AP Calculus BC Quiz: Riemann Sums And Notation
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Riemann Sums And NotationQuestion 1 of 20

A quantity is approximated by ∑k=120011+(4+k50) 150\sum_{k=1}^{200} \frac{1}{\sqrt{1+\left(4+\frac{k}{50}\right)}}\,\frac{1}{50}. Which integral matches?

∫4811+x dx\displaystyle \int_{4}^{8} \frac{1}{\sqrt{1+x}}\,dx
∫0411+(4+x) dx\displaystyle \int_{0}^{4} \frac{1}{\sqrt{1+(4+x)}}\,dx
∫420011+(4+x50) dx\displaystyle \int_{4}^{200} \frac{1}{\sqrt{1+\left(4+\frac{x}{50}\right)}}\,dx
∫4811+x 150 dx\displaystyle \int_{4}^{8} \frac{1}{\sqrt{1+x}}\,\frac{1}{50}\,dx
∫4.02811+x dx\displaystyle \int_{4.02}^{8} \frac{1}{\sqrt{1+x}}\,dx
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AP Calculus BC Quiz

AP Calculus BC Quiz: Riemann Sums And Notation

Practice Riemann Sums And Notation in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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 BC.

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.

All questions

Question 1

A quantity is approximated by ∑k=120011+(4+k50) 150\sum_{k=1}^{200} \frac{1}{\sqrt{1+\left(4+\frac{k}{50}\right)}}\,\frac{1}{50}. Which integral matches?

  1. ∫4811+x dx\displaystyle \int_{4}^{8} \frac{1}{\sqrt{1+x}}\,dx (correct answer)
  2. ∫0411+(4+x) dx\displaystyle \int_{0}^{4} \frac{1}{\sqrt{1+(4+x)}}\,dx
  3. ∫420011+(4+x50) dx\displaystyle \int_{4}^{200} \frac{1}{\sqrt{1+\left(4+\frac{x}{50}\right)}}\,dx
  4. ∫4811+x 150 dx\displaystyle \int_{4}^{8} \frac{1}{\sqrt{1+x}}\,\frac{1}{50}\,dx
  5. ∫4.02811+x dx\displaystyle \int_{4.02}^{8} \frac{1}{\sqrt{1+x}}\,dx
Explanation: 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 + k50\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.

Question 2

Which definite integral is represented by ∑i=1n(2−5in)35n\sum_{i=1}^{n} \left(2-\frac{5i}{n}\right)^3\frac{5}{n}?

  1. ∫05(2−5x)3 dx\displaystyle \int_{0}^{5} (2-5x)^3\,dx
  2. ∫−32x3 dx\displaystyle \int_{-3}^{2} x^3\,dx
  3. ∫05(2−x)3 dx\displaystyle \int_{0}^{5} (2-x)^3\,dx (correct answer)
  4. ∫05(2−x)3 5 dx\displaystyle \int_{0}^{5} (2-x)^3\,5\,dx
  5. ∫05(2−5xn)35n dx\displaystyle \int_{0}^{5} \left(2-\frac{5x}{n}\right)^3\frac{5}{n}\,dx
Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The (5n\frac{5}{n}) acts as (Δ\Delta x), indicating an interval from 0 to 5. The cubed term (2 - 5in\frac{5i}{n}) corresponds to (f(x_i) = (2 - x_i)^3) with (x_i = 5in\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 (Δ\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = lim⁡\lim n ⋅\cdot Δ\Delta x), and rewrite the summand as (f(x_i)).

Question 3

Which integral is represented by ∑i=1n(in)cos⁡ ⁣(2in)1n\sum_{i=1}^{n} \left(\frac{i}{n}\right)\cos\!\left(2\frac{i}{n}\right)\frac{1}{n}?

  1. ∫01xcos⁡(2x) dx\displaystyle \int_{0}^{1} x\cos(2x)\,dx (correct answer)
  2. ∫02xcos⁡(2x) dx\displaystyle \int_{0}^{2} x\cos(2x)\,dx
  3. ∫01xcos⁡(x) dx\displaystyle \int_{0}^{1} x\cos(x)\,dx
  4. ∫01nxcos⁡(2x) dx\displaystyle \int_{0}^{1} nx\cos(2x)\,dx
  5. ∫01(xn)cos⁡ ⁣(2xn) dx\displaystyle \int_{0}^{1} \left(\frac{x}{n}\right)\cos\!\left(2\frac{x}{n}\right)\,dx
Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The (1n\frac{1}{n}) is (Δ\Delta x), pointing to an interval from 0 to 1. The product (\frac{i}{n} \cos(2 in\frac{i}{n})) maps to (f(x_i) = x_i \cos(2 x_i)) where (x_i = in\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 (Δ\Delta x = 1n\frac{1}{n}), not (2n\frac{2}{n}). To translate Riemann sums generally, identify (Δ\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = lim⁡\lim n ⋅\cdot Δ\Delta x), and rewrite the summand as (f(x_i)).

Question 4

Which integral corresponds to ∑i=1nln⁡ ⁣(4+2in)2n\sum_{i=1}^{n} \ln\!\left(4+\frac{2i}{n}\right)\frac{2}{n}?

  1. ∫46ln⁡x dx\displaystyle \int_{4}^{6} \ln x\,dx
  2. ∫02ln⁡(4+x) dx\displaystyle \int_{0}^{2} \ln(4+x)\,dx (correct answer)
  3. ∫02ln⁡(4+2x) dx\displaystyle \int_{0}^{2} \ln(4+2x)\,dx
  4. ∫02ln⁡(4+x) 2 dx\displaystyle \int_{0}^{2} \ln(4+x)\,2\,dx
  5. ∫1nln⁡ ⁣(4+2xn)2n dx\displaystyle \int_{1}^{n} \ln\!\left(4+\frac{2x}{n}\right)\frac{2}{n}\,dx
Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The (2n\frac{2}{n}) is (Δ\Delta x), so the interval is from 0 to 2. The logarithm (\ln(4 + 2in\frac{2i}{n})) matches (f(x_i) = \ln(4 + x_i)) where (x_i = 2in\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 (Δ\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = lim⁡\lim n ⋅\cdot Δ\Delta x), and rewrite the summand as (f(x_i)).

Question 5

Which definite integral matches the sum ∑i=1n(5−in)1n\sum_{i=1}^{n} \left(5-\frac{i}{n}\right)\frac{1}{n}?

  1. ∫01(5−x) dx\displaystyle \int_{0}^{1} (5-x)\,dx (correct answer)
  2. ∫05(5−x) dx\displaystyle \int_{0}^{5} (5-x)\,dx
  3. ∫45x dx\displaystyle \int_{4}^{5} x\,dx
  4. ∫01(5−nx) dx\displaystyle \int_{0}^{1} (5-nx)\,dx
  5. ∫01(5−x) 1n dx\displaystyle \int_{0}^{1} (5-x)\,\frac{1}{n}\,dx
Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The (1n\frac{1}{n}) serves as (Δ\Delta x), indicating limits from 0 to 1. The expression (5 - in\frac{i}{n}) corresponds to (f(x_i) = 5 - x_i) with (x_i = in\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 (Δ\Delta x). To translate Riemann sums generally, identify (Δ\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = lim⁡\lim n ⋅\cdot Δ\Delta x), and rewrite the summand as (f(x_i)).

Question 6

The work is approximated by ∑i=1100(5+1+0.02i)(0.02)\sum_{i=1}^{100} \left(5+\sqrt{1+0.02i}\right)(0.02). Which integral matches this sum?

  1. ∫02(5+1+x) dx\displaystyle \int_{0}^{2} \left(5+\sqrt{1+x}\right)\,dx (correct answer)
  2. ∫02(5+1+0.02x) dx\displaystyle \int_{0}^{2} \left(5+\sqrt{1+0.02x}\right)\,dx
  3. ∫0100(5+1+0.02x) dx\displaystyle \int_{0}^{100} \left(5+\sqrt{1+0.02x}\right)\,dx
  4. ∫0.022(5+1+x) dx\displaystyle \int_{0.02}^{2} \left(5+\sqrt{1+x}\right)\,dx
  5. ∫02(5+1+x)(0.02) dx\displaystyle \int_{0}^{2} \left(5+\sqrt{1+x}\right)(0.02)\,dx
Explanation: 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.

Question 7

A Riemann sum is ∑k=180(2k80)3 280\sum_{k=1}^{80} \left(\frac{2k}{80}\right)^3\,\frac{2}{80}. Which integral does it represent?

  1. ∫02x3 dx\displaystyle \int_{0}^{2} x^3\,dx (correct answer)
  2. ∫080(2x80)3 dx\displaystyle \int_{0}^{80} \left(\frac{2x}{80}\right)^3\,dx
  3. ∫02(2x80)3 dx\displaystyle \int_{0}^{2} \left(\frac{2x}{80}\right)^3\,dx
  4. ∫02x3 280 dx\displaystyle \int_{0}^{2} x^3\,\frac{2}{80}\,dx
  5. ∫080x3 dx\displaystyle \int_{0}^{80} x^3\,dx
Explanation: This problem tests the skill of translating a Riemann sum into its corresponding definite integral notation. The sum ∑k=180(2k80)3280\sum_{k=1}^{80} \left( \frac{2k}{80} \right)^3 \frac{2}{80} has Δx=280=0.025\Delta x = \frac{2}{80} = 0.025, xk=kΔx=2k80x_k = k \Delta x = \frac{2k}{80}. The term is (xk)3(x_k)^3, so function x3x^3. Limits from 0 to 2. A tempting distractor is ∫080(2x80)3dx\int_0^{80} \left( \frac{2x}{80} \right)^3 dx, which incorrectly sets the upper limit to 80 without adjusting Δx\Delta x. To translate any Riemann sum to an integral, identify Δx\Delta x, express the argument in terms of xk=a+kΔxx_k = a + k \Delta x, determine the limits from the range of xkx_k, and form the integral of the resulting function over those limits.

Question 8

A midpoint Riemann sum is ∑k=1nln⁡ ⁣(2+3n(k−12))3n\sum_{k=1}^{n} \ln\!\left(2+\frac{3}{n}\left(k-\frac12\right)\right)\frac{3}{n}. Which integral matches it?

  1. ∫25ln⁡(x) dx\displaystyle \int_{2}^{5} \ln(x)\,dx (correct answer)
  2. ∫23ln⁡(x) dx\displaystyle \int_{2}^{3} \ln(x)\,dx
  3. ∫03ln⁡(x) dx\displaystyle \int_{0}^{3} \ln(x)\,dx
  4. ∫253ln⁡(x) dx\displaystyle \int_{2}^{5} 3\ln(x)\,dx
  5. ∫25ln⁡(2+3x) dx\displaystyle \int_{2}^{5} \ln(2+3x)\,dx
Explanation: This problem asks us to translate a midpoint Riemann sum to integral notation. The sum ∑k=1nln⁡ ⁣(2+3n(k−12))3n\sum_{k=1}^{n} \ln\!\left(2+\frac{3}{n}\left(k-\frac12\right)\right)\frac{3}{n} samples at midpoints. When k=1k=1, we get x=2+3n⋅12x=2+\frac{3}{n}\cdot\frac{1}{2}, and when k=nk=n, we get approximately x=2+3(1−12n)x=2+3\left(1-\frac{1}{2n}\right), which approaches 55 as nn increases. The width 3n\frac{3}{n} confirms interval length 5−2=35-2=3. Choice E might seem appealing with ln⁡(2+3x)\ln(2+3x), but this misinterprets the sampling points as a function transformation. The key is recognizing that midpoint sums still approximate ∫abf(x) dx\int_a^b f(x)\,dx, not a transformed function.

Question 9

Which definite integral matches ∑i=1nsin⁡ ⁣(πin)πn\sum_{i=1}^{n} \sin\!\left(\frac{\pi i}{n}\right)\frac{\pi}{n}?

  1. ∫0πsin⁡x dx\displaystyle \int_{0}^{\pi} \sin x\,dx (correct answer)
  2. ∫01sin⁡(πx) dx\displaystyle \int_{0}^{1} \sin(\pi x)\,dx
  3. ∫0πsin⁡(πx) dx\displaystyle \int_{0}^{\pi} \sin(\pi x)\,dx
  4. ∫1nsin⁡ ⁣(πxn)πn dx\displaystyle \int_{1}^{n} \sin\!\left(\frac{\pi x}{n}\right)\frac{\pi}{n}\,dx
  5. ∫0πcos⁡x dx\displaystyle \int_{0}^{\pi} \cos x\,dx
Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The term (\frac{\pi}{n}) is (Δ\Delta x), indicating an interval length of (π\pi), typically from 0 to (π\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 (∫0\int_{0}^{π\pi} sin⁡\sin x , dx) as (n) grows large. Choice C, (\int_{0}^{\pi} \sin(π\pi x) , dx), is a common distractor but fails because it mistakenly places (π\pi) inside the sine instead of recognizing the direct mapping to (sin⁡\sin x). To translate Riemann sums generally, identify (Δ\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = lim⁡\lim n ⋅\cdot Δ\Delta x), and rewrite the summand as (f(x_i)).

Question 10

Which definite integral is represented by ∑i=1n2+3in 3n\sum_{i=1}^{n} \sqrt{2+\frac{3i}{n}}\,\frac{3}{n}?

  1. ∫032+3x dx\displaystyle \int_{0}^{3} \sqrt{2+3x}\,dx
  2. ∫25x dx\displaystyle \int_{2}^{5} \sqrt{x}\,dx
  3. ∫032+x dx\displaystyle \int_{0}^{3} \sqrt{2+x}\,dx (correct answer)
  4. ∫032+x 3 dx\displaystyle \int_{0}^{3} \sqrt{2+x}\,3\,dx
  5. ∫032+3xn 3n dx\displaystyle \int_{0}^{3} \sqrt{2+\frac{3x}{n}}\,\frac{3}{n}\,dx
Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. Here, (3n\frac{3}{n}) serves as (Δ\Delta x), implying an interval length of 3, from 0 to 3. The expression (2+3in\sqrt{2 + \frac{3i}{n}}) corresponds to (f(x_i) = \sqrt{2 + x_i}) with (x_i = 3in\frac{3i}{n}). As such, the sum converges to (∫0\int_{0}^{3} 2+x\sqrt{2 + x} , dx). Choice D, (∫0\int_{0}^{3} 2+x\sqrt{2 + x} ⋅\cdot 3 , dx), is a distractor that wrongly multiplies the integrand by 3, confusing the role of (Δ\Delta x)'s numerator. To translate Riemann sums generally, identify (Δ\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = lim⁡\lim n ⋅\cdot Δ\Delta x), and rewrite the summand as (f(x_i)).

Question 11

Which definite integral is represented by the Riemann sum ∑i=1n(3+2in)22n\sum_{i=1}^{n}\left(3+\frac{2i}{n}\right)^2\frac{2}{n}?

  1. ∫35x2 dx\displaystyle \int_{3}^{5} x^2\,dx
  2. ∫02(3+x)2 dx\displaystyle \int_{0}^{2} (3+x)^2\,dx (correct answer)
  3. ∫1n(3+2xn)22n dx\displaystyle \int_{1}^{n} \left(3+\frac{2x}{n}\right)^2\frac{2}{n}\,dx
  4. ∫02(3+2x)2 dx\displaystyle \int_{0}^{2} (3+2x)^2\,dx
  5. ∫02(3+x)2 2 dx\displaystyle \int_{0}^{2} (3+x)^2\,2\,dx
Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The term (2n\frac{2}{n}) represents the width of each subinterval, (Δ\Delta x), so the total interval length is 2, suggesting limits from 0 to 2. The expression inside the sum, (3 + 2in\frac{2i}{n}), corresponds to the sample point (x_i = 2in\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 (Δ\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = lim⁡\lim n ⋅\cdot Δ\Delta x), and rewrite the summand as (f(x_i)).

Question 12

The sum ∑k=1ncos⁡ ⁣(π+πkn)πn\sum_{k=1}^{n} \cos\!\left(\pi+\frac{\pi k}{n}\right)\frac{\pi}{n} is a Riemann sum. Which integral matches it?

  1. ∫0πcos⁡(x) dx\displaystyle \int_{0}^{\pi} \cos(x)\,dx
  2. ∫π2πcos⁡(x) dx\displaystyle \int_{\pi}^{2\pi} \cos(x)\,dx (correct answer)
  3. ∫π2ππcos⁡(x) dx\displaystyle \int_{\pi}^{2\pi} \pi\cos(x)\,dx
  4. ∫π2πcos⁡(πx) dx\displaystyle \int_{\pi}^{2\pi} \cos(\pi x)\,dx
  5. ∫2ππcos⁡(x) dx\displaystyle \int_{2\pi}^{\pi} \cos(x)\,dx
Explanation: This problem requires translating a Riemann sum into integral notation. The sum ∑k=1ncos⁡ ⁣(π+πkn)πn\sum_{k=1}^{n} \cos\!\left(\pi+\frac{\pi k}{n}\right)\frac{\pi}{n} samples cosine at points π+πkn\pi+\frac{\pi k}{n}. When k=0k=0, we would get x=πx=\pi, and when k=nk=n, we get x=π+π=2πx=\pi+\pi=2\pi, establishing interval [π,2π][\pi,2\pi]. The width πn\frac{\pi}{n} confirms the interval length is 2π−π=π2\pi-\pi=\pi. Choice D with cos⁡(πx)\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.

Question 13

Which integral matches the sum ∑i=1ne1+in1n\sum_{i=1}^{n} e^{1+\frac{i}{n}}\frac{1}{n}?

  1. ∫01e1+x dx\displaystyle \int_{0}^{1} e^{1+x}\,dx (correct answer)
  2. ∫12ex dx\displaystyle \int_{1}^{2} e^{x}\,dx
  3. ∫01e1+nx dx\displaystyle \int_{0}^{1} e^{1+nx}\,dx
  4. ∫01e1+x n dx\displaystyle \int_{0}^{1} e^{1+x}\,n\,dx
  5. ∫01ex+1n dx\displaystyle \int_{0}^{1} e^{x+\frac{1}{n}}\,dx
Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The (1n\frac{1}{n}) is (Δ\Delta x), suggesting limits from 0 to 1. The term (e^{1 + in\frac{i}{n}}) maps to (f(x_i) = e^{1 + x_i}) where (x_i = in\frac{i}{n}). Thus, the sum represents (∫0\int_{0}^{1} e1+xe^{1 + x} , dx). Choice D, (∫0\int_{0}^{1} e1+xe^{1 + x} ⋅\cdot n , dx), might tempt but fails by incorrectly multiplying by (n), which inverts (Δ\Delta x)'s role. To translate Riemann sums generally, identify (Δ\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = lim⁡\lim n ⋅\cdot Δ\Delta x), and rewrite the summand as (f(x_i)).

Question 14

Which integral corresponds to ∑i=1n(1+4in)4n\sum_{i=1}^{n} \left(1+\frac{4i}{n}\right)\frac{4}{n}?

  1. ∫04(1+4x) dx\displaystyle \int_{0}^{4} (1+4x)\,dx
  2. ∫15x dx\displaystyle \int_{1}^{5} x\,dx
  3. ∫04(1+x) dx\displaystyle \int_{0}^{4} (1+x)\,dx (correct answer)
  4. ∫04(1+x) 4 dx\displaystyle \int_{0}^{4} (1+x)\,4\,dx
  5. ∫04(1+4xn)4n dx\displaystyle \int_{0}^{4} \left(1+\frac{4x}{n}\right)\frac{4}{n}\,dx
Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The (4n\frac{4}{n}) term is (Δ\Delta x), pointing to an interval from 0 to 4. The summand (1 + 4in\frac{4i}{n}) matches (f(x_i) = 1 + x_i) where (x_i = 4in\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 (Δ\Delta x)'s numerator inside the integrand instead of as the differential. To translate Riemann sums generally, identify (Δ\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = lim⁡\lim n ⋅\cdot Δ\Delta x), and rewrite the summand as (f(x_i)).

Question 15

Which definite integral matches ∑i=1n11+(3in)23n\sum_{i=1}^{n} \frac{1}{1+\left(\frac{3i}{n}\right)^2}\frac{3}{n}?

  1. ∫0311+x2 dx\displaystyle \int_{0}^{3} \frac{1}{1+x^2}\,dx (correct answer)
  2. ∫0111+9x2 dx\displaystyle \int_{0}^{1} \frac{1}{1+9x^2}\,dx
  3. ∫0311+9x2 dx\displaystyle \int_{0}^{3} \frac{1}{1+9x^2}\,dx
  4. ∫0311+(3xn)2 dx\displaystyle \int_{0}^{3} \frac{1}{1+\left(\frac{3x}{n}\right)^2}\,dx
  5. ∫0331+x2 dx\displaystyle \int_{0}^{3} \frac{3}{1+x^2}\,dx
Explanation: This problem tests your ability to translate a Riemann sum into its corresponding definite integral notation. The term (3n\frac{3}{n}) represents (Δ\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 = 3in\frac{3i}{n}). Hence, the sum approximates (∫0\int_{0}^{3} 11+x2\frac{1}{1 + x^2} , dx). Choice E, (∫0\int_{0}^{3} 31+x2\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 (Δ\Delta x) as the coefficient outside, the interval from (a) to (a + b) where (b = lim⁡\lim n ⋅\cdot Δ\Delta x), and rewrite the summand as (f(x_i)).

Question 16

A sum ∑i=1309−(i10)2 110\sum_{i=1}^{30} \sqrt{9-\left(\frac{i}{10}\right)^2}\,\frac{1}{10} represents area. Which integral matches it?

  1. ∫039−x2 dx\displaystyle \int_{0}^{3} \sqrt{9-x^2}\,dx (correct answer)
  2. ∫0309−(x10)2 dx\displaystyle \int_{0}^{30} \sqrt{9-\left(\frac{x}{10}\right)^2}\,dx
  3. ∫039−(x10)2 dx\displaystyle \int_{0}^{3} \sqrt{9-\left(\frac{x}{10}\right)^2}\,dx
  4. ∫11039−x2 dx\displaystyle \int_{\frac{1}{10}}^{3} \sqrt{9-x^2}\,dx
  5. ∫039−x2 110 dx\displaystyle \int_{0}^{3} \sqrt{9-x^2}\,\frac{1}{10}\,dx
Explanation: 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} ight)^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.

Question 17

The sum ∑j=1n((3+2jn)2)2n\sum_{j=1}^{n} \left(\left(3+\frac{2j}{n}\right)^2\right)\frac{2}{n} is a Riemann sum. Which integral matches it?

  1. ∫35x2 dx\displaystyle \int_{3}^{5} x^2\,dx (correct answer)
  2. ∫35(3+2x)2 dx\displaystyle \int_{3}^{5} (3+2x)^2\,dx
  3. ∫02(3+x)2 dx\displaystyle \int_{0}^{2} (3+x)^2\,dx
  4. ∫352x2 dx\displaystyle \int_{3}^{5} 2x^2\,dx
  5. ∫02x2 dx\displaystyle \int_{0}^{2} x^2\,dx
Explanation: This problem requires translating a Riemann sum into integral notation. The sum ∑j=1n((3+2jn)2)2n\sum_{j=1}^{n} \left(\left(3+\frac{2j}{n}\right)^2\right)\frac{2}{n} evaluates the function at points 3+2jn3+\frac{2j}{n}. When j=0j=0, we would get x=3x=3, and when j=nj=n, we get x=3+2=5x=3+2=5, establishing interval [3,5][3,5]. The width 2n\frac{2}{n} confirms the interval length is 5−3=25-3=2. Choice B might tempt you with (3+2x)2(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.

Question 18

The sum ∑k=0n−17+2kn 2n\sum_{k=0}^{n-1} \sqrt{7+\frac{2k}{n}}\,\frac{2}{n} is a Riemann sum. Which integral matches it?

  1. ∫79x dx\displaystyle \int_{7}^{9} \sqrt{x}\,dx (correct answer)
  2. ∫027+x dx\displaystyle \int_{0}^{2} \sqrt{7+x}\,dx
  3. ∫797+x dx\displaystyle \int_{7}^{9} \sqrt{7+x}\,dx
  4. ∫792x dx\displaystyle \int_{7}^{9} 2\sqrt{x}\,dx
  5. ∫02x dx\displaystyle \int_{0}^{2} \sqrt{x}\,dx
Explanation: This problem requires identifying the integral matching a Riemann sum. The sum ∑k=0n−17+2kn 2n\sum_{k=0}^{n-1} \sqrt{7+\frac{2k}{n}}\,\frac{2}{n} evaluates at points 7+2kn7+\frac{2k}{n}. When k=0k=0, we get x=7x=7, and when k=n−1k=n-1, we approach x=7+2=9x=7+2=9, establishing interval [7,9][7,9]. The width 2n\frac{2}{n} matches the interval length 9−7=29-7=2. Choice B shows 7+x\sqrt{7+x}, which incorrectly interprets the argument as a function composition rather than sampling points. The transferable principle is that f(a+(b−a)kn)f(a+\frac{(b-a)k}{n}) samples f(x)f(x) on [a,b][a,b], not f(a+x)f(a+x).

Question 19

A Riemann sum for area is ∑i=1ng ⁣(−1+6in)6n\sum_{i=1}^{n} g\!\left(-1+\frac{6i}{n}\right)\frac{6}{n}. Which integral matches it?

  1. ∫−15g(x) dx\displaystyle \int_{-1}^{5} g(x)\,dx (correct answer)
  2. ∫−16g(x) dx\displaystyle \int_{-1}^{6} g(x)\,dx
  3. ∫06g(x) dx\displaystyle \int_{0}^{6} g(x)\,dx
  4. ∫−156g(x) dx\displaystyle \int_{-1}^{5} 6g(x)\,dx
  5. ∫−15g(−1+6x) dx\displaystyle \int_{-1}^{5} g(-1+6x)\,dx
Explanation: This problem asks us to match a Riemann sum to its integral representation. The sum ∑i=1ng ⁣(−1+6in)6n\sum_{i=1}^{n} g\!\left(-1+\frac{6i}{n}\right)\frac{6}{n} samples at points −1+6in-1+\frac{6i}{n}. When i=0i=0, we would get x=−1x=-1, and when i=ni=n, we get x=−1+6=5x=-1+6=5, defining interval [−1,5][-1,5]. The width 6n\frac{6}{n} confirms the interval length is 5−(−1)=65-(-1)=6. Choice B extends to x=6x=6, which overshoots our endpoint. The key insight is that the coefficient of i/ni/n gives the interval length, and adding it to the constant term gives the right endpoint.

Question 20

A particle's velocity is approximated by ∑k=1nv ⁣(0.5+1.5kn)1.5n\sum_{k=1}^{n} v\!\left(0.5+\frac{1.5k}{n}\right)\frac{1.5}{n}. Which integral matches this displacement estimate?

  1. ∫0.52v(t) dt\displaystyle \int_{0.5}^{2} v(t)\,dt (correct answer)
  2. ∫01.5v(t) dt\displaystyle \int_{0}^{1.5} v(t)\,dt
  3. ∫0.521.5 v(t) dt\displaystyle \int_{0.5}^{2} 1.5\,v(t)\,dt
  4. ∫0.52v ⁣(0.5+1.5tn) dt\displaystyle \int_{0.5}^{2} v\!\left(0.5+\frac{1.5t}{n}\right)\,dt
  5. ∫0.51.5v(t) dt\displaystyle \int_{0.5}^{1.5} v(t)\,dt
Explanation: This problem asks us to match a Riemann sum to its integral representation. The sum ∑k=1nv ⁣(0.5+1.5kn)1.5n\sum_{k=1}^{n} v\!\left(0.5+\frac{1.5k}{n}\right)\frac{1.5}{n} evaluates velocity at points 0.5+1.5kn0.5+\frac{1.5k}{n}. When k=1k=1, we get approximately x=0.5x=0.5, and when k=nk=n, we get x=0.5+1.5=2x=0.5+1.5=2, defining the interval [0.5,2][0.5,2]. The width 1.5n\frac{1.5}{n} matches the interval length 2−0.5=1.52-0.5=1.5. Choice C incorrectly places the factor 1.51.5 inside the integral as a coefficient of v(t)v(t). Remember that in Riemann sums, the pattern a+(b−a)kna+\frac{(b-a)k}{n} with width b−an\frac{b-a}{n} corresponds to ∫abf(x) dx\int_a^b f(x)\,dx.