AP Calculus AB Quiz: Riemann Sums And Notation

What this quiz covers

This quiz focuses on Riemann Sums And Notation, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\displaystyle \int_{0}^{\pi/2}\sin(x)\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{1}\sin\left(\frac{\pi}{2}x\right)\,dx$
3. $\displaystyle \int_{0}^{\pi}\sin(x)\,dx$
4. $\displaystyle \int_{0}^{\pi/2}\sin(2x)\,dx$
5. $\displaystyle \int_{0}^{\pi/2}\sin\left(\frac{x}{2}\right)\,dx$

Explanation: This problem tests the skill of translating a Riemann sum into its corresponding definite integral. The factor \frac{\pi}{2n} represents \Delta x, indicating the total interval length is \pi/2. The argument \frac{\pi i}{2n} maps to x_i = i \Delta x, from near 0 to \pi/2. The function is \sin(x), with limits from 0 to \pi/2. A tempting distractor is choice D, \int_{0}^{\pi/2}\sin(2x),dx, which fails by incorrectly doubling the argument of sine. A general strategy for translating Riemann sums is to identify \Delta x = \frac{b-a}{n}, the form of x_i = a + i \Delta x, and the function f(x_i), so the sum approximates \int_a^b f(x) dx as n approaches infinity.

Question 2

1. $\displaystyle \int_{6}^{12} s(x)\,dx$ (correct answer)
2. $\displaystyle \int_{6.75}^{12.75} s(x)\,dx$
3. $\displaystyle \int_{6}^{12} s(x)(0.75)\,dx$
4. $\displaystyle \int_{0}^{8} s(6+0.75x)\,dx$
5. $\displaystyle \int_{6}^{12.75} s(x)\,dx$

Explanation: This question assesses the skill of translating a Riemann sum into its corresponding definite integral. The Riemann sum is (\sum_{n=1}^{8} s(6+0.75n)(0.75)), with (\Delta x = 0.75) as the width. The points are 6 + 0.75n for n from 1 to 8, from 6.75 to 12, covering 8 terms. This right Riemann sum corresponds to (\int_{6}^{12} s(x),dx), total length 8 × 0.75 = 6, from 6 to 12. A tempting distractor like choice B, (\int_{6.75}^{12.75} s(x),dx), fails by using points as limits and adding extra width. A general strategy for translating Riemann sums is to identify (\Delta x), count the terms to find the total length, and determine the limits by aligning the starting point with the overall interval.

Question 3

1. $\displaystyle \int_{1.5}^{3.5} F(x)\,dx$ (correct answer)
2. $\displaystyle \int_{1.55}^{3.55} F(x)\,dx$
3. $\displaystyle \int_{1.5}^{3.5} F(1.5+0.05x)\,dx$
4. $\displaystyle \int_{0}^{40} F(1.5+0.05x)\,dx$
5. $\displaystyle \int_{1.5}^{3.5} F(x)(0.05)\,dx$

Explanation: This question assesses the skill of translating a Riemann sum into its corresponding definite integral. The Riemann sum is (\sum_{i=1}^{40} F(1.5+0.05i)(0.05)), where (\Delta x = 0.05) is the subinterval width. The evaluation points are 1.5 + 0.05i for i from 1 to 40, from 1.55 to 3.5, with 40 terms. This right Riemann sum approximates (\int_{1.5}^{3.5} F(x),dx), total length 40 × 0.05 = 2, from 1.5 to 3.5. A tempting distractor like choice B, (\int_{1.55}^{3.55} F(x),dx), fails by shifting limits to points without maintaining the interval length. A general strategy for translating Riemann sums is to identify (\Delta x), count the terms to find the total length, and determine the limits by aligning the starting point with the overall interval.

Question 4

1. $\displaystyle \int_{0}^{3}x^2\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{1}(3x)^2\,dx$
3. $\displaystyle \int_{0}^{3}(3x)^2\,dx$
4. $\displaystyle \int_{0}^{1}x^2\,dx$
5. $\displaystyle \int_{1}^{3}x^2\,dx$

Explanation: This problem tests the skill of translating a Riemann sum into its corresponding definite integral. The factor \frac{3}{n} represents \Delta x, indicating the total interval length is 3. The expression \left(\frac{3k}{n}\right)^2 maps to x_k = k \Delta x, with function x^2. The limits are from 0 to 3, as points go from near 0 to 3. A tempting distractor is choice C, \int_{0}^{3}(3x)^2,dx, which fails by incorrectly scaling the integrand with 3 inside. A general strategy for translating Riemann sums is to identify \Delta x = \frac{b-a}{n}, the form of x_i = a + i \Delta x, and the function f(x_i), so the sum approximates \int_a^b f(x) dx as n approaches infinity.

Question 5

1. $\displaystyle \int_{5}^{15} h(x)\,dx$ (correct answer)
2. $\displaystyle \int_{5}^{15} h(x)\,(0.2)\,dx$
3. $\displaystyle \int_{5}^{15} h(5+0.2x)\,dx$
4. $\displaystyle \int_{5.2}^{15.2} h(x)\,dx$
5. $\displaystyle \int_{0}^{50} h(5+0.2x)\,dx$

Explanation: This question assesses the skill of translating a Riemann sum into its corresponding definite integral. The Riemann sum is (\sum_{i=1}^{50} h(5+0.2i)(0.2)), where (\Delta x = 0.2) is the width of each subinterval. The evaluation points are 5 + 0.2i for i from 1 to 50, ranging from 5.2 to 15, with 50 terms. This corresponds to a right Riemann sum for (\int_{5}^{15} h(x),dx), with total length 50 × 0.2 = 10, from 5 to 15. A tempting distractor like choice D, (\int_{5.2}^{15.2} h(x),dx), fails by shifting both limits to the evaluation points without preserving the original interval length. A general strategy for translating Riemann sums is to identify (\Delta x), count the terms to find the total length, and determine the limits by aligning the starting point with the overall interval.

Question 6

1. $\displaystyle \int_{0}^{2}e^{x}\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{1}e^{2x}\,dx$
3. $\displaystyle \int_{1}^{2}e^{x}\,dx$
4. $\displaystyle \int_{0}^{2}e^{2x}\,dx$
5. $\displaystyle \int_{0}^{4}e^{x}\,dx$

Explanation: This problem tests the skill of translating a Riemann sum into its corresponding definite integral. The factor \frac{2}{n} represents \Delta x, indicating the total interval length is 2. The exponent \frac{2i}{n} maps to x_i = i \Delta x, with function e^x. The limits are from 0 to 2, as the sample points cover near 0 to 2. A tempting distractor is choice D, \int_{0}^{2}e^{2x},dx, which fails by incorrectly doubling the exponent instead of recognizing the scaling in \Delta x. A general strategy for translating Riemann sums is to identify \Delta x = \frac{b-a}{n}, the form of x_i = a + i \Delta x, and the function f(x_i), so the sum approximates \int_a^b f(x) dx as n approaches infinity.

Question 7

1. $\displaystyle \int_{2}^{4} f(x)\,dx$ (correct answer)
2. $\displaystyle \int_{2}^{4} f(2+0.1x)\,dx$
3. $\displaystyle \int_{2}^{4} f(x)(0.1)\,dx$
4. $\displaystyle \int_{2.1}^{4.1} f(x)\,dx$
5. $\displaystyle \int_{2}^{4} f(x)\,(0.2)\,dx$

Explanation: This question assesses the skill of translating a Riemann sum into its corresponding definite integral. The Riemann sum is (\sum_{i=1}^{20} f(2+0.1i)(0.1)), where (\Delta x = 0.1) represents the width of each subinterval. The evaluation points are 2 + 0.1i for i from 1 to 20, starting at 2.1 and ending at 4, covering 20 subintervals. This setup corresponds to a right Riemann sum approximating the integral (\int_{2}^{4} f(x),dx), as the total length is 20 × 0.1 = 2, spanning from 2 to 4. A tempting distractor like choice D, (\int_{2.1}^{4.1} f(x),dx), fails because it incorrectly uses the evaluation points as limits without accounting for the full interval covered by the sum. A general strategy for translating Riemann sums is to identify (\Delta x), count the terms to find the total length, and determine the limits by aligning the starting point with the overall interval.

Question 8

1. $\displaystyle \int_{0}^{4}x^3\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{1}(4x)^3\,dx$
3. $\displaystyle \int_{0}^{4}(4x)^3\,dx$
4. $\displaystyle \int_{0}^{1}x^3\,dx$
5. $\displaystyle \int_{1}^{4}x^3\,dx$

Explanation: This question tests your ability to translate a Riemann sum into its corresponding definite integral notation. The Riemann sum uses $Δx = 4/n$, indicating an interval width of 4. The argument $4i/n$ increases from approximately 0 to 4 as i goes from 1 to n, mapping to the integral limits from 0 to 4. The integrand $x^3$ corresponds directly to the cubed expression in the sum. A tempting distractor is choice B, which fails because using $(4x)^3$ over 0 to 1 scales by $4^3 = 64$ but integrates over width 1, resulting in 64 times the integral $\int_0^1 x^3 \, dx$, which is too large. A transferable translation strategy is to identify $Δx$ as $(b - a)/n$, determine a and b from the range of the argument as the index varies from 1 to n, and set the integrand to match the function of that argument.

Question 9

1. $\displaystyle \int_{3}^{9} p(x)\,dx$ (correct answer)
2. $\displaystyle \int_{3}^{9} p(x)\,(0.5)\,dx$
3. $\displaystyle \int_{3.5}^{9.5} p(x)\,dx$
4. $\displaystyle \int_{0}^{12} p(3+0.5x)\,dx$
5. $\displaystyle \int_{3}^{9} p(3+0.5x)\,dx$

Explanation: This question assesses the skill of translating a Riemann sum into its corresponding definite integral. The Riemann sum is (\sum_{j=1}^{12} p(3+0.5j)(0.5)), with (\Delta x = 0.5) as the subinterval width. The points are 3 + 0.5j for j from 1 to 12, from 3.5 to 9, covering 12 terms. This is a right Riemann sum approximating (\int_{3}^{9} p(x),dx), as the total length is 12 × 0.5 = 6, spanning 3 to 9. A tempting distractor like choice C, (\int_{3.5}^{9.5} p(x),dx), fails because it adjusts the limits to the points but adds an extra half-width incorrectly. A general strategy for translating Riemann sums is to identify (\Delta x), count the terms to find the total length, and determine the limits by aligning the starting point with the overall interval.

Question 10

1. $\displaystyle \int_{-2}^{1} q(x)\,dx$ (correct answer)
2. $\displaystyle \int_{-1.9}^{1.1} q(x)\,dx$
3. $\displaystyle \int_{-2}^{1} q(-2+0.1x)\,dx$
4. $\displaystyle \int_{-2}^{1} q(x)(0.1)\,dx$
5. $\displaystyle \int_{0}^{30} q(-2+0.1x)\,dx$

Explanation: This question assesses the skill of translating a Riemann sum into its corresponding definite integral. The Riemann sum is (\sum_{i=1}^{30} q(-2+0.1i)(0.1)), with (\Delta x = 0.1) as the width. The points are -2 + 0.1i for i from 1 to 30, from -1.9 to 1, with 30 terms. This is a right Riemann sum for (\int_{-2}^{1} q(x),dx), total length 30 × 0.1 = 3, from -2 to 1. A tempting distractor like choice B, (\int_{-1.9}^{1.1} q(x),dx), fails by shifting limits to points and adding extra width incorrectly. A general strategy for translating Riemann sums is to identify (\Delta x), count the terms to find the total length, and determine the limits by aligning the starting point with the overall interval.

Question 11

1. $\displaystyle \int_{4}^{7.75} r(x)\,dx$
2. $\displaystyle \int_{4}^{8} r(x)\,dx$ (correct answer)
3. $\displaystyle \int_{4.25}^{8.25} r(x)\,dx$
4. $\displaystyle \int_{0}^{15} r(4+0.25x)\,dx$
5. $\displaystyle \int_{4}^{8} r(4+0.25x)\,dx$

Explanation: This question assesses the skill of translating a Riemann sum into its corresponding definite integral. The Riemann sum is (\sum_{m=0}^{15} r(4+0.25m)(0.25)), where (\Delta x = 0.25) is the subinterval width. The points are 4 + 0.25m for m from 0 to 15, from 4 to 7.75, with 16 terms. This left Riemann sum approximates (\int_{4}^{8} r(x),dx), with total length 16 × 0.25 = 4, from 4 to 8. A tempting distractor like choice A, (\int_{4}^{7.75} r(x),dx), fails by limiting the upper bound to the last point without the full interval. A general strategy for translating Riemann sums is to identify (\Delta x), count the terms to find the total length, and determine the limits by aligning the starting point with the overall interval.

Question 12

1. $\displaystyle \int_{0}^{2} (1+x)^{-2}\,dx$
2. $\displaystyle \int_{1}^{3} x^{-2}\,dx$ (correct answer)
3. $\displaystyle \int_{1}^{\frac{23}{8}} x^{-2}\,dx$
4. $\displaystyle \int_{1}^{3} (1+x)^{-2}\,dx$
5. $\displaystyle \int_{0}^{15} \left(1+\frac{x}{8}\right)^{-2}\,dx$

Explanation: This problem asks us to translate $\sum_{k=0}^{15} (1+\frac{k}{8})^{-2}(\frac{1}{8})$ into definite integral notation. The sum has form $\sum f(x_k)\Delta x$ where $f(x_k) = (1+\frac{k}{8})^{-2}$ and $\Delta x = \frac{1}{8}$. When $k=0$, we have $x_0 = 1+\frac{0}{8} = 1$, and when $k=15$, we have $x_{15} = 1+\frac{15}{8} = \frac{23}{8}$. Since this is a left Riemann sum (starting at $k=0$), the integral runs from $x_0 = 1$ to $x_{16} = 1+\frac{16}{8} = 3$. The function is $f(x) = x^{-2}$ because when $x = 1+\frac{k}{8}$, we have $(1+\frac{k}{8})^{-2} = x^{-2}$. A common error would be to write $(1+\frac{x}{8})^{-2}$ as the integrand, failing to recognize that $1+\frac{k}{8}$ represents the $x$-values themselves. The translation strategy is to identify what $x$-values correspond to your indices and express the function purely in terms of $x$.

Question 13

1. $\displaystyle \int_{3}^{5} x\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{2} (5-0.1x)\,dx$
3. $\displaystyle \int_{3}^{5} (5-0.1x)\,dx$
4. $\displaystyle \int_{3}^{5} x\,dx\cdot 0.1$
5. $\displaystyle \int_{5}^{3} x\,dx$

Explanation: This problem requires translating $\sum_{i=1}^{20} (5-0.1i)(0.1)$ into integral notation. The sum follows $\sum f(x_i)\Delta x$ where $f(x_i) = 5-0.1i$ and $\Delta x = 0.1$. When $i=1$, we get $x_1 = 5-0.1(1) = 4.9$, and when $i=20$, we get $x_{20} = 5-0.1(20) = 3$. For a right Riemann sum, we need to find where the rectangles start: at $x_0 = 5-0.1(0) = 5$. So the integral runs from $x_{20} = 3$ to $x_0 = 5$, or equivalently from 3 to 5. The function is $f(x) = x$ because when we solve $x = 5-0.1i$ for the relationship, we get $5-0.1i = x$, so the function value at each $x$ is simply $x$. A mistake would be to use $(5-0.1x)$ as the integrand, which confuses indices with variables. The key insight is that even though the sum uses a decreasing sequence, we still integrate $f(x) = x$ over the interval [3,5].

Question 14

1. $\displaystyle \int_{0}^{2\pi} \cos x\,dx$
2. $\displaystyle \int_{\pi}^{3\pi} \cos x\,dx$ (correct answer)
3. $\displaystyle \int_{\pi}^{3\pi} \cos\!\left(\pi+\frac{x\pi}{10}\right)\,dx$
4. $\displaystyle \int_{\pi}^{3\pi} \cos x\left(\frac{\pi}{10}\right)\,dx$
5. $\displaystyle \int_{\pi}^{2\pi} \cos x\,dx$

Explanation: This problem requires converting $\sum_{m=1}^{20} \cos\!\left(\pi+\frac{m\pi}{10}\right)\left(\frac{\pi}{10}\right)$ to integral form. The argument $\pi+\frac{m\pi}{10}$ tells us the $x$-values where we evaluate cosine. As $m$ goes from 1 to 20, these values range from $\pi+\frac{\pi}{10} = \frac{11\pi}{10}$ to $\pi+\frac{20\pi}{10} = 3\pi$. With $\Delta x = \frac{\pi}{10}$ and 20 rectangles, the interval spans $20 \times \frac{\pi}{10} = 2\pi$, so we integrate from $\frac{11\pi}{10} - \frac{\pi}{10} = \pi$ to $3\pi$. The integrand is simply $\cos x$ evaluated at these points. Choice C incorrectly keeps the linear expression inside the cosine function, not recognizing that $\pi+\frac{m\pi}{10}$ specifies where to evaluate cosine. To convert any trigonometric Riemann sum, identify the evaluation points from the argument expression and integrate the base trig function over that domain.

Question 15

1. $\displaystyle \int_{0}^{1} \frac{1}{1+x^2}\,dx$
2. $\displaystyle \int_{1}^{2} \frac{1}{1+x^2}\,dx$ (correct answer)
3. $\displaystyle \int_{1}^{1.9} \frac{1}{1+x^2}\,dx$
4. $\displaystyle \int_{0}^{9} \frac{1}{1+(1+0.1x)^2}\,dx$
5. $\displaystyle \int_{1}^{2} \frac{1}{1+(1+0.1x)^2}\,dx$

Explanation: This problem requires translating $\sum_{i=0}^{9} \frac{1}{1+(1+0.1i)^2}(0.1)$ into integral notation. The sum follows $\sum f(x_i)\Delta x$ where $f(x_i) = \frac{1}{1+(1+0.1i)^2}$ and $\Delta x = 0.1$. When $i=0$, we get $x_0 = 1+0.1(0) = 1$, and when $i=9$, we get $x_9 = 1+0.1(9) = 1.9$. Since this is a left Riemann sum (starting at $i=0$), the integral runs from $x_0 = 1$ to $x_{10} = 1+0.1(10) = 2$. The function is $f(x) = \frac{1}{1+x^2}$ because when $x = 1+0.1i$, we have $\frac{1}{1+(1+0.1i)^2} = \frac{1}{1+x^2}$. A mistake would be to write $\frac{1}{1+(1+0.1x)^2}$ as the integrand, which incorrectly substitutes the index for the variable. The key insight is that $1+0.1i$ represents the $x$-value itself, so $(1+0.1i)^2 = x^2$.

Question 16

1. $\displaystyle \int_{0}^{3}(x+1)\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{3}(3x+1)\,dx$
3. $\displaystyle \int_{0}^{1}(3x+1)\,dx$
4. $\displaystyle \int_{1}^{4}(x+1)\,dx$
5. $\displaystyle \int_{0}^{1}(x+1)\,dx$

Explanation: This question tests your ability to translate a Riemann sum into its corresponding definite integral notation. The Riemann sum uses Δx = 3/n, indicating an interval width of 3. The argument 3k/n + 1 increases from approximately 1 to 4 as k goes from 1 to n, but the choice maps it to x + 1 over 0 to 3 via substitution. The integrand x + 1 corresponds to the linear expression in the sum after adjusting limits. A tempting distractor is choice B, which fails because using 3x + 1 over 0 to 3 triples the slope, resulting in a larger integral value that doesn't match the sum. A transferable translation strategy is to identify Δx as (b - a)/n, determine a and b from the range of the argument as the index varies from 1 to n, and set the integrand to match the function of that argument.

Question 17

1. $\displaystyle \int_{0}^{1}(4-x)^3\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{4}(1-x)^3\,dx$
3. $\displaystyle \int_{0}^{1}(4-4x)^3\,dx$
4. $\displaystyle \int_{1}^{4}x^3\,dx$
5. $\displaystyle \int_{0}^{1}(4+x)^3\,dx$

Explanation: This problem tests the skill of translating a Riemann sum into its corresponding definite integral. The factor \frac{1}{n} represents \Delta x, indicating the total interval length is 1. The expression 4 - \frac{k}{n} corresponds to sample points decreasing from near 4 to 3, with the function (x)^3. To match, we can set limits from 0 to 1 with integrand (4 - x)^3, capturing the reverse progression. A tempting distractor is choice D, \int_{1}^{4}x^3,dx, which fails because it uses incorrect limits of 1-4 instead of adjusting for the sum's range. A general strategy for translating Riemann sums is to identify \Delta x = \frac{b-a}{n}, the form of x_i = a + i \Delta x, and the function f(x_i), so the sum approximates \int_a^b f(x) dx as n approaches infinity.

Question 18

1. $\displaystyle \int_{0}^{1}x^4\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{4}x\,dx$
3. $\displaystyle \int_{1}^{n}\left(\frac{x}{n}\right)^4\frac{1}{n}\,dx$
4. $\displaystyle \int_{0}^{1}x^5\,dx$
5. $\displaystyle \int_{0}^{1}(4x)^4\,dx$

Explanation: This problem tests the skill of translating a Riemann sum into its corresponding definite integral. The factor \frac{1}{n} represents \Delta x, indicating the total interval length is 1. The expression \left(\frac{k}{n}\right)^4 maps to x_k = k \Delta x, with function x^4. The limits are from 0 to 1, covering the sample points from near 0 to 1. A tempting distractor is choice E, \int_{0}^{1}(4x)^4,dx, which fails by incorrectly introducing a factor of 4 in the integrand. A general strategy for translating Riemann sums is to identify \Delta x = \frac{b-a}{n}, the form of x_i = a + i \Delta x, and the function f(x_i), so the sum approximates \int_a^b f(x) dx as n approaches infinity.

Question 19

1. $\displaystyle \int_{0}^{\pi}x\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{1}\pi x\,dx$
3. $\displaystyle \int_{0}^{\pi}\pi x\,dx$
4. $\displaystyle \int_{0}^{1}x\,dx$
5. $\displaystyle \int_{1}^{\pi}x\,dx$

Explanation: This question tests your ability to translate a Riemann sum into its corresponding definite integral notation. The Riemann sum uses Δx = π/n, indicating an interval width of π. The argument π k/n increases from approximately 0 to π as k goes from 1 to n, mapping to the integral limits from 0 to π. The integrand x corresponds directly to the linear expression in the sum. A tempting distractor is choice B, which fails because integrating π x from 0 to 1 misses an additional π factor, resulting in a value π times smaller than the sum's limit. A transferable translation strategy is to identify Δx as (b - a)/n, determine a and b from the range of the argument as the index varies from 1 to n, and set the integrand to match the function of that argument.

Question 20

1. $\displaystyle \int_{1}^{3.7} g(x)\,dx$
2. $\displaystyle \int_{1}^{4} g(x)\,dx$ (correct answer)
3. $\displaystyle \int_{1}^{4} g(1+0.3x)\,dx$
4. $\displaystyle \int_{1.3}^{4.3} g(x)\,dx$
5. $\displaystyle \int_{0}^{9} g(1+0.3x)\,dx$

Explanation: This question assesses the skill of translating a Riemann sum into its corresponding definite integral. The Riemann sum is (\sum_{k=0}^{9} g(1+0.3k)(0.3)), where (\Delta x = 0.3) is the subinterval width. The evaluation points are 1 + 0.3k for k from 0 to 9, starting at 1 and ending at 3.7, with 10 terms. This forms a left Riemann sum for (\int_{1}^{4} g(x),dx), as the total length is 10 × 0.3 = 3, extending from 1 to 4. A tempting distractor like choice A, (\int_{1}^{3.7} g(x),dx), fails by mistakenly limiting the upper bound to the last evaluation point instead of adding the full width. A general strategy for translating Riemann sums is to identify (\Delta x), count the terms to find the total length, and determine the limits by aligning the starting point with the overall interval.