AP Calculus BC Quiz: Approximating Areas With Riemann Sums
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Approximating Areas With Riemann SumsQuestion 1 of 15

The left Riemann sum ∑i=0n−1f(a+iΔx)Δx\sum_{i=0}^{n-1} f(a + i\Delta x) \Delta x approximates ∫abf(x) dx\int_a^b f(x) \, dx where Δx=b−an\Delta x = \frac{b-a}{n}. If a=2a = 2, b=10b = 10, and n=4n = 4, what are the xx-values used in the sum?

2,4,6,82, 4, 6, 8
4,6,8,104, 6, 8, 10
2,4,6,8,102, 4, 6, 8, 10
3,5,7,93, 5, 7, 9
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AP Calculus BC Quiz

AP Calculus BC Quiz: Approximating Areas With Riemann Sums

Practice Approximating Areas With Riemann Sums 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 Approximating Areas With Riemann Sums, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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Question 1

The left Riemann sum ∑i=0n−1f(a+iΔx)Δx\sum_{i=0}^{n-1} f(a + i\Delta x) \Delta x approximates ∫abf(x) dx\int_a^b f(x) \, dx where Δx=b−an\Delta x = \frac{b-a}{n}. If a=2a = 2, b=10b = 10, and n=4n = 4, what are the xx-values used in the sum?

  1. 2,4,6,82, 4, 6, 8 (correct answer)
  2. 4,6,8,104, 6, 8, 10
  3. 2,4,6,8,102, 4, 6, 8, 10
  4. 3,5,7,93, 5, 7, 9
Explanation: With a=2a = 2, b=10b = 10, n=4n = 4, we have Δx=2\Delta x = 2. The left endpoints are a+iΔxa + i\Delta x for i=0,1,2,3i = 0, 1, 2, 3: 2+0(2)=22 + 0(2) = 2, 2+1(2)=42 + 1(2) = 4, 2+2(2)=62 + 2(2) = 6, 2+3(2)=82 + 3(2) = 8. Choice B gives right endpoints, choice C includes the right boundary, and choice D gives midpoints.

Question 2

If ∫08f(x) dx\int_0^8 f(x) \, dx is approximated using both a left Riemann sum and a right Riemann sum with n=4n = 4 subintervals, and ff is a continuous increasing function, then:

  1. Both approximations will be exactly equal to the true integral value
  2. The left sum will be greater than the right sum by a predictable amount
  3. The right sum will be greater than the left sum by some positive amount (correct answer)
  4. The midpoint sum will be exactly halfway between the left and right sums
Explanation: For an increasing function, the left sum uses minimum values on each subinterval (underestimate) while the right sum uses maximum values (overestimate). Therefore, the right sum will be greater than the left sum. Choice A is wrong since Riemann sums are approximations, choice B has the inequality backwards, and choice D is incorrect since the midpoint sum position depends on the function's concavity, not just monotonicity.

Question 3

A right Riemann sum approximation of ∫17f(x) dx\int_1^7 f(x) \, dx with n=3n = 3 subintervals can be written as:

  1. 2[f(1)+f(3)+f(5)]2[f(1) + f(3) + f(5)]
  2. 2[f(3)+f(5)+f(7)]2[f(3) + f(5) + f(7)] (correct answer)
  3. 23[f(3)+f(5)+f(7)]\frac{2}{3}[f(3) + f(5) + f(7)]
  4. 6[f(3)+f(5)+f(7)]6[f(3) + f(5) + f(7)]
Explanation: With n=3n = 3 on [1,7][1,7], Δx=2\Delta x = 2. The subintervals are [1,3],[3,5],[5,7][1,3], [3,5], [5,7] with right endpoints 3,5,73, 5, 7. The right sum is Δx[f(3)+f(5)+f(7)]=2[f(3)+f(5)+f(7)]\Delta x[f(3) + f(5) + f(7)] = 2[f(3) + f(5) + f(7)]. Choice A uses left endpoints, choice C has wrong Δx=23\Delta x = \frac{2}{3}, and choice D has wrong Δx=6\Delta x = 6.

Question 4

If f(x)=sin⁡xf(x) = \sin x and we use a left Riemann sum with n=6n = 6 equal subintervals to approximate ∫0πf(x) dx\int_0^\pi f(x) \, dx, which of the following represents the approximation?

  1. π6[sin⁡(0)+sin⁡(π6)+sin⁡(π3)+sin⁡(π2)+sin⁡(2π3)+sin⁡(5π6)]\frac{\pi}{6}[\sin(0) + \sin(\frac{\pi}{6}) + \sin(\frac{\pi}{3}) + \sin(\frac{\pi}{2}) + \sin(\frac{2\pi}{3}) + \sin(\frac{5\pi}{6})] (correct answer)
  2. π6[sin⁡(π6)+sin⁡(π3)+sin⁡(π2)+sin⁡(2π3)+sin⁡(5π6)+sin⁡(π)]\frac{\pi}{6}[\sin(\frac{\pi}{6}) + \sin(\frac{\pi}{3}) + \sin(\frac{\pi}{2}) + \sin(\frac{2\pi}{3}) + \sin(\frac{5\pi}{6}) + \sin(\pi)]
  3. π3[sin⁡(0)+sin⁡(π6)+sin⁡(π3)+sin⁡(π2)+sin⁡(2π3)+sin⁡(5π6)]\frac{\pi}{3}[\sin(0) + \sin(\frac{\pi}{6}) + \sin(\frac{\pi}{3}) + \sin(\frac{\pi}{2}) + \sin(\frac{2\pi}{3}) + \sin(\frac{5\pi}{6})]
  4. π[sin⁡(0)+sin⁡(π6)+sin⁡(π3)+sin⁡(π2)+sin⁡(2π3)+sin⁡(5π6)]\pi[\sin(0) + \sin(\frac{\pi}{6}) + \sin(\frac{\pi}{3}) + \sin(\frac{\pi}{2}) + \sin(\frac{2\pi}{3}) + \sin(\frac{5\pi}{6})]
Explanation: With n=6n = 6 on [0,π][0,\pi], Δx=π6\Delta x = \frac{\pi}{6}. Left endpoints are 0,π6,π3,π2,2π3,5π60, \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{5\pi}{6}. The left sum uses these six points. Choice B uses right endpoints, choice C has wrong Δx=π3\Delta x = \frac{\pi}{3}, and choice D has wrong Δx=π\Delta x = \pi.

Question 5

The trapezoidal approximation of ∫04(3x2+1) dx\int_0^4 (3x^2 + 1) \, dx using n=4n = 4 subintervals equals:

  1. 12[1+2(4)+2(13)+2(28)+49]\frac{1}{2}[1 + 2(4) + 2(13) + 2(28) + 49]
  2. 1⋅12[1+2(4)+2(13)+2(28)+49]1 \cdot \frac{1}{2}[1 + 2(4) + 2(13) + 2(28) + 49] (correct answer)
  3. 12[1+4+13+28+49]\frac{1}{2}[1 + 4 + 13 + 28 + 49]
  4. 2⋅12[1+4+13+28+49]2 \cdot \frac{1}{2}[1 + 4 + 13 + 28 + 49]
Explanation: With n=4n = 4 on [0,4][0,4], Δx=1\Delta x = 1. Points are 0,1,2,3,40, 1, 2, 3, 4 with function values 1,4,13,28,491, 4, 13, 28, 49. The trapezoidal rule is Δx2[f(x0)+2f(x1)+2f(x2)+2f(x3)+f(x4)]=12[1+2(4)+2(13)+2(28)+49]\frac{\Delta x}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4)] = \frac{1}{2}[1 + 2(4) + 2(13) + 2(28) + 49]. Choice A omits Δx\Delta x, choice C omits the factor of 2, choice D uses wrong Δx\Delta x.

Question 6

A midpoint Riemann sum with n=5n = 5 equal subintervals is used to approximate ∫111ln⁡x dx\int_1^{11} \ln x \, dx. The midpoints of the subintervals are:

  1. 2,4,6,8,102, 4, 6, 8, 10 (correct answer)
  2. 1,3,5,7,91, 3, 5, 7, 9
  3. 3,5,7,9,113, 5, 7, 9, 11
  4. 1,3,5,7,9,111, 3, 5, 7, 9, 11
Explanation: With n=5n = 5 on [1,11][1,11], Δx=2\Delta x = 2. The subintervals are [1,3],[3,5],[5,7],[7,9],[9,11][1,3], [3,5], [5,7], [7,9], [9,11] with midpoints 2,4,6,8,102, 4, 6, 8, 10. Choice B gives left endpoints, choice C gives right endpoints, and choice D incorrectly includes both endpoints with an extra point.

Question 7

Using the midpoint rule with n=2n = 2 subintervals, the approximation of ∫15x dx\int_1^5 \sqrt{x} \, dx is:

  1. 2(2+4)2(\sqrt{2} + \sqrt{4}) (correct answer)
  2. 2(1+3)2(\sqrt{1} + \sqrt{3})
  3. 4(2+4)4(\sqrt{2} + \sqrt{4})
  4. 2(3+5)2(\sqrt{3} + \sqrt{5})
Explanation: With n=2n = 2 on [1,5][1,5], Δx=2\Delta x = 2. The subintervals are [1,3][1,3] and [3,5][3,5] with midpoints 22 and 44. The approximation is 2[2+4]2[\sqrt{2} + \sqrt{4}]. Choice B uses wrong midpoints 11 and 33, choice C has wrong Δx=4\Delta x = 4, and choice D uses wrong midpoints 33 and 55.

Question 8

The trapezoidal rule with n=4n = 4 subintervals is used to approximate ∫261x dx\int_2^6 \frac{1}{x} \, dx. What is the value of Δx\Delta x and how many function evaluations are needed?

  1. Δx=1\Delta x = 1 and 44 function evaluations are needed
  2. Δx=1\Delta x = 1 and 55 function evaluations are needed (correct answer)
  3. Δx=2\Delta x = 2 and 44 function evaluations are needed
  4. Δx=2\Delta x = 2 and 55 function evaluations are needed
Explanation: For n=4n = 4 subintervals on [2,6][2,6], Δx=6−24=1\Delta x = \frac{6-2}{4} = 1. The trapezoidal rule requires evaluation at all endpoints: x0=2,x1=3,x2=4,x3=5,x4=6x_0 = 2, x_1 = 3, x_2 = 4, x_3 = 5, x_4 = 6, which is 55 points total. Choice A has the wrong number of evaluations, while choices C and D have incorrect Δx\Delta x values.

Question 9

A right Riemann sum with n=8n = 8 subintervals approximates ∫412(x2−3x) dx\int_4^{12} (x^2 - 3x) \, dx. The width of each subinterval is:

  1. 0.50.5
  2. 11 (correct answer)
  3. 1.51.5
  4. 22
Explanation: For any Riemann sum with nn equal subintervals on [a,b][a,b], the width is Δx=b−an\Delta x = \frac{b-a}{n}. Here, Δx=12−48=88=1\Delta x = \frac{12-4}{8} = \frac{8}{8} = 1. The other choices result from incorrect calculations: choice A gives 48\frac{4}{8}, choice C gives 128\frac{12}{8}, and choice D gives 168\frac{16}{8}.

Question 10

For a decreasing function gg on [0,4][0,4], which statement about Riemann sum approximations of ∫04g(x) dx\int_0^4 g(x) \, dx is correct?

  1. The left Riemann sum provides an underestimate of the integral value
  2. The right Riemann sum provides an overestimate of the integral value
  3. The left Riemann sum provides an overestimate of the integral value (correct answer)
  4. The midpoint sum is always exactly equal to the integral value
Explanation: For a decreasing function, the left endpoints give the maximum value on each subinterval, making the left Riemann sum an overestimate. The right endpoints give minimum values, making the right sum an underestimate. Choice A is backwards, choice B is backwards, and choice D is incorrect since midpoint sums are approximations, not exact values.

Question 11

Using the midpoint rule with n=4n = 4 to approximate ∫−22x3 dx\int_{-2}^2 x^3 \, dx, the approximation equals:

  1. 1⋅[(−1.5)3+(−0.5)3+(0.5)3+(1.5)3]1 \cdot [(-1.5)^3 + (-0.5)^3 + (0.5)^3 + (1.5)^3] (correct answer)
  2. 1⋅[(−2)3+(−1)3+(0)3+(1)3]1 \cdot [(-2)^3 + (-1)^3 + (0)^3 + (1)^3]
  3. 2⋅[(−1.5)3+(−0.5)3+(0.5)3+(1.5)3]2 \cdot [(-1.5)^3 + (-0.5)^3 + (0.5)^3 + (1.5)^3]
  4. 1⋅[(−1)3+(0)3+(1)3+(2)3]1 \cdot [(-1)^3 + (0)^3 + (1)^3 + (2)^3]
Explanation: With n=4n = 4 on [−2,2][-2,2], Δx=1\Delta x = 1. The subintervals are [−2,−1],[−1,0],[0,1],[1,2][-2,-1], [-1,0], [0,1], [1,2] with midpoints −1.5,−0.5,0.5,1.5-1.5, -0.5, 0.5, 1.5. The approximation is 1⋅[(−1.5)3+(−0.5)3+(0.5)3+(1.5)3]1 \cdot [(-1.5)^3 + (-0.5)^3 + (0.5)^3 + (1.5)^3]. Choice B uses left endpoints, choice C has wrong Δx=2\Delta x = 2, choice D uses different points entirely.

Question 12

A function ff is increasing on [a,b][a,b]. If LnL_n, RnR_n, and MnM_n represent left, right, and midpoint Riemann sums respectively with nn subintervals, then:

  1. Ln≤Mn≤Rn≤∫abf(x) dxL_n \leq M_n \leq R_n \leq \int_a^b f(x) \, dx
  2. Ln≤∫abf(x) dx≤Mn≤RnL_n \leq \int_a^b f(x) \, dx \leq M_n \leq R_n
  3. Ln≤Mn≤∫abf(x) dx≤RnL_n \leq M_n \leq \int_a^b f(x) \, dx \leq R_n (correct answer)
  4. Mn≤Ln≤∫abf(x) dx≤RnM_n \leq L_n \leq \int_a^b f(x) \, dx \leq R_n
Explanation: For an increasing function, the left sum underestimates (uses minimum values), the right sum overestimates (uses maximum values), and the midpoint sum is generally more accurate than both, falling between the left sum and the true integral value. Thus Ln≤Mn≤∫abf(x) dx≤RnL_n \leq M_n \leq \int_a^b f(x) \, dx \leq R_n. The other choices incorrectly order these approximations.

Question 13

The sum ∑k=1434⋅sin⁡(3k4)\sum_{k=1}^4 \frac{3}{4} \cdot \sin\left(\frac{3k}{4}\right) represents which type of Riemann sum approximation?

  1. Left Riemann sum for ∫03sin⁡x dx\int_0^3 \sin x \, dx with n=4n = 4
  2. Right Riemann sum for ∫03sin⁡x dx\int_0^3 \sin x \, dx with n=4n = 4 (correct answer)
  3. Midpoint Riemann sum for ∫03sin⁡x dx\int_0^3 \sin x \, dx with n=4n = 4
  4. Left Riemann sum for ∫04sin⁡x dx\int_0^4 \sin x \, dx with n=3n = 3
Explanation: The sum has Δx=34\Delta x = \frac{3}{4} and evaluates at 3⋅14,3⋅24,3⋅34,3⋅44=34,32,94,3\frac{3 \cdot 1}{4}, \frac{3 \cdot 2}{4}, \frac{3 \cdot 3}{4}, \frac{3 \cdot 4}{4} = \frac{3}{4}, \frac{3}{2}, \frac{9}{4}, 3. With n=4n = 4 on [0,3][0,3], Δx=34\Delta x = \frac{3}{4}, and these are the right endpoints of subintervals [0,34],[34,32],[32,94],[94,3][0,\frac{3}{4}], [\frac{3}{4},\frac{3}{2}], [\frac{3}{2},\frac{9}{4}], [\frac{9}{4},3]. Choice A would start at k=0k=0, choice C would use midpoints, choice D has wrong interval.

Question 14

A midpoint Riemann sum with n=3n = 3 subintervals approximates ∫03ex dx\int_0^3 e^x \, dx. Which expression represents this approximation?

  1. 1⋅[e0.5+e1.5+e2.5]1 \cdot [e^{0.5} + e^{1.5} + e^{2.5}] (correct answer)
  2. 1⋅[e0+e1+e2]1 \cdot [e^0 + e^1 + e^2]
  3. 1⋅[e1+e2+e3]1 \cdot [e^1 + e^2 + e^3]
  4. 3⋅[e0.5+e1.5+e2.5]3 \cdot [e^{0.5} + e^{1.5} + e^{2.5}]
Explanation: With n=3n = 3 on [0,3][0,3], Δx=1\Delta x = 1. The subintervals are [0,1],[1,2],[2,3][0,1], [1,2], [2,3] with midpoints 0.5,1.5,2.50.5, 1.5, 2.5. The midpoint sum is Δx∑f(midpoints)=1[e0.5+e1.5+e2.5]\Delta x \sum f(\text{midpoints}) = 1[e^{0.5} + e^{1.5} + e^{2.5}]. Choice B uses left endpoints, choice C uses right endpoints, and choice D has the wrong coefficient.

Question 15

The trapezoidal rule approximation of ∫02ex2 dx\int_0^2 e^{x^2} \, dx with n=2n = 2 subintervals is:

  1. 12[e0+e1+e4]\frac{1}{2}[e^0 + e^1 + e^4]
  2. 12[e0+2e1+e4]\frac{1}{2}[e^0 + 2e^1 + e^4] (correct answer)
  3. 1⋅12[e0+2e1+e4]1 \cdot \frac{1}{2}[e^0 + 2e^1 + e^4]
  4. 2⋅12[e0+2e1+e4]2 \cdot \frac{1}{2}[e^0 + 2e^1 + e^4]
Explanation: With n=2n = 2 on [0,2][0,2], Δx=1\Delta x = 1. Points are 0,1,20, 1, 2 with function values e0,e1,e4e^0, e^1, e^4. The trapezoidal rule is Δx2[f(0)+2f(1)+f(2)]=12[e0+2e1+e4]\frac{\Delta x}{2}[f(0) + 2f(1) + f(2)] = \frac{1}{2}[e^0 + 2e^1 + e^4]. Choice A omits the factor of 2 for the middle term, choices C and D incorrectly separate out Δx\Delta x when it's already incorporated in the 12\frac{1}{2} factor.