Final answer:
To express the given integral as a limit of Riemann sums using right endpoints, determine the width of each subinterval as 2/n units and then set up the Riemann sum as the limit of the sum of the areas of the rectangles with height √(3 + (4 + i * (2/n))^2) and width 2/n as n approaches infinity.
Step-by-step explanation:
To express the integral of f(x) = √(3 + x²) from x = 4 to x = 6 as a limit of Riemann sums using right endpoints, we must first define a partition of the interval [4, 6] into n subintervals of equal width.
The width of each subinterval, Δx, is obtained by subtracting the lower limit of integration from the upper limit and then dividing by the number of subintervals: Δx = (b - a) / n = (6 - 4) / n = 2 / n units. The right endpoint for the i-th interval is x_i = a + i * Δx. Therefore, the Riemann sum using right endpoints is the sum from i = 1 to n of f(a + i * Δx) * Δx, which becomes Σ_{i=1}^{n} √(3 + (4 + i * (2/n))^2) * (2/n). The expression as a limit of Riemann sums without evaluation is λ = lim_{n → ∞} Σ_{i=1}^{n} √(3 + (4 + i * (2/n))^2) * (2/n).