1. **Problem Statement:** We approximate the integral $$\int_1^5 \sqrt{x} \, dx$$ using 100 rectangles and right-hand sums. 2. **Step 1: Define the interval and width of each rectangle.** The interval is from 1 to 5, so the total length is $$5 - 1 = 4$$. With 100 rectangles, the width of each rectangle is $$\Delta x = \frac{4}{100} = 0.04$$. 3. **Step 2: Right-hand sum means the height of each rectangle is the function value at the right endpoint of each subinterval.** The first rectangle corresponds to the interval $$[1, 1.04]$$, so its height is $$f(1.04) = \sqrt{1.04}$$. The last rectangle corresponds to the interval $$[4.96, 5]$$, so its height is $$f(5) = \sqrt{5}$$. 4. **Step 3: Calculate the area of the first rectangle.** Area $$= \text{height} \times \text{width} = \sqrt{1.04} \times 0.04$$. Calculate $$\sqrt{1.04} \approx 1.0198$$. So, area $$\approx 1.0198 \times 0.04 = 0.04079$$. 5. **Step 4: Calculate the area of the last rectangle.** Area $$= \sqrt{5} \times 0.04$$. Calculate $$\sqrt{5} \approx 2.2361$$. So, area $$\approx 2.2361 \times 0.04 = 0.08944$$. 6. **Step 5: Determine if the approximation is an overestimate or underestimate.** Since $$f(x) = \sqrt{x}$$ is an increasing and concave down function on $$[1,5]$$, the right-hand sum overestimates the integral because the function values at the right endpoints are always greater than or equal to the function values at the left endpoints. **Final answers:** (a) Area of the first rectangle $$\approx 0.04079$$. (b) Area of the last rectangle $$\approx 0.08944$$. (c) The approximation is an overestimate because $$\sqrt{x}$$ is increasing, so right endpoints give larger heights than left endpoints.