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

Let gg be a function that is strictly decreasing on the interval [a,b][a, b]. Which of the following statements provides the best comparison between the right Riemann sum approximation (RnR_n) and the true value of the integral ∫abg(x)dx\int_a^b g(x) dx?

RnR_n is an overestimate of the integral.
RnR_n is an underestimate of the integral.
RnR_n is equal to the value of the integral.
The relationship cannot be determined from the given information.
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AP Calculus AB Quiz

AP Calculus AB Quiz: Approximating Areas With Riemann Sums

Practice Approximating Areas With Riemann Sums in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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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 AB.

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Question 1

Let gg be a function that is strictly decreasing on the interval [a,b][a, b]. Which of the following statements provides the best comparison between the right Riemann sum approximation (RnR_n) and the true value of the integral ∫abg(x)dx\int_a^b g(x) dx?

  1. RnR_n is an overestimate of the integral.
  2. RnR_n is an underestimate of the integral. (correct answer)
  3. RnR_n is equal to the value of the integral.
  4. The relationship cannot be determined from the given information.
Explanation: For a strictly decreasing function on a given subinterval, the function's value at the right endpoint is the minimum value on that subinterval. Therefore, a right Riemann sum, which uses the function values at the right endpoints, will be an underestimate of the true value of the integral.

Question 2

Let ff be a function such that f′(x)<0f'(x) < 0 and f′′(x)<0f''(x) < 0 for all xx in the interval [a,b][a, b]. Let I=∫abf(x)dxI = \int_a^b f(x) dx. For a given number of subintervals nn, which of the following must be true about the left Riemann sum (LnL_n) and trapezoidal sum (TnT_n) approximations?

  1. Tn<I<LnT_n < I < L_n (correct answer)
  2. Ln<I<TnL_n < I < T_n
  3. I<Tn<LnI < T_n < L_n
  4. Tn<Ln<IT_n < L_n < I
Explanation: Since f′(x)<0f'(x) < 0, the function ff is decreasing. For a decreasing function, the left Riemann sum LnL_n is an overestimate of the integral II. Since f′′(x)<0f''(x) < 0, the function is concave down. For a concave down function, the trapezoidal sum TnT_n is an underestimate of the integral II. Combining these facts, we have Tn<I<LnT_n < I < L_n.

Question 3

The approximation for ∫abf(x)dx\int_a^b f(x) dx using a left Riemann sum is ALA_L and using a right Riemann sum is ARA_R. If the trapezoidal approximation is ATA_T, which of the following gives an expression for ATA_T in terms of ALA_L and ARA_R, assuming equal subintervals?

  1. AL+AR2\frac{A_L + A_R}{2} (correct answer)
  2. AL+ARA_L + A_R
  3. ALAR\sqrt{A_L A_R}
  4. AR−ALA_R - A_L
Explanation: For equal subintervals of width Δx\Delta x, the left Riemann sum is AL=Δx∑i=0n−1f(xi)A_L = \Delta x \sum_{i=0}^{n-1} f(x_i) and the right Riemann sum is AR=Δx∑i=1nf(xi)A_R = \Delta x \sum_{i=1}^{n} f(x_i). The trapezoidal sum is AT=Δx2∑i=1n(f(xi−1)+f(xi))=Δx2[(f(x0)+f(x1))+...+(f(xn−1)+f(xn))]=Δx2[f(x0)+2f(x1)+...+2f(xn−1)+f(xn)]=12[Δx(f(x0)+...+f(xn−1))+Δx(f(x1)+...+f(xn))]=AL+AR2A_T = \frac{\Delta x}{2} \sum_{i=1}^{n} (f(x_{i-1}) + f(x_i)) = \frac{\Delta x}{2} [ (f(x_0)+f(x_1)) + ... + (f(x_{n-1})+f(x_n)) ] = \frac{\Delta x}{2} [f(x_0) + 2f(x_1) + ... + 2f(x_{n-1}) + f(x_n)] = \frac{1}{2} [\Delta x(f(x_0)+...+f(x_{n-1})) + \Delta x(f(x_1)+...+f(x_n))] = \frac{A_L+A_R}{2}.

Question 4

The rate at which water flows into a reservoir is given by a continuous function R(t)R(t), where tt is in hours and R(t)R(t) is in cubic meters per hour. At t=0,2,4,6t=0, 2, 4, 6 hours, the rates are R(0)=50,R(2)=70,R(4)=80,R(6)=60R(0)=50, R(2)=70, R(4)=80, R(6)=60 cubic meters per hour.

Using a left Riemann sum with three subintervals of equal width, what is the approximation of the total volume of water that flowed into the reservoir during the 6-hour period?

  1. 200 cubic meters
  2. 400 cubic meters (correct answer)
  3. 420 cubic meters
  4. 540 cubic meters
Explanation: The total volume is approximated by the integral of the rate, ∫06R(t)dt\int_0^6 R(t) dt. The interval is [0,6][0, 6] with 3 subintervals, so the width is Δt=6−03=2\Delta t = \frac{6-0}{3}=2 hours. A left Riemann sum uses the left endpoints t=0,2,4t=0, 2, 4. The approximation is Δt[R(0)+R(2)+R(4)]=2[50+70+80]=2[200]=400\Delta t [R(0) + R(2) + R(4)] = 2[50+70+80] = 2[200] = 400. The units are (hours) * (cubic meters/hour) = cubic meters.

Question 5

The values of a continuous function ff for selected values of xx are given as follows: f(0)=5,f(1)=8,f(2)=13,f(3)=20f(0)=5, f(1)=8, f(2)=13, f(3)=20. What is the value of a right Riemann sum approximation of ∫03f(x)dx\int_0^3 f(x) dx with 3 equal subintervals?

  1. 26
  2. 36
  3. 41 (correct answer)
  4. 46
Explanation: The interval of integration is [0,3][0, 3] with n=3n=3 subintervals, so the width of each subinterval is Δx=3−03=1\Delta x = \frac{3-0}{3} = 1. The subintervals are [0,1],[1,2],[2,3][0,1], [1,2], [2,3]. A right Riemann sum uses the right endpoints x=1,2,3x=1, 2, 3. The approximation is Δx[f(1)+f(2)+f(3)]=1[8+13+20]=41\Delta x [f(1) + f(2) + f(3)] = 1[8 + 13 + 20] = 41.

Question 6

A function g(x)g(x) is continuous and its values at several points are g(1)=5,g(3)=8,g(5)=12,g(7)=15g(1)=5, g(3)=8, g(5)=12, g(7)=15.

Using the given values, approximate the definite integral ∫17g(x)dx\int_1^7 g(x) dx with a right Riemann sum using 3 subintervals of equal width.

  1. 50
  2. 60
  3. 70 (correct answer)
  4. 35
Explanation: The interval is [1,7][1, 7] with n=3n=3 subintervals, so the width of each subinterval is Δx=7−13=2\Delta x = \frac{7-1}{3} = 2. The subintervals are [1,3],[3,5],[5,7][1,3], [3,5], [5,7]. For a right Riemann sum, we use the right endpoints 3,5,73, 5, 7. The approximation is Δx[g(3)+g(5)+g(7)]=2[8+12+15]=2[35]=70\Delta x [g(3) + g(5) + g(7)] = 2[8 + 12 + 15] = 2[35] = 70.

Question 7

Use a midpoint Riemann sum with 2 equal subintervals to approximate the value of ∫02x3dx\int_0^2 x^3 dx.

  1. 1.0
  2. 3.5 (correct answer)
  3. 5.0
  4. 9.0
Explanation: The interval is [0,2][0, 2] with n=2n=2 subintervals, so the width is Δx=2−02=1\Delta x = \frac{2-0}{2} = 1. The subintervals are [0,1][0,1] and [1,2][1,2]. The midpoints are 0.50.5 and 1.51.5. The midpoint Riemann sum is Δx[f(0.5)+f(1.5)]=1[(0.5)3+(1.5)3]=1[0.125+3.375]=3.5\Delta x [f(0.5) + f(1.5)] = 1[(0.5)^3 + (1.5)^3] = 1[0.125 + 3.375] = 3.5.

Question 8

Let ff be a function that is strictly increasing on the interval [a,b][a, b]. Which of the following statements must be true about the approximations for ∫abf(x)dx\int_a^b f(x) dx using a left Riemann sum (LnL_n) and a right Riemann sum (RnR_n) with nn subintervals?

  1. Ln<∫abf(x)dx<RnL_n < \int_a^b f(x) dx < R_n (correct answer)
  2. Rn<∫abf(x)dx<LnR_n < \int_a^b f(x) dx < L_n
  3. Ln<Rn<∫abf(x)dxL_n < R_n < \int_a^b f(x) dx
  4. ∫abf(x)dx<Ln<Rn\int_a^b f(x) dx < L_n < R_n
Explanation: For a strictly increasing function on a given subinterval, the minimum value occurs at the left endpoint and the maximum value occurs at the right endpoint. Therefore, a left Riemann sum (LnL_n) will be an underestimate of the true integral, and a right Riemann sum (RnR_n) will be an overestimate. This leads to the inequality Ln<∫abf(x)dx<RnL_n < \int_a^b f(x) dx < R_n.

Question 9

The function h(x)h(x) is twice differentiable and h′′(x)>0h''(x) > 0 for all xx in the interval [0,4][0, 4]. Let T4T_4 be the trapezoidal sum approximation with 4 equal subintervals for ∫04h(x)dx\int_0^4 h(x) dx. Which statement about T4T_4 must be true?

  1. T4T_4 is an underestimate of ∫04h(x)dx\int_0^4 h(x) dx.
  2. T4T_4 is an overestimate of ∫04h(x)dx\int_0^4 h(x) dx. (correct answer)
  3. T4T_4 is equal to ∫04h(x)dx\int_0^4 h(x) dx.
  4. The relationship cannot be determined without knowing if h(x)h(x) is increasing or decreasing.
Explanation: The condition h′′(x)>0h''(x) > 0 means that the graph of h(x)h(x) is concave up. For a concave up function, the secant line segment connecting the endpoints of any subinterval lies above the curve. Since the trapezoidal rule uses these secant line segments to form the tops of the trapezoids, the area of each trapezoid is greater than the area under the curve on that subinterval. Thus, the trapezoidal sum T4T_4 is an overestimate.

Question 10

The function ff is continuous on [0,6][0, 6]. Using three subintervals of equal width and right endpoints, the right Riemann sum approximation for ∫06f(x)dx\int_0^6 f(x) dx is calculated. Which of the following expressions represents this approximation?

  1. 2(f(0)+f(2)+f(4))2(f(0)+f(2)+f(4))
  2. 3(f(2)+f(4)+f(6))3(f(2)+f(4)+f(6))
  3. 2(f(2)+f(4)+f(6))2(f(2)+f(4)+f(6)) (correct answer)
  4. 3(f(0)+f(3)+f(6))3(f(0)+f(3)+f(6))
Explanation: The interval is [0,6][0, 6] with n=3n=3 subintervals of equal width. The width of each subinterval is Δx=6−03=2\Delta x = \frac{6-0}{3} = 2. The subintervals are [0,2],[2,4],[4,6][0,2], [2,4], [4,6]. For a right Riemann sum, we use the function values at the right endpoints of these subintervals, which are x=2,x=4,x=6x=2, x=4, x=6. The approximation is given by the sum of the areas of the three rectangles: f(2)⋅Δx+f(4)⋅Δx+f(6)⋅Δx=Δx(f(2)+f(4)+f(6))=2(f(2)+f(4)+f(6))f(2) \cdot \Delta x + f(4) \cdot \Delta x + f(6) \cdot \Delta x = \Delta x (f(2)+f(4)+f(6)) = 2(f(2)+f(4)+f(6)).

Question 11

The speed of a runner during the first 4 seconds of a race is given by a strictly increasing, differentiable function s(t)s(t), where tt is in seconds and ss is in meters per second.

A right Riemann sum is used to estimate the distance the runner travels during the first 4 seconds. How does this estimate compare to the actual distance traveled?

  1. The estimate is an overestimate because the function is strictly increasing. (correct answer)
  2. The estimate is an underestimate because the function is strictly increasing.
  3. The estimate is exact because the function is differentiable.
  4. The relationship cannot be determined without knowing the concavity of the function.
Explanation: The distance traveled is the integral of the speed function, ∫04s(t)dt\int_0^4 s(t) dt. Since the speed function s(t)s(t) is strictly increasing, the value at the right endpoint of any subinterval is the maximum value on that subinterval. Therefore, a right Riemann sum will use the maximum value for each rectangle's height, resulting in an overestimate of the actual distance traveled.

Question 12

A right Riemann sum with 5 equal subintervals is used to approximate ∫212f(x)dx\int_2^{12} f(x) dx. Which of the following is the width of each rectangle?

  1. 1
  2. 2 (correct answer)
  3. 5
  4. 10
Explanation: The width of each subinterval (rectangle) in a Riemann sum with equal subintervals is given by the formula Δx=b−an\Delta x = \frac{b-a}{n}, where [a,b][a, b] is the interval of integration and nn is the number of subintervals. In this case, a=2,b=12a=2, b=12, and n=5n=5. Therefore, the width is Δx=12−25=105=2\Delta x = \frac{12-2}{5} = \frac{10}{5} = 2.

Question 13

For a certain continuous function f(x)f(x), it is known that a left Riemann sum is always an overestimate and a right Riemann sum is always an underestimate for ∫abf(x)dx\int_a^b f(x) dx for any number of subintervals. Which of the following must be true about f(x)f(x) on [a,b][a, b]?

  1. f(x)f(x) is increasing
  2. f(x)f(x) is decreasing (correct answer)
  3. f(x)f(x) is concave up
  4. f(x)f(x) is concave down
Explanation: If a left Riemann sum is an overestimate, it means the function's value at the left endpoint of each subinterval is the maximum value in that subinterval. This occurs when a function is decreasing. If a right Riemann sum is an underestimate, it means the function's value at the right endpoint is the minimum value, which also occurs when a function is decreasing. Therefore, f(x)f(x) must be a decreasing function on the interval.

Question 14

Use a left Riemann sum with 4 equal subintervals to approximate the area of the region bounded by the graph of f(x)=x2f(x) = x^2, the x-axis, from x=0x=0 to x=4x=4.

  1. 14 (correct answer)
  2. 21
  3. 22
  4. 30
Explanation: The interval is [0,4][0, 4] with n=4n=4 subintervals, so the width of each subinterval is Δx=4−04=1\Delta x = \frac{4-0}{4} = 1. The subintervals are [0,1],[1,2],[2,3],[3,4][0,1], [1,2], [2,3], [3,4]. For a left Riemann sum, we use the left endpoints of these subintervals, which are 0,1,2,30, 1, 2, 3. The approximation is Δx[f(0)+f(1)+f(2)+f(3)]=1[02+12+22+32]=1[0+1+4+9]=14\Delta x [f(0) + f(1) + f(2) + f(3)] = 1[0^2 + 1^2 + 2^2 + 3^2] = 1[0+1+4+9] = 14.

Question 15

Use the trapezoidal rule with 4 equal subintervals to approximate ∫02ex2dx\int_0^2 e^{x^2} dx.

  1. 14(e0+2e0.25+2e1+2e2.25+e4)\frac{1}{4}(e^0 + 2e^{0.25} + 2e^1 + 2e^{2.25} + e^4) (correct answer)
  2. 12(e0+e0.25+e1+e2.25)\frac{1}{2}(e^0 + e^{0.25} + e^1 + e^{2.25})
  3. 12(e0.25+e1+e2.25+e4)\frac{1}{2}(e^{0.25} + e^1 + e^{2.25} + e^4)
  4. 14(e0.25+e1+e2.25+e4)\frac{1}{4}(e^{0.25} + e^1 + e^{2.25} + e^4)
Explanation: The interval is [0,2][0, 2] with n=4n=4 subintervals, so the width is Δx=2−04=0.5\Delta x = \frac{2-0}{4} = 0.5. The endpoints are 0,0.5,1,1.5,20, 0.5, 1, 1.5, 2. The trapezoidal rule formula is Δx2[f(x0)+2f(x1)+2f(x2)+2f(x3)+f(x4)]\frac{\Delta x}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4)]. Plugging in the values: 0.52[e02+2e0.52+2e12+2e1.52+e22]\frac{0.5}{2}[e^{0^2} + 2e^{0.5^2} + 2e^{1^2} + 2e^{1.5^2} + e^{2^2}]. This simplifies to 14[e0+2e0.25+2e1+2e2.25+e4]\frac{1}{4}[e^0 + 2e^{0.25} + 2e^1 + 2e^{2.25} + e^4].