Riemann Sums and Notation - AP Calculus BC

1. A Riemann sum is $\sum_{k=1}^{80} \left(\frac{2k}{80}\right)^3\,\frac{2}{80}$. Which integral does it represent?
   Answer: $\displaystyle \int_{0}^{2} x^3\,dx$ 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.
2. Which integral corresponds to $\sum_{i=1}^{n} \ln\!\left(4+\frac{2i}{n}\right)\frac{2}{n}$?
   Answer: $\displaystyle \int_{0}^{2} \ln(4+x)\,dx$ 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($x_i$)).
3. Which definite integral is represented by $\sum_{i=1}^{n} \sqrt{2+\frac{3i}{n}}\,\frac{3}{n}$?
   Answer: $\displaystyle \int_{0}^{3} \sqrt{2+x}\,dx$ 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($x_i$)).
4. The work is approximated by $\sum_{i=1}^{100} \left(5+\sqrt{1+0.02i}\right)(0.02)$. Which integral matches this sum?
   Answer: $\displaystyle \int_{0}^{2} \left(5+\sqrt{1+x}\right)\,dx$ 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.
5. A sum $\sum_{i=1}^{30} \sqrt{9-\left(\frac{i}{10}\right)^2}\,\frac{1}{10}$ represents area. Which integral matches it?
   Answer: $\displaystyle \int_{0}^{3} \sqrt{9-x^2}\,dx$ This problem tests the skill of translating a Riemann sum into its corresponding definite integral notation. The sum ∑_{i=1}^{30} sqrt{9 - left( \frac{i}{10} \right)^2 } \frac{1}{10} has Δx=0.1, $x_i$ = i Δx = i/10, function sqrt{9 - x^2}. Limits from 0 to 3. Thus, ∫_0^3 sqrt{9 - x^2} dx. A tempting distractor is ∫_0^3 sqrt{9 - x^2} \frac{1}{10} dx, mistakenly placing Δx inside the integral. 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.
6. A midpoint Riemann sum is $\sum_{k=1}^{n} \ln\!\left(2+\frac{3}{n}\left(k-\frac12\right)\right)\frac{3}{n}$. Which integral matches it?
   Answer: $\displaystyle \int_{2}^{5} \ln(x)\,dx$ 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.
7. Which integral matches the sum $\sum_{i=1}^{n} e^{1+\frac{i}{n}}\frac{1}{n}$?
   Answer: $\displaystyle \int_{0}^{1} e^{1+x}\,dx$ This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The ($\frac{1}{n}$) is ($\Delta x$), suggesting limits from 0 to 1. The term ($e^{1 + \frac{i}{n}}$) maps to (f($x_i$) = $e^{1 + x_i}$) where ($x_i = \frac{i}{n}$). Thus, the sum represents ($\int_{0}^{1} e^{1 + x} , dx$). Choice D, ($\int_{0}^{1} e^{1 + x} \cdot n , dx$), might tempt but fails by incorrectly multiplying by (n), which inverts ($\Delta x$)'s role. 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($x_i$)).
8. The sum $\sum_{k=1}^{n} \cos\!\left(\pi+\frac{\pi k}{n}\right)\frac{\pi}{n}$ is a Riemann sum. Which integral matches it?
   Answer: $\displaystyle \int_{\pi}^{2\pi} \cos(x)\,dx$ This problem requires translating a Riemann sum into integral notation. The sum $\sum_{k=1}^{n} \cos\!\left(\pi+\frac{\pi k}{n}\right)\frac{\pi}{n}$ samples cosine at points $\pi+\frac{\pi k}{n}$. When $k=0$, we would get $x=\pi$, and when $k=n$, we get $x=\pi+\pi=2\pi$, establishing interval $[\pi,2\pi]$. The width $\frac{\pi}{n}$ confirms the interval length is $2\pi-\pi=\pi$. Choice D with $\cos(\pi x)$ misinterprets the argument as a function transformation. Remember that the pattern inside the function tells us where to evaluate, not how to transform the function itself.
9. A Riemann sum for area is $\sum_{i=1}^{n} g\!\left(-1+\frac{6i}{n}\right)\frac{6}{n}$. Which integral matches it?
   Answer: $\displaystyle \int_{-1}^{5} g(x)\,dx$ This problem asks us to match a Riemann sum to its integral representation. The sum $\sum_{i=1}^{n} g\!\left(-1+\frac{6i}{n}\right)\frac{6}{n}$ samples at points $-1+\frac{6i}{n}$. When $i=0$, we would get $x=-1$, and when $i=n$, we get $x=-1+6=5$, defining interval $[-1,5]$. The width $\frac{6}{n}$ confirms the interval length is $5-(-1)=6$. Choice B extends to $x=6$, which overshoots our endpoint. The key insight is that the coefficient of $i/n$ gives the interval length, and adding it to the constant term gives the right endpoint.
10. The sum $\sum_{k=0}^{n-1} \sqrt{7+\frac{2k}{n}}\,\frac{2}{n}$ is a Riemann sum. Which integral matches it?
    Answer: $\displaystyle \int_{7}^{9} \sqrt{x}\,dx$ This problem requires identifying the integral matching a Riemann sum. The sum $\sum_{k=0}^{n-1} \sqrt{7+\frac{2k}{n}}\,\frac{2}{n}$ evaluates at points $7+\frac{2k}{n}$. When $k=0$, we get $x=7$, and when $k=n-1$, we approach $x=7+2=9$, establishing interval $[7,9]$. The width $\frac{2}{n}$ matches the interval length $9-7=2$. Choice B shows $\sqrt{7+x}$, which incorrectly interprets the argument as a function composition rather than sampling points. The transferable principle is that $f(a+\frac{(b-a)k}{n})$ samples $f(x)$ on $[a,b]$, not $f(a+x)$.
11. The sum $\sum_{i=1}^{n} e^{\,1+\frac{5i}{n}}\frac{5}{n}$ is a Riemann sum. Which integral matches it?
    Answer: $\displaystyle \int_{1}^{6} e^x\,dx$ This problem requires matching a Riemann sum to its integral form. In $\sum_{i=1}^{n} e^{\,1+\frac{5i}{n}}\frac{5}{n}$, the exponent follows the pattern $1+\frac{5i}{n}$. When $i=0$, we would get $x=1$, and when $i=n$, we get $x=1+5=6$, defining interval $[1,6]$. The width factor $\frac{5}{n}$ confirms the interval length is $6-1=5$. Choice D incorrectly interprets the exponent as $e^{1+5x}$, treating the sampling formula as part of the function composition. Remember that in Riemann sums, the argument tells us where to sample, not how to transform the function.
12. The sum $\sum_{i=0}^{n-1} \sin\!\left(2+\frac{3i}{n}\right)\frac{3}{n}$ represents a Riemann sum. Which integral matches it?
    Answer: $\displaystyle \int_{2}^{5} \sin(x)\,dx$ This problem requires translating a Riemann sum to integral notation. In $\sum_{i=0}^{n-1} \sin\!\left(2+\frac{3i}{n}\right)\frac{3}{n}$, the argument follows the pattern $2+\frac{3i}{n}$. When $i=0$, we get $x=2$, and when $i=n-1$, we approach $x=2+3=5$, establishing the interval $[2,5]$. The width factor $\frac{3}{n}$ confirms the interval length is $5-2=3$. Choice D might seem appealing with $\sin(2+x)$, but this misinterprets the sample points as a function composition. The transferable strategy is recognizing that $a+\frac{(b-a)i}{n}$ samples the interval $[a,b]$.
13. A quantity is approximated by $\sum_{k=1}^{200} \frac{1}{\sqrt{1+\left(4+\frac{k}{50}\right)}}\,\frac{1}{50}$. Which integral matches?
    Answer: $\displaystyle \int_{4}^{8} \frac{1}{\sqrt{1+x}}\,dx$ This problem tests the skill of translating a Riemann sum into its corresponding definite integral notation. The sum ∑_{k=1}^{200} \frac{1}{sqrt{1 + ($4 + \frac{k}{50}$) }} \frac{1}{50} has Δx=1/50=0.02, the argument 4 + k/50 = 4 + k Δx. Letting $y_k$ = 4 + k Δx, f(y) = 1/sqrt{1 + y}, limits from 4 to 8. Thus, the sum approximates ∫_4^8 \frac{1}{sqrt{1 + x}} dx. A tempting distractor is ∫_4^8 \frac{1}{sqrt{1 + x}} \frac{1}{50} dx, incorrectly including Δx within the integrand. 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.
14. Which integral is represented by $\sum_{i=1}^{n} \left(\frac{i}{n}\right)\cos\!\left(2\frac{i}{n}\right)\frac{1}{n}$?
    Answer: $\displaystyle \int_{0}^{1} x\cos(2x)\,dx$ 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($x_i$)).
15. Which integral corresponds to $\sum_{i=1}^{n} \left(1+\frac{4i}{n}\right)\frac{4}{n}$?
    Answer: $\displaystyle \int_{0}^{4} (1+x)\,dx$ This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The ($\frac{4}{n}$) term is ($\Delta x$), pointing to an interval from 0 to 4. The summand ($1 + \frac{4i}{n}$) matches (f($x_i$) = 1 + $x_i$) where ($x_i = \frac{4i}{n}$). Therefore, the Riemann sum approximates ($\int_{0}^{4} (1 + x) , dx$). Choice D, ($\int_{0}^{4} (1 + x) \cdot 4 , dx$), tempts by multiplying by 4 but fails as it incorrectly incorporates ($\Delta x$)'s numerator inside the integrand instead of as 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($x_i$)).
16. Which definite integral is represented by $\sum_{i=1}^{10}\left(3+0.2i\right)^2(0.2)$?
    Answer: $\displaystyle \int_{3}^{5}\!x^2\,dx$ This problem requires translating a Riemann sum into its corresponding definite integral notation. The sum $\sum_{i=1}^{10}\left(3+0.2i\right)^2(0.2)$ has the form $\sum f(x_i)\Delta x$ where $f(x_i) = (3+0.2i)^2$ and $\Delta x = 0.2$. Since $i$ ranges from 1 to 10, the sample points are $x_i = 3+0.2i$, giving us $x_1 = 3.2$ and $x_{10} = 5$. With $\Delta x = 0.2$ and 10 subintervals, the interval length is 2, so we integrate from $a = 3$ to $b = 5$. The function being integrated is $f(x) = x^2$ since when $x = 3+0.2i$, we get $(3+0.2i)^2 = x^2$. Choice B incorrectly shifts the function by adding 0.2 inside the squared term. The key strategy is to identify the pattern $x_i = a + i\Delta x$ and recognize that the function of $x_i$ becomes the integrand $f(x)$.
17. The sum $\sum_{j=1}^{n} \left(\left(3+\frac{2j}{n}\right)^2\right)\frac{2}{n}$ is a Riemann sum. Which integral matches it?
    Answer: $\displaystyle \int_{3}^{5} x^2\,dx$ This problem requires translating a Riemann sum into integral notation. The sum $\sum_{j=1}^{n} \left(\left(3+\frac{2j}{n}\right)^2\right)\frac{2}{n}$ evaluates the function at points $3+\frac{2j}{n}$. When $j=0$, we would get $x=3$, and when $j=n$, we get $x=3+2=5$, establishing interval $[3,5]$. The width $\frac{2}{n}$ confirms the interval length is $5-3=2$. Choice B might tempt you with $(3+2x)^2$, but this misinterprets the sampling points as part of the function itself. The key insight is that the expression inside the function gives us sample points, not the function's formula.
18. Which definite integral matches $\sum_{i=1}^{n} \frac{1}{1+\left(\frac{3i}{n}\right)^2}\frac{3}{n}$?
    Answer: $\displaystyle \int_{0}^{3} \frac{1}{1+x^2}\,dx$ This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The term ($\frac{3}{n}$) represents ($\Delta x$), implying limits from 0 to 3. The fraction ($\frac{1}{1 + (\frac{3i}{n})^2}$) corresponds to ($f(x_i) = \frac{1}{1 + x_i^2}$) with ($x_i = \frac{3i}{n}$). Hence, the sum approximates ($\int_{0}^{3} \frac{1}{1 + x^2} , dx$). Choice E, ($\int_{0}^{3} \frac{3}{1 + x^2} , dx$), distracts by placing the 3 in the numerator but fails to recognize it belongs in 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($x_i$)).
19. Which definite integral matches $\sum_{i=1}^{n} \sin\!\left(\frac{\pi i}{n}\right)\frac{\pi}{n}$?
    Answer: $\displaystyle \int_{0}^{\pi} \sin x\,dx$ 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($x_i$)).
20. Which definite integral is represented by $\sum_{i=1}^{n} \left(2-\frac{5i}{n}\right)^3\frac{5}{n}$?
    Answer: $\displaystyle \int_{0}^{5} (2-x)^3\,dx$ 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($x_i$)).

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