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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.

Question 1 / 15

0 of 15 answered

The trapezoidal rule with n=4n = 4n=4 subintervals is used to approximate ∫261x dx\int_2^6 \frac{1}{x} \, dx∫26​x1​dx. What is the value of Δx\Delta xΔx and how many function evaluations are needed?

Select an answer to continue

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.

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

The trapezoidal rule with n=4n = 4n=4 subintervals is used to approximate ∫261x dx\int_2^6 \frac{1}{x} \, dx∫26​x1​dx. What is the value of Δx\Delta xΔx and how many function evaluations are needed?

  1. Δx=1\Delta x = 1Δx=1 and 444 function evaluations are needed
  2. Δx=1\Delta x = 1Δx=1 and 555 function evaluations are needed (correct answer)
  3. Δx=2\Delta x = 2Δx=2 and 444 function evaluations are needed
  4. Δx=2\Delta x = 2Δx=2 and 555 function evaluations are needed

Explanation: For n=4n = 4n=4 subintervals on [2,6][2,6][2,6], Δx=6−24=1\Delta x = \frac{6-2}{4} = 1Δx=46−2​=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 = 6x0​=2,x1​=3,x2​=4,x3​=5,x4​=6, which is 555 points total. Choice A has the wrong number of evaluations, while choices C and D have incorrect Δx\Delta xΔx values.

Question 2

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

  1. 0.50.50.5
  2. 111 (correct answer)
  3. 1.51.51.5
  4. 222

Explanation: For any Riemann sum with nnn equal subintervals on [a,b][a,b][a,b], the width is Δx=b−an\Delta x = \frac{b-a}{n}Δx=nb−a​. Here, Δx=12−48=88=1\Delta x = \frac{12-4}{8} = \frac{8}{8} = 1Δx=812−4​=88​=1. The other choices result from incorrect calculations: choice A gives 48\frac{4}{8}84​, choice C gives 128\frac{12}{8}812​, and choice D gives 168\frac{16}{8}816​.

Question 3

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

  1. 2(2+4)2(\sqrt{2} + \sqrt{4})2(2​+4​) (correct answer)
  2. 2(1+3)2(\sqrt{1} + \sqrt{3})2(1​+3​)
  3. 4(2+4)4(\sqrt{2} + \sqrt{4})4(2​+4​)
  4. 2(3+5)2(\sqrt{3} + \sqrt{5})2(3​+5​)

Explanation: With n=2n = 2n=2 on [1,5][1,5][1,5], Δx=2\Delta x = 2Δx=2. The subintervals are [1,3][1,3][1,3] and [3,5][3,5][3,5] with midpoints 222 and 444. The approximation is 2[2+4]2[\sqrt{2} + \sqrt{4}]2[2​+4​]. Choice B uses wrong midpoints 111 and 333, choice C has wrong Δx=4\Delta x = 4Δx=4, and choice D uses wrong midpoints 333 and 555.

Question 4

The trapezoidal approximation of ∫04(3x2+1) dx\int_0^4 (3x^2 + 1) \, dx∫04​(3x2+1)dx using n=4n = 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]21​[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]1⋅21​[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]21​[1+4+13+28+49]
  4. 2⋅12[1+4+13+28+49]2 \cdot \frac{1}{2}[1 + 4 + 13 + 28 + 49]2⋅21​[1+4+13+28+49]

Explanation: With n=4n = 4n=4 on [0,4][0,4][0,4], Δx=1\Delta x = 1Δx=1. Points are 0,1,2,3,40, 1, 2, 3, 40,1,2,3,4 with function values 1,4,13,28,491, 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]2Δx​[f(x0​)+2f(x1​)+2f(x2​)+2f(x3​)+f(x4​)]=21​[1+2(4)+2(13)+2(28)+49]. Choice A omits Δx\Delta xΔx, choice C omits the factor of 2, choice D uses wrong Δx\Delta xΔx.

Question 5

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

  1. 2,4,6,82, 4, 6, 82,4,6,8 (correct answer)
  2. 4,6,8,104, 6, 8, 104,6,8,10
  3. 2,4,6,8,102, 4, 6, 8, 102,4,6,8,10
  4. 3,5,7,93, 5, 7, 93,5,7,9

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

Question 6

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

  1. 2,4,6,8,102, 4, 6, 8, 102,4,6,8,10 (correct answer)
  2. 1,3,5,7,91, 3, 5, 7, 91,3,5,7,9
  3. 3,5,7,9,113, 5, 7, 9, 113,5,7,9,11
  4. 1,3,5,7,9,111, 3, 5, 7, 9, 111,3,5,7,9,11

Explanation: With n=5n = 5n=5 on [1,11][1,11][1,11], Δx=2\Delta x = 2Δx=2. The subintervals are [1,3],[3,5],[5,7],[7,9],[9,11][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, 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

For a decreasing function ggg on [0,4][0,4][0,4], which statement about Riemann sum approximations of ∫04g(x) dx\int_0^4 g(x) \, dx∫04​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 8

Using the midpoint rule with n=4n = 4n=4 to approximate ∫−22x3 dx\int_{-2}^2 x^3 \, dx∫−22​x3dx, 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]1⋅[(−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]1⋅[(−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]2⋅[(−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]1⋅[(−1)3+(0)3+(1)3+(2)3]

Explanation: With n=4n = 4n=4 on [−2,2][-2,2][−2,2], Δx=1\Delta x = 1Δx=1. The subintervals are [−2,−1],[−1,0],[0,1],[1,2][-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, 1.5−1.5,−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]1⋅[(−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 = 2Δx=2, choice D uses different points entirely.

Question 9

A function fff is increasing on [a,b][a,b][a,b]. If LnL_nLn​, RnR_nRn​, and MnM_nMn​ represent left, right, and midpoint Riemann sums respectively with nnn subintervals, then:

  1. Ln≤Mn≤Rn≤∫abf(x) dxL_n \leq M_n \leq R_n \leq \int_a^b f(x) \, dxLn​≤Mn​≤Rn​≤∫ab​f(x)dx
  2. Ln≤∫abf(x) dx≤Mn≤RnL_n \leq \int_a^b f(x) \, dx \leq M_n \leq R_nLn​≤∫ab​f(x)dx≤Mn​≤Rn​
  3. Ln≤Mn≤∫abf(x) dx≤RnL_n \leq M_n \leq \int_a^b f(x) \, dx \leq R_nLn​≤Mn​≤∫ab​f(x)dx≤Rn​ (correct answer)
  4. Mn≤Ln≤∫abf(x) dx≤RnM_n \leq L_n \leq \int_a^b f(x) \, dx \leq R_nMn​≤Ln​≤∫ab​f(x)dx≤Rn​

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_nLn​≤Mn​≤∫ab​f(x)dx≤Rn​. The other choices incorrectly order these approximations.

Question 10

The sum ∑k=1434⋅sin⁡(3k4)\sum_{k=1}^4 \frac{3}{4} \cdot \sin\left(\frac{3k}{4}\right)∑k=14​43​⋅sin(43k​) represents which type of Riemann sum approximation?

  1. Left Riemann sum for ∫03sin⁡x dx\int_0^3 \sin x \, dx∫03​sinxdx with n=4n = 4n=4
  2. Right Riemann sum for ∫03sin⁡x dx\int_0^3 \sin x \, dx∫03​sinxdx with n=4n = 4n=4 (correct answer)
  3. Midpoint Riemann sum for ∫03sin⁡x dx\int_0^3 \sin x \, dx∫03​sinxdx with n=4n = 4n=4
  4. Left Riemann sum for ∫04sin⁡x dx\int_0^4 \sin x \, dx∫04​sinxdx with n=3n = 3n=3

Explanation: The sum has Δx=34\Delta x = \frac{3}{4}Δx=43​ 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}, 343⋅1​,43⋅2​,43⋅3​,43⋅4​=43​,23​,49​,3. With n=4n = 4n=4 on [0,3][0,3][0,3], Δx=34\Delta x = \frac{3}{4}Δx=43​, 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][0,43​],[43​,23​],[23​,49​],[49​,3]. Choice A would start at k=0k=0k=0, choice C would use midpoints, choice D has wrong interval.

Question 11

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

  1. 1⋅[e0.5+e1.5+e2.5]1 \cdot [e^{0.5} + e^{1.5} + e^{2.5}]1⋅[e0.5+e1.5+e2.5] (correct answer)
  2. 1⋅[e0+e1+e2]1 \cdot [e^0 + e^1 + e^2]1⋅[e0+e1+e2]
  3. 1⋅[e1+e2+e3]1 \cdot [e^1 + e^2 + e^3]1⋅[e1+e2+e3]
  4. 3⋅[e0.5+e1.5+e2.5]3 \cdot [e^{0.5} + e^{1.5} + e^{2.5}]3⋅[e0.5+e1.5+e2.5]

Explanation: With n=3n = 3n=3 on [0,3][0,3][0,3], Δx=1\Delta x = 1Δx=1. The subintervals are [0,1],[1,2],[2,3][0,1], [1,2], [2,3][0,1],[1,2],[2,3] with midpoints 0.5,1.5,2.50.5, 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}]Δx∑f(midpoints)=1[e0.5+e1.5+e2.5]. Choice B uses left endpoints, choice C uses right endpoints, and choice D has the wrong coefficient.

Question 12

If ∫08f(x) dx\int_0^8 f(x) \, dx∫08​f(x)dx is approximated using both a left Riemann sum and a right Riemann sum with n=4n = 4n=4 subintervals, and fff 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 13

The trapezoidal rule approximation of ∫02ex2 dx\int_0^2 e^{x^2} \, dx∫02​ex2dx with n=2n = 2n=2 subintervals is:

  1. 12[e0+e1+e4]\frac{1}{2}[e^0 + e^1 + e^4]21​[e0+e1+e4]
  2. 12[e0+2e1+e4]\frac{1}{2}[e^0 + 2e^1 + e^4]21​[e0+2e1+e4] (correct answer)
  3. 1⋅12[e0+2e1+e4]1 \cdot \frac{1}{2}[e^0 + 2e^1 + e^4]1⋅21​[e0+2e1+e4]
  4. 2⋅12[e0+2e1+e4]2 \cdot \frac{1}{2}[e^0 + 2e^1 + e^4]2⋅21​[e0+2e1+e4]

Explanation: With n=2n = 2n=2 on [0,2][0,2][0,2], Δx=1\Delta x = 1Δx=1. Points are 0,1,20, 1, 20,1,2 with function values e0,e1,e4e^0, e^1, e^4e0,e1,e4. 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]2Δx​[f(0)+2f(1)+f(2)]=21​[e0+2e1+e4]. Choice A omits the factor of 2 for the middle term, choices C and D incorrectly separate out Δx\Delta xΔx when it's already incorporated in the 12\frac{1}{2}21​ factor.

Question 14

If f(x)=sin⁡xf(x) = \sin xf(x)=sinx and we use a left Riemann sum with n=6n = 6n=6 equal subintervals to approximate ∫0πf(x) dx\int_0^\pi f(x) \, dx∫0π​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})]6π​[sin(0)+sin(6π​)+sin(3π​)+sin(2π​)+sin(32π​)+sin(65π​)] (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)]6π​[sin(6π​)+sin(3π​)+sin(2π​)+sin(32π​)+sin(65π​)+sin(π)]
  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})]3π​[sin(0)+sin(6π​)+sin(3π​)+sin(2π​)+sin(32π​)+sin(65π​)]
  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})]π[sin(0)+sin(6π​)+sin(3π​)+sin(2π​)+sin(32π​)+sin(65π​)]

Explanation: With n=6n = 6n=6 on [0,π][0,\pi][0,π], Δx=π6\Delta x = \frac{\pi}{6}Δx=6π​. 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}0,6π​,3π​,2π​,32π​,65π​. The left sum uses these six points. Choice B uses right endpoints, choice C has wrong Δx=π3\Delta x = \frac{\pi}{3}Δx=3π​, and choice D has wrong Δx=π\Delta x = \piΔx=π.

Question 15

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

  1. 2[f(1)+f(3)+f(5)]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)]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)]32​[f(3)+f(5)+f(7)]
  4. 6[f(3)+f(5)+f(7)]6[f(3) + f(5) + f(7)]6[f(3)+f(5)+f(7)]

Explanation: With n=3n = 3n=3 on [1,7][1,7][1,7], Δx=2\Delta x = 2Δx=2. The subintervals are [1,3],[3,5],[5,7][1,3], [3,5], [5,7][1,3],[3,5],[5,7] with right endpoints 3,5,73, 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)]Δx[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}Δx=32​, and choice D has wrong Δx=6\Delta x = 6Δx=6.