1. **State the problem:** Find the area of the region between the graphs of $$f(x)=\frac{4}{x}$$ and $$g(x)=5$$ from $$x=-6$$ to $$x=-2$$. 2. **Formula and rules:** The area between two curves $$y=f(x)$$ and $$y=g(x)$$ from $$x=a$$ to $$x=b$$ is given by: $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx$$ Since $$f(x) = \frac{4}{x}$$ and $$g(x) = 5$$, we find which function is on top in the interval $$[-6,-2]$$. 3. **Determine which function is greater:** For $$x<0$$, $$\frac{4}{x}$$ is negative (since numerator positive, denominator negative), and $$5$$ is positive. Thus, $$g(x) = 5$$ is always greater than $$f(x) = \frac{4}{x}$$ on $$[-6,-2]$$. 4. **Set up the integral:** $$\text{Area} = \int_{-6}^{-2} (5 - \frac{4}{x}) \, dx$$ 5. **Integrate:** $$\int (5 - \frac{4}{x}) \, dx = \int 5 \, dx - \int \frac{4}{x} \, dx = 5x - 4 \ln|x| + C$$ 6. **Evaluate definite integral:** $$\text{Area} = \left[5x - 4 \ln|x|\right]_{-6}^{-2} = \left(5(-2) - 4 \ln|-2|\right) - \left(5(-6) - 4 \ln|-6|\right)$$ 7. **Simplify:** $$= (-10 - 4 \ln 2) - (-30 - 4 \ln 6) = -10 - 4 \ln 2 + 30 + 4 \ln 6 = 20 + 4(\ln 6 - \ln 2)$$ 8. **Use log properties:** $$\ln 6 - \ln 2 = \ln \frac{6}{2} = \ln 3$$ 9. **Final expression:** $$\text{Area} = 20 + 4 \ln 3$$ 10. **Check answer choices:** Note that $$\ln 81 = \ln (3^4) = 4 \ln 3$$, so $$20 + 4 \ln 3 = 20 + \ln 81$$ **Answer: D**