1. **Problem statement:** Calculate the area of the region bounded by the lines and curve: $$y=5$$ (top), $$y=\sqrt{x}$$ (bottom), $$x=1$$ (left), and $$x=4$$ (right). 2. **Formula and approach:** The area between two curves $$y=f(x)$$ and $$y=g(x)$$ from $$x=a$$ to $$x=b$$ is given by: $$\text{Area} = \int_a^b [f(x) - g(x)] \, dx$$ Here, $$f(x) = 5$$ and $$g(x) = \sqrt{x}$$, with $$a=1$$ and $$b=4$$. 3. **Set up the integral:** $$\text{Area} = \int_1^4 \left(5 - \sqrt{x}\right) dx$$ 4. **Calculate the integral:** $$\int_1^4 5 \, dx = 5x \Big|_1^4 = 5(4) - 5(1) = 20 - 5 = 15$$ $$\int_1^4 \sqrt{x} \, dx = \int_1^4 x^{1/2} \, dx = \frac{2}{3} x^{3/2} \Big|_1^4 = \frac{2}{3} (4^{3/2} - 1^{3/2})$$ Calculate powers: $$4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$$ So, $$\int_1^4 \sqrt{x} \, dx = \frac{2}{3} (8 - 1) = \frac{2}{3} \times 7 = \frac{14}{3}$$ 5. **Subtract integrals:** $$\text{Area} = 15 - \frac{14}{3} = \frac{45}{3} - \frac{14}{3} = \frac{31}{3}$$ 6. **Final answer:** $$\boxed{\frac{31}{3}}$$ This is the area of the region bounded by the given curves and lines.