1. **State the problem:** Find the total area of the region bounded by the curve $f(x) = \sqrt{x^2 - 2x - 4}$, the x-axis, and the vertical lines $x=2$ and $x=10$. The zero of the function near $x=5$ must be found and rounded to 4 decimals. 2. **Find the zero of the function:** Solve $x^2 - 2x - 4 = 0$ because the square root is zero when the inside is zero. 3. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-4)}}{2} = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm \sqrt{20}}{2}$$ 4. Simplify: $$x = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5}$$ 5. Calculate the roots: $$1 - \sqrt{5} \approx 1 - 2.2361 = -1.2361$$ $$1 + \sqrt{5} \approx 1 + 2.2361 = 3.2361$$ 6. The zero near $x=5$ mentioned is actually $x \approx 3.2361$ (rounded to 4 decimals). 7. **Determine the domain of $f(x)$:** The expression inside the root must be non-negative: $$x^2 - 2x - 4 \geq 0$$ 8. The parabola opens upward, so $f(x)$ is real for $x \leq -1.2361$ or $x \geq 3.2361$. 9. Since the interval is $[2,10]$, the function is defined only for $x \geq 3.2361$ in this interval. 10. **Set up the integral for the total area:** The function is zero between $2$ and $3.2361$ (since inside root is negative), so the area from $2$ to $3.2361$ is zero. 11. The total area is the integral of $f(x)$ from $3.2361$ to $10$: $$\text{Area} = \int_{3.2361}^{10} \sqrt{x^2 - 2x - 4} \, dx$$ 12. **Simplify the integrand:** Complete the square inside the root: $$x^2 - 2x - 4 = (x^2 - 2x + 1) - 1 - 4 = (x - 1)^2 - 5$$ 13. So the integral becomes: $$\int_{3.2361}^{10} \sqrt{(x - 1)^2 - 5} \, dx$$ 14. **Use substitution:** Let $t = x - 1$, then when $x=3.2361$, $t=2.2361$, and when $x=10$, $t=9$. 15. The integral is: $$\int_{2.2361}^{9} \sqrt{t^2 - 5} \, dt$$ 16. **Integral formula:** $$\int \sqrt{t^2 - a^2} \, dt = \frac{t}{2} \sqrt{t^2 - a^2} - \frac{a^2}{2} \ln \left| t + \sqrt{t^2 - a^2} \right| + C$$ where $a^2 = 5$. 17. **Evaluate the definite integral:** $$\left[ \frac{t}{2} \sqrt{t^2 - 5} - \frac{5}{2} \ln \left| t + \sqrt{t^2 - 5} \right| \right]_{2.2361}^{9}$$ 18. Calculate at $t=9$: $$\frac{9}{2} \sqrt{81 - 5} - \frac{5}{2} \ln(9 + \sqrt{76}) = 4.5 \times \sqrt{76} - 2.5 \ln(9 + 8.7178)$$ $$= 4.5 \times 8.7178 - 2.5 \ln(17.7178) = 39.2301 - 2.5 \times 2.875 = 39.2301 - 7.1875 = 32.0426$$ 19. Calculate at $t=2.2361$: $$\frac{2.2361}{2} \sqrt{(2.2361)^2 - 5} - \frac{5}{2} \ln \left| 2.2361 + \sqrt{(2.2361)^2 - 5} \right|$$ Note that $(2.2361)^2 = 5.0000$ approximately, so inside the root is $5 - 5 = 0$. So the first term is: $$\frac{2.2361}{2} \times 0 = 0$$ The logarithm term: $$\ln(2.2361 + 0) = \ln(2.2361) = 0.805$$ So the second term is: $$- \frac{5}{2} \times 0.805 = -2.0125$$ 20. The value at $t=2.2361$ is: $$0 - 2.0125 = -2.0125$$ 21. **Subtract to find the area:** $$32.0426 - (-2.0125) = 32.0426 + 2.0125 = 34.0551$$ 22. **Round to the nearest thousandth:** $$\boxed{34.055}$$ **Final answer:** The total area of the red region is approximately **34.055**.