This problem tests the skill of translating a Riemann sum into its corresponding definite integral notation. The sum ∑${i=1}^{100}$ (5 + sqrt(1 + 0.02 i)) (0.02) has Δx = 0.02, x_i = i Δx = 0.02 i, so the term inside sqrt is 1 + x_i, function 5 + sqrt(1 + x). The limits are from 0 to 2, as i=1 to 100, x from 0.02 to 2. Thus, the sum is a right Riemann sum for ∫$0^2$ (5 + sqrt(1 + x)) dx. A tempting distractor is ∫_$0^2$ (5 + sqrt(1 + x)) (0.02) dx, but this incorrectly includes the Δx inside the integral instead of recognizing it as the width multiplier in the sum. To translate any Riemann sum to an integral, identify Δx, express the argument in terms of x_k = a + k Δx, determine the limits from the range of x_k, and form the integral of the resulting function over those limits.

This problem tests the skill of translating a Riemann sum into its corresponding definite integral notation. The sum $ \sum_{k=1}^{80} \left( \frac{2k}{80} \right)^3 \frac{2}{80} $ has $ \Delta x = \frac{2}{80} = 0.025 $, $ x_k = k \Delta x = \frac{2k}{80} $. The term is $ (x_k)^3 $, so function $ x^3 $. Limits from 0 to 2. A tempting distractor is $ \int_0^{80} \left( \frac{2x}{80} \right)^3 dx $, which incorrectly sets the upper limit to 80 without adjusting $ \Delta x $. To translate any Riemann sum to an integral, identify $ \Delta x $, express the argument in terms of $ x_k = a + k \Delta x $, determine the limits from the range of $ x_k $, and form the integral of the resulting function over those limits.

This problem asks us to translate a midpoint Riemann sum to integral notation. The sum $\sum_{k=1}^{n} \ln!\left(2+\frac{3}{n}\left(k-\frac12\right)\right)\frac{3}{n}$ samples at midpoints. When $k=1$, we get $x=2+\frac{3}{n}\cdot\frac{1}{2}$, and when $k=n$, we get approximately $x=2+3\left(1-\frac{1}{2n}\right)$, which approaches $5$ as $n$ increases. The width $\frac{3}{n}$ confirms interval length $5-2=3$. Choice E might seem appealing with $\ln(2+3x)$, but this misinterprets the sampling points as a function transformation. The key is recognizing that midpoint sums still approximate $\int_a^b f(x),dx$, not a transformed function.

This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The term ($\frac{\pi}{n}$) is (\Delta x), indicating an interval length of (\pi), typically from 0 to (\pi). The argument of the sine, ($\frac{\pi i}{n}$), is the sample point (x_i = $\frac{\pi i}{n}$), so the function is (f(x) = \sin x). The sum thus represents the integral (\int_${0}^{\pi}$ \sin x , dx) as (n) grows large. Choice C, (\int_${0}^{\pi}$ \sin(\pi x) , dx), is a common distractor but fails because it mistakenly places (\pi) inside the sine instead of recognizing the direct mapping to (\sin x). To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. Here, ($\frac{3}{n}$) serves as (\Delta x), implying an interval length of 3, from 0 to 3. The expression ($\sqrt{2 + \frac{3i}{n}$}) corresponds to (f(x_i) = $\sqrt{2 + x_i}$) with (x_i = $\frac{3i}{n}$). As such, the sum converges to (\int_${0}^{3}$ $\sqrt{2 + x}$ , dx). Choice D, (\int_${0}^{3}$ $\sqrt{2 + x}$ \cdot 3 , dx), is a distractor that wrongly multiplies the integrand by 3, confusing the role of (\Delta x)'s numerator. To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The term ($\frac{2}{n}$) represents the width of each subinterval, (\Delta x), so the total interval length is 2, suggesting limits from 0 to 2. The expression inside the sum, (3 + $\frac{2i}{n}$), corresponds to the sample point (x_i = $\frac{2i}{n}$) evaluated in the function (f(x) = (3 + $x)^2$). As (n) approaches infinity, this right Riemann sum approximates the integral (\int_${0}^{2}$ (3 + $x)^2$ , dx). A tempting distractor like choice D, (\int_${0}^{2}$ (3 + $2x)^2$ , dx), fails because it incorrectly doubles the coefficient of (x) instead of matching the sum's structure. To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The ($\frac{5}{n}$) acts as (\Delta x), indicating an interval from 0 to 5. The cubed term (2 - $\frac{5i}{n}$) corresponds to (f(x_i) = (2 - x_$i)^3$) with (x_i = $\frac{5i}{n}$). Consequently, the Riemann sum approximates (\int_${0}^{5}$ (2 - $x)^3$ , dx). A distractor like choice D, (\int_${0}^{5}$ (2 - $x)^3$ \cdot 5 , dx), fails by erroneously including the 5 inside the integrand rather than as part of the differential. To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The ($\frac{1}{n}$) is (\Delta x), pointing to an interval from 0 to 1. The product ($\frac{i}{n}$ \cos(2 $\frac{i}{n}$)) maps to (f(x_i) = x_i \cos(2 x_i)) where (x_i = $\frac{i}{n}$). The sum thus represents (\int_${0}^{1}$ x \cos(2x) , dx). Choice B, (\int_${0}^{2}$ x \cos(2x) , dx), might tempt by extending the upper limit to 2 but fails as the interval is determined by (\Delta x = $\frac{1}{n}$), not ($\frac{2}{n}$). To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The ($\frac{2}{n}$) is (\Delta x), so the interval is from 0 to 2. The logarithm (\ln(4 + $\frac{2i}{n}$)) matches (f(x_i) = \ln(4 + x_i)) where (x_i = $\frac{2i}{n}$). This sum therefore converges to (\int_${0}^{2}$ \ln(4 + x) , dx). Choice D, (\int_${0}^{2}$ \ln(4 + x) \cdot 2 , dx), is tempting but incorrect as it multiplies the integrand by 2 instead of using it in the dx term. To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).

This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The ($\frac{1}{n}$) serves as (\Delta x), indicating limits from 0 to 1. The expression (5 - $\frac{i}{n}$) corresponds to (f(x_i) = 5 - x_i) with (x_i = $\frac{i}{n}$). Therefore, the Riemann sum approximates (\int_${0}^{1}$ (5 - x) , dx). Choice B, (\int_${0}^{5}$ (5 - x) , dx), is a distractor that incorrectly expands the interval to 5, missing that the interval length is 1 based on (\Delta x). To translate Riemann sums generally, identify (\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = \lim n \cdot \Delta x), and rewrite the summand as (f(x_i)).