - 4 \cdot 4 \right) - \left( \frac{(-1)^3}{3} - \frac{3 \cdot (-1)^2}{2} - 4 \cdot (-1) \right)$$ Calculate each part: $$\frac{4^3}{3} = \frac{64}{3}$$ $$\frac{3 \cdot 4^2}{2} = \frac{3 \cdot 16}{2} = 24$$ $$4 \cdot 4 = 16$$ So the first part: $$\frac{64}{3} - 24 - 16 = \frac{64}{3} - 40 = \frac{64 - 120}{3} = \frac{-56}{3}$$ Calculate the second part: $$\frac{(-1)^3}{3} = \frac{-1}{3}$$ $$\frac{3 \cdot (-1)^2}{2} = \frac{3 \cdot 1}{2} = \frac{3}{2}$$ $$4 \cdot (-1) = -4$$ So the second part: $$-\frac{1}{3} - \frac{3}{2} + 4 = -\frac{1}{3} - \frac{3}{2} + 4$$ Find common denominator 6: $$-\frac{2}{6} - \frac{9}{6} + \frac{24}{6} = \frac{-2 - 9 + 24}{6} = \frac{13}{6}$$ 6. **Subtract:** $$\frac{-56}{3} - \frac{13}{6} = \frac{-112}{6} - \frac{13}{6} = \frac{-125}{6}$$ 7. **Calculate the area:** Since the integral is negative (area below x-axis), multiply by $-1$: $$\text{Area} = - \left( \frac{-125}{6} \right) = \frac{125}{6} \approx 20.833$$ 8. **Round to the nearest thousandth:** $$20.833$$ **Final answer:** $$\boxed{20.833}$$