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

The limit lim⁡n→∞∑i=1ni3n4\lim_{n\to\infty} \sum_{i=1}^n \frac{i^3}{n^4} can be evaluated by converting it to a definite integral. What is the value of the limit?

0
14\frac{1}{4}
13\frac{1}{3}
1
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Calculus 1 Quiz

Calculus 1 Quiz: Riemann Sums And Notation

Practice Riemann Sums And Notation in Calculus 1 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 Calculus 1.

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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.

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Question 1

The limit lim⁡n→∞∑i=1ni3n4\lim_{n\to\infty} \sum_{i=1}^n \frac{i^3}{n^4} can be evaluated by converting it to a definite integral. What is the value of the limit?

  1. 0
  2. 14\frac{1}{4} (correct answer)
  3. 13\frac{1}{3}
  4. 1
Explanation: We can rewrite the expression inside the limit to match the form of a Riemann sum. ∑i=1ni3n4=∑i=1n(in)31n\sum_{i=1}^n \frac{i^3}{n^4} = \sum_{i=1}^n \left(\frac{i}{n}\right)^3 \frac{1}{n}. This is a right Riemann sum for a function on the interval [0,1][0, 1]. We can identify Δx=1n\Delta x = \frac{1}{n}, the sample points xi=inx_i = \frac{i}{n}, and the function f(x)=x3f(x) = x^3. The limit is therefore equal to the definite integral ∫01x3 dx\int_0^1 x^3 \,dx. Evaluating the integral gives [x44]01=144−044=14\left[\frac{x^4}{4}\right]_0^1 = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}.

Question 2

If ∑k=1nf(a+k⋅b−an)⋅b−an\sum_{k=1}^{n} f\left(a + k \cdot \frac{b-a}{n}\right) \cdot \frac{b-a}{n} represents a right Riemann sum, which definite integral does this sum approximate as n→∞n \to \infty?

  1. ∫abf(x) dx\int_a^b f(x) \, dx (correct answer)
  2. ∫0b−af(x) dx\int_0^{b-a} f(x) \, dx
  3. ∫a+b−anb+b−anf(x) dx\int_{a+\frac{b-a}{n}}^{b+\frac{b-a}{n}} f(x) \, dx
  4. ∫ab−af(x) dx\int_a^{b-a} f(x) \, dx
Explanation: The given sum has interval width Δx=b−an\Delta x = \frac{b-a}{n} and sample points xk=a+k⋅b−anx_k = a + k \cdot \frac{b-a}{n} for k=1,2,…,nk = 1, 2, \ldots, n. These are the right endpoints of subintervals on [a,b][a,b]. As n→∞n \to \infty, this right Riemann sum converges to ∫abf(x) dx\int_a^b f(x) \, dx. Choice B has wrong limits of integration. Choice C incorrectly shifts the integration bounds. Choice D uses b−ab-a as the upper limit instead of bb.

Question 3

The limit lim⁡n→∞1n∑k=1n(kn)4\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left( \frac{k}{n} \right)^4 represents a definite integral. What is the value of this integral?

  1. 0
  2. 15\frac{1}{5} (correct answer)
  3. 14\frac{1}{4}
  4. 1
Explanation: This limit is the definition of a definite integral. We identify the components: Δx=1n\Delta x = \frac{1}{n} and the right endpoints are xk=a+kΔxx_k = a + k\Delta x. Since the term inside the function is kn\frac{k}{n}, we can set a=0a=0. Then xk=knx_k = \frac{k}{n}. The interval is [0,1][0, 1] since b−a=nΔx=n(1n)=1b-a = n\Delta x = n(\frac{1}{n})=1. The function is f(xk)=(kn)4f(x_k) = (\frac{k}{n})^4, so f(x)=x4f(x) = x^4. The limit is equivalent to the integral ∫01x4 dx\int_0^1 x^4 \,dx. Evaluating this gives [x55]01=155−055=15\left[\frac{x^5}{5}\right]_0^1 = \frac{1^5}{5} - \frac{0^5}{5} = \frac{1}{5}.

Question 4

For the definite integral ∫15(3x+2) dx\int_1^5 (3x + 2) \, dx, what is the width of each subinterval when using n=8n = 8 equal subintervals?

  1. 34\frac{3}{4}
  2. 14\frac{1}{4}
  3. 58\frac{5}{8}
  4. 12\frac{1}{2} (correct answer)
Explanation: When you encounter questions about subinterval width in definite integrals, you're dealing with the foundation of numerical integration methods like Riemann sums. The key insight is that subinterval width depends only on the interval boundaries and number of subdivisions—it's completely independent of the function being integrated. To find the width of each subinterval, use the formula: Δx=b−an\Delta x = \frac{b-a}{n}, where [a,b][a,b] is the interval and nn is the number of equal subintervals. Here, you have the interval [1,5][1,5] divided into n=8n = 8 subintervals, so: Δx=5−18=48=12\Delta x = \frac{5-1}{8} = \frac{4}{8} = \frac{1}{2} This confirms answer choice D is correct. Let's examine why the other options are wrong. Choice A (34\frac{3}{4}) likely comes from incorrectly using the coefficient 3 from the function 3x+23x + 2—remember, the function's coefficients never affect subinterval width. Choice B (14\frac{1}{4}) suggests someone doubled the number of subintervals, perhaps confusing n=8n = 8 with n=16n = 16. Choice C (58\frac{5}{8}) appears to use the upper limit 5 as the numerator instead of the interval length 4. Study tip: Always remember that subinterval width equals interval lengthnumber of subintervals\frac{\text{interval length}}{\text{number of subintervals}}. The function being integrated is irrelevant to this calculation. Focus on the limits of integration and the value of nn only.

Question 5

Which summation correctly represents the left Riemann sum for ∫03ex dx\int_0^3 e^x \, dx using nn subintervals?

  1. ∑i=0n−1e3(i+1)n⋅3n\sum_{i=0}^{n-1} e^{\frac{3(i+1)}{n}} \cdot \frac{3}{n}
  2. ∑i=1ne3in⋅3n\sum_{i=1}^{n} e^{\frac{3i}{n}} \cdot \frac{3}{n}
  3. ∑i=0n−1e3in⋅3n\sum_{i=0}^{n-1} e^{\frac{3i}{n}} \cdot \frac{3}{n} (correct answer)
  4. ∑i=1ne3(i−1)n⋅3n\sum_{i=1}^{n} e^{\frac{3(i-1)}{n}} \cdot \frac{3}{n}
Explanation: When you encounter Riemann sum questions, the key is systematically identifying three components: the interval width (Δx\Delta x), the sample points within each subinterval, and how the summation indices work together. For a left Riemann sum of ∫03ex dx\int_0^3 e^x \, dx with nn subintervals, start by finding Δx=b−an=3−0n=3n\Delta x = \frac{b-a}{n} = \frac{3-0}{n} = \frac{3}{n}. Next, divide [0,3][0,3] into nn equal subintervals: [0,3n],[3n,6n],[6n,9n],…,[3(n−1)n,3][0, \frac{3}{n}], [\frac{3}{n}, \frac{6}{n}], [\frac{6}{n}, \frac{9}{n}], \ldots, [\frac{3(n-1)}{n}, 3]. For a left Riemann sum, you evaluate the function at the left endpoint of each subinterval: x0=0,x1=3n,x2=6n,…,xn−1=3(n−1)nx_0 = 0, x_1 = \frac{3}{n}, x_2 = \frac{6}{n}, \ldots, x_{n-1} = \frac{3(n-1)}{n}. These points follow the pattern xi=3inx_i = \frac{3i}{n} where ii runs from 0 to n−1n-1. Answer C is correct: ∑i=0n−1e3in⋅3n\sum_{i=0}^{n-1} e^{\frac{3i}{n}} \cdot \frac{3}{n}. The index starts at i=0i=0 (giving the first left endpoint x=0x=0) and ends at i=n−1i=n-1 (giving the last left endpoint x=3(n−1)nx=\frac{3(n-1)}{n}). Answer A uses 3(i+1)n\frac{3(i+1)}{n}, which gives right endpoints instead of left endpoints. Answer B starts the index at i=1i=1, missing the leftmost endpoint at x=0x=0. Answer D uses the correct left endpoints but with the wrong indexing pattern. Remember: left Riemann sums use left endpoints, so your index should start at i=0i=0 and the xx-values should be 3in\frac{3i}{n}, not shifted versions.

Question 6

Which of the following expressions represents a left Riemann sum approximation for ∫37ln⁡(x) dx\int_3^7 \ln(x) \,dx with n=8n=8 subintervals?

  1. 0.5∑i=07ln⁡(3+0.5i)0.5 \sum_{i=0}^{7} \ln(3 + 0.5i) (correct answer)
  2. 0.5∑i=18ln⁡(3+0.5i)0.5 \sum_{i=1}^{8} \ln(3 + 0.5i)
  3. 0.5∑i=18ln⁡(3.25+0.5(i−1))0.5 \sum_{i=1}^{8} \ln(3.25 + 0.5(i-1))
  4. 4∑i=07ln⁡(3+4i)4 \sum_{i=0}^{7} \ln(3 + 4i)
Explanation: For the integral ∫37ln⁡(x) dx\int_3^7 \ln(x) \,dx with n=8n=8, the width of each subinterval is Δx=7−38=48=0.5\Delta x = \frac{7-3}{8} = \frac{4}{8} = 0.5. The partition points are x0=3,x1=3.5,x2=4,…,x8=7x_0=3, x_1=3.5, x_2=4, \dots, x_8=7. A left Riemann sum uses the left endpoints of the subintervals: x0,x1,…,x7x_0, x_1, \dots, x_7. The formula for the left endpoints is xi=a+iΔx=3+0.5ix_i = a + i\Delta x = 3 + 0.5i. The sum is ∑i=0n−1f(xi)Δx=∑i=07ln⁡(xi)(0.5)=0.5∑i=07ln⁡(3+0.5i)\sum_{i=0}^{n-1} f(x_i) \Delta x = \sum_{i=0}^{7} \ln(x_i) (0.5) = 0.5 \sum_{i=0}^{7} \ln(3 + 0.5i). B) represents the right Riemann sum. C) represents the midpoint Riemann sum. D) incorrectly uses the total interval width (4) as the multiplier and also as part of the function argument, confusing the setup.

Question 7

A Riemann sum is used to approximate ∫210(x+1) dx\int_2^{10} (x+1) \,dx using the partition P={2,4,7,10}P = \{2, 4, 7, 10\} and right endpoints as sample points. What is the value of the approximation?

  1. 45
  2. 56
  3. 67 (correct answer)
  4. 72
Explanation: The partition P={2,4,7,10}P = \{2, 4, 7, 10\} defines three subintervals: [2,4][2, 4], [4,7][4, 7], and [7,10][7, 10]. The widths of these subintervals are not equal. Δx1=4−2=2\Delta x_1 = 4-2=2, Δx2=7−4=3\Delta x_2 = 7-4=3, and Δx3=10−7=3\Delta x_3 = 10-7=3. The problem specifies using right endpoints, which are x1=4x_1=4, x2=7x_2=7, and x3=10x_3=10. The function is f(x)=x+1f(x)=x+1. The Riemann sum is f(4)Δx1+f(7)Δx2+f(10)Δx3=(4+1)(2)+(7+1)(3)+(10+1)(3)=(5)(2)+(8)(3)+(11)(3)=10+24+33=67f(4)\Delta x_1 + f(7)\Delta x_2 + f(10)\Delta x_3 = (4+1)(2) + (7+1)(3) + (10+1)(3) = (5)(2) + (8)(3) + (11)(3) = 10 + 24 + 33 = 67.

Question 8

If nn is a positive integer, which of the following summations is a right Riemann sum for ∫121x dx\int_1^2 \frac{1}{x} \,dx?

  1. ∑i=0n−11n11+i/n\sum_{i=0}^{n-1} \frac{1}{n} \frac{1}{1+i/n}
  2. ∑i=1n1n11+i/n\sum_{i=1}^{n} \frac{1}{n} \frac{1}{1+i/n} (correct answer)
  3. ∑i=1n2n11+2i/n\sum_{i=1}^{n} \frac{2}{n} \frac{1}{1+2i/n}
  4. ∑i=1n1n1i/n\sum_{i=1}^{n} \frac{1}{n} \frac{1}{i/n}
Explanation: For the integral ∫121x dx\int_1^2 \frac{1}{x} \,dx, we have a=1,b=2a=1, b=2, and f(x)=1xf(x) = \frac{1}{x}. The width of each subinterval is Δx=b−an=2−1n=1n\Delta x = \frac{b-a}{n} = \frac{2-1}{n} = \frac{1}{n}. For a right Riemann sum, the sample points are xi=a+iΔx=1+inx_i = a + i\Delta x = 1 + \frac{i}{n}. The sum is ∑i=1nf(xi)Δx=∑i=1n11+i/n⋅1n\sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} \frac{1}{1+i/n} \cdot \frac{1}{n}. This matches option B. Option A is a left Riemann sum. Option C has an incorrect Δx\Delta x. Option D represents the integral ∫011x dx\int_0^1 \frac{1}{x} \,dx, which is an improper integral.

Question 9

The midpoint Riemann sum for ∫−13x3 dx\int_{-1}^{3} x^3 \, dx using 4 subintervals has sample points at which xx-values?

  1. x=0,1,2,3x = 0, 1, 2, 3
  2. x=−1,0,1,2x = -1, 0, 1, 2
  3. x=−0.5,0.5,1.5,2.5x = -0.5, 0.5, 1.5, 2.5 (correct answer)
  4. x=−0.5,1,2,2.5x = -0.5, 1, 2, 2.5
Explanation: When you encounter a midpoint Riemann sum problem, you need to systematically find where the sample points fall within each subinterval. The key is understanding that midpoint means exactly halfway between the left and right endpoints of each subinterval. For this integral from x=−1x = -1 to x=3x = 3 with 4 subintervals, start by finding the width: Δx=3−(−1)4=1\Delta x = \frac{3 - (-1)}{4} = 1. This gives you four subintervals: [−1,0],[0,1],[1,2][-1, 0],[0, 1],[1, 2], and [2,3][2, 3]. The midpoint of each subinterval is found by averaging its endpoints. For [−1,0][-1, 0]: −1+02=−0.5\frac{-1 + 0}{2} = -0.5. For [0,1][0, 1]: 0+12=0.5\frac{0 + 1}{2} = 0.5. For [1,2][1, 2]: 1+22=1.5\frac{1 + 2}{2} = 1.5. For [2,3][2, 3]: 2+32=2.5\frac{2 + 3}{2} = 2.5. This confirms answer C. Choice A gives you x=0,1,2,3x = 0, 1, 2, 3, which are the right endpoints of each subinterval—this would be a right Riemann sum. Choice B gives you x=−1,0,1,2x = -1, 0, 1, 2, which are the left endpoints—this would be a left Riemann sum. Choice D mixes some correct midpoints (−0.5-0.5 and 2.5) with incorrect values, representing a common error where students confuse different Riemann sum types. Remember: midpoint Riemann sums always use sample points that are exactly halfway between subinterval boundaries. Calculate Δx\Delta x first, then find the midpoint of each subinterval by averaging its endpoints.

Question 10

Let A=∫04(x−1) dxA = \int_0^4 (x-1)\,dx. A midpoint Riemann sum with n=2n=2 subintervals is used to approximate AA. What is the value of this approximation?

  1. 0
  2. 4 (correct answer)
  3. 6
  4. 8
Explanation: The interval is [0,4][0, 4] and n=2n=2, so the subintervals are [0,2][0, 2] and [2,4][2, 4]. The width of each subinterval is Δx=4−02=2\Delta x = \frac{4-0}{2}=2. The midpoint rule uses the midpoints of these subintervals as sample points. The midpoint of [0,2][0, 2] is c1=1c_1 = 1. The midpoint of [2,4][2, 4] is c2=3c_2 = 3. The function is f(x)=x−1f(x) = x-1. The midpoint sum is M2=Δx[f(c1)+f(c2)]=2⋅[f(1)+f(3)]M_2 = \Delta x [f(c_1) + f(c_2)] = 2 \cdot [f(1) + f(3)]. f(1)=1−1=0f(1) = 1-1 = 0. f(3)=3−1=2f(3) = 3-1 = 2. The sum is 2⋅[0+2]=42 \cdot [0 + 2] = 4. For this linear function, the midpoint sum gives the exact value of the integral.

Question 11

Suppose ff is an integrable function and lim⁡n→∞∑i=1nf(ci)Δxi=I\lim_{n\to\infty} \sum_{i=1}^n f(c_i) \Delta x_i = I for any choice of points cic_i in the ii-th subinterval. If the interval of integration is [2,6][2, 6] and a midpoint Riemann sum with 4 equal subintervals gives a value of 10, what is Δx\Delta x for this sum?

  1. 1/21/2
  2. 1 (correct answer)
  3. 4
  4. 10
Explanation: This question tests the understanding of the components of a Riemann sum, specifically Δx\Delta x. The other information provided (the value of the sum is 10, it's a midpoint sum, the limit exists) is extraneous to the question being asked. The interval is [2,6][2, 6] and there are n=4n=4 equal subintervals. The width of each subinterval is calculated as Δx=b−an=6−24=44=1\Delta x = \frac{b-a}{n} = \frac{6-2}{4} = \frac{4}{4} = 1.

Question 12

The limit L=lim⁡n→∞∑i=1n1n+iL = \lim_{n\to\infty} \sum_{i=1}^{n} \frac{1}{n+i} is equivalent to which definite integral?

  1. ∫0111+xdx\int_{0}^{1} \frac{1}{1+x} dx (correct answer)
  2. ∫121xdx\int_{1}^{2} \frac{1}{x} dx
  3. ∫011xdx\int_{0}^{1} \frac{1}{x} dx
  4. ∫121n+xdx\int_{1}^{2} \frac{1}{n+x} dx
Explanation: To identify the definite integral, we need to rewrite the sum in the form ∑f(xi)Δx\sum f(x_i) \Delta x. We can factor out 1n\frac{1}{n} from the term in the summand: 1n+i=1n(1+i/n)=1n⋅11+i/n\frac{1}{n+i} = \frac{1}{n(1+i/n)} = \frac{1}{n} \cdot \frac{1}{1+i/n}. The sum becomes lim⁡n→∞∑i=1n11+i/n⋅1n\lim_{n\to\infty} \sum_{i=1}^{n} \frac{1}{1+i/n} \cdot \frac{1}{n}. Now it matches the form of a right Riemann sum. We can identify Δx=1n\Delta x = \frac{1}{n}. This means b−a=1b-a=1. The sample points are xi=a+iΔxx_i = a+i\Delta x. From the term 11+i/n\frac{1}{1+i/n}, we can identify f(xi)=11+xif(x_i) = \frac{1}{1+x_i} with xi=i/nx_i = i/n. If xi=i/nx_i = i/n, then a=0a=0 and the interval is [0,1][0, 1]. The function is f(x)=11+xf(x) = \frac{1}{1+x}. Therefore, the limit is equivalent to the integral ∫0111+xdx\int_{0}^{1} \frac{1}{1+x} dx. Note that ∫121xdx\int_{1}^{2} \frac{1}{x} dx has the same value, ln⁡(2)\ln(2), but it corresponds to the sum lim⁡n→∞∑i=1n11+i/n⋅1n\lim_{n\to\infty} \sum_{i=1}^{n} \frac{1}{1+i/n} \cdot \frac{1}{n}, which is not the original expression.

Question 13

Let Rn=∑i=1nf(ci)ΔxR_n = \sum_{i=1}^{n} f(c_i) \Delta x be a Riemann sum for a function ff on [a,b][a, b], with Δx=b−an\Delta x = \frac{b-a}{n} and cic_i being any point in the ii-th subinterval. Which of the following is a sufficient condition on the function ff to guarantee that lim⁡n→∞Rn=∫abf(x)dx\lim_{n\to\infty} R_n = \int_a^b f(x) dx?

  1. ff must be a polynomial function.
  2. ff must be a differentiable function.
  3. ff must be a continuous function. (correct answer)
  4. ff must be a strictly monotonic function.
Explanation: A fundamental theorem in calculus states that if a function ff is continuous on a closed interval [a,b][a, b], then it is Riemann integrable on [a,b][a, b]. This means that the limit of the Riemann sum exists as n→∞n \to \infty and is equal to the definite integral ∫abf(x)dx\int_a^b f(x) dx, regardless of the choice of sample points cic_i in each subinterval. While polynomial, differentiable, and monotonic functions are all integrable on closed intervals (because they are continuous), continuity is the most general sufficient condition among the choices.

Question 14

Let f(x)f(x) be a strictly decreasing and concave down function on the interval [a,b][a, b]. Let LnL_n, RnR_n, MnM_n, and TnT_n be the left-hand, right-hand, midpoint, and trapezoidal rule approximations for ∫abf(x)dx\int_a^b f(x) dx with nn subintervals, respectively. Which of the following inequalities must be true?

  1. Ln<Mn<∫abf(x)dx<TnL_n < M_n < \int_a^b f(x) dx < T_n
  2. Rn<Tn<∫abf(x)dx<MnR_n < T_n < \int_a^b f(x) dx < M_n (correct answer)
  3. Mn<∫abf(x)dx<Tn<LnM_n < \int_a^b f(x) dx < T_n < L_n
  4. Rn<Mn<∫abf(x)dx<LnR_n < M_n < \int_a^b f(x) dx < L_n
Explanation: For a strictly decreasing function, the left-hand sum uses the maximum value in each subinterval, making it an overestimate. The right-hand sum uses the minimum value, making it an underestimate. Thus, Rn<∫abf(x)dx<LnR_n < \int_a^b f(x) dx < L_n. For a concave down function, the secant line segment used by the trapezoidal rule lies below the curve, so TnT_n is an underestimate. The tangent line at the midpoint used for the midpoint rule lies above the curve, so MnM_n is an overestimate. Thus, Tn<∫abf(x)dx<MnT_n < \int_a^b f(x) dx < M_n. Combining these, we know both RnR_n and TnT_n are underestimates, and MnM_n is an overestimate. It is a known property that for a function that is both decreasing and concave down, the trapezoidal approximation TnT_n is a better approximation than the right-hand sum RnR_n. Therefore, the correct ordering is Rn<Tn<∫abf(x)dx<MnR_n < T_n < \int_a^b f(x) dx < M_n.

Question 15

An approximation of ∫04(x2+1)dx\int_{0}^{4} (x^2 + 1) dx is made using a right Riemann sum with nn subintervals of equal width, denoted by Sn=∑i=1nf(xi)ΔxS_n = \sum_{i=1}^{n} f(x_i) \Delta x. Which of the following expressions is equivalent to SnS_n?

  1. 16n2∑i=1ni2+4\frac{16}{n^2} \sum_{i=1}^{n} i^2 + 4
  2. 64n3∑i=1ni2+4n\frac{64}{n^3} \sum_{i=1}^{n} i^2 + \frac{4}{n}
  3. 64n3(∑i=1ni)2+4\frac{64}{n^3} \left( \sum_{i=1}^{n} i \right)^2 + 4
  4. 64n3∑i=1ni2+4\frac{64}{n^3} \sum_{i=1}^{n} i^2 + 4 (correct answer)
Explanation: For the integral ∫04(x2+1)dx\int_{0}^{4} (x^2 + 1) dx with nn subintervals, Δx=4−0n=4n\Delta x = \frac{4-0}{n} = \frac{4}{n}. For a right Riemann sum, the sample points are xi=a+iΔx=0+i4n=4inx_i = a + i\Delta x = 0 + i\frac{4}{n} = \frac{4i}{n}. The function is f(x)=x2+1f(x) = x^2 + 1, so f(xi)=(4in)2+1=16i2n2+1f(x_i) = \left(\frac{4i}{n}\right)^2 + 1 = \frac{16i^2}{n^2} + 1. The Riemann sum is Sn=∑i=1nf(xi)Δx=∑i=1n(16i2n2+1)4nS_n = \sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} \left(\frac{16i^2}{n^2} + 1\right) \frac{4}{n}. Using properties of summation, we can distribute: Sn=4n∑i=1n(16i2n2+1)=4n(∑i=1n16i2n2+∑i=1n1)S_n = \frac{4}{n} \sum_{i=1}^{n} \left(\frac{16i^2}{n^2} + 1\right) = \frac{4}{n} \left( \sum_{i=1}^{n} \frac{16i^2}{n^2} + \sum_{i=1}^{n} 1 \right). We can pull out constants from the sums: Sn=4n(16n2∑i=1ni2+n)S_n = \frac{4}{n} \left( \frac{16}{n^2} \sum_{i=1}^{n} i^2 + n \right). Distributing the 4n\frac{4}{n} gives Sn=64n3∑i=1ni2+4n⋅n=64n3∑i=1ni2+4S_n = \frac{64}{n^3} \sum_{i=1}^{n} i^2 + \frac{4}{n} \cdot n = \frac{64}{n^3} \sum_{i=1}^{n} i^2 + 4.

Question 16

Let ff be a function that is strictly decreasing and concave up on the interval [a,b][a, b]. Let Ln,Rn,Mn,TnL_n, R_n, M_n, T_n be the left-hand, right-hand, midpoint, and trapezoidal rule approximations, respectively, for ∫abf(x) dx\int_a^b f(x) \,dx with nn subintervals. Which of the following inequalities must be true?

  1. Ln<∫abf(x) dxL_n < \int_a^b f(x) \,dx
  2. Rn>∫abf(x) dxR_n > \int_a^b f(x) \,dx
  3. Mn>∫abf(x) dxM_n > \int_a^b f(x) \,dx
  4. Tn>∫abf(x) dxT_n > \int_a^b f(x) \,dx (correct answer)
Explanation: For a function that is strictly decreasing on [a,b][a, b], the left Riemann sum LnL_n is an overestimate and the right Riemann sum RnR_n is an underestimate. Thus, Rn<∫abf(x) dx<LnR_n < \int_a^b f(x) \,dx < L_n. This eliminates choices A and B. For a function that is concave up on [a,b][a, b], the trapezoidal rule TnT_n is an overestimate and the midpoint rule MnM_n is an underestimate. Thus, Mn<∫abf(x) dx<TnM_n < \int_a^b f(x) \,dx < T_n. This eliminates choice C and confirms choice D is correct. The information that the function is decreasing is used to eliminate A and B, while the information about concavity is used for C and D.

Question 17

Let Sn=∑i=1nf(ci)ΔxS_n = \sum_{i=1}^n f(c_i) \Delta x be a Riemann sum for a continuous function ff on [a,b][a, b] with Δx=(b−a)/n\Delta x = (b-a)/n and ci∈[xi−1,xi]c_i \in [x_{i-1}, x_i]. Which of the following statements is not necessarily true?

  1. lim⁡n→∞Sn\lim_{n \to \infty} S_n exists and is equal to ∫abf(x) dx\int_a^b f(x) \,dx.
  2. If RnR_n is the right Riemann sum, then lim⁡n→∞Rn=∫abf(x) dx\lim_{n \to \infty} R_n = \int_a^b f(x) \,dx.
  3. If ff is strictly increasing, then the left Riemann sum LnL_n is a strict underestimate of ∫abf(x) dx\int_a^b f(x) \,dx.
  4. If ff is strictly increasing, then the sequence of left Riemann sums LnL_n is strictly increasing, i.e., Ln<Ln+1L_n < L_{n+1} for all n≥1n \ge 1. (correct answer)
Explanation: A) This is the definition of the definite integral for a continuous (and thus integrable) function; it is true regardless of the choice of sample points cic_i. B) The right Riemann sum is a specific type of Riemann sum, so its limit is also the definite integral. This is true. C) For a strictly increasing function, the minimum value on any subinterval [xi−1,xi][x_{i-1}, x_i] is at the left endpoint xi−1x_{i-1}. Therefore, the left Riemann sum LnL_n will be a strict underestimate of the integral. This is true. D) While LnL_n converges to the integral from below, the sequence {Ln}\{L_n\} itself is not guaranteed to be monotonically increasing. Adding more rectangles can sometimes slightly decrease the sum value before it continues to increase toward the limit. This statement is not necessarily true.

Question 18

The rate of consumption of a resource is given by C(t)=10−e−0.1tC(t) = 10 - e^{-0.1t} units per year, where tt is the number of years from the present. A right Riemann sum with 5 equal subintervals is used to estimate the total consumption over the next 10 years. This estimate is:

  1. an overestimate, because C(t)C(t) is increasing. (correct answer)
  2. an underestimate, because C(t)C(t) is increasing.
  3. an overestimate, because C(t)C(t) is concave down.
  4. an underestimate, because C(t)C(t) is concave down.
Explanation: The relationship between a Riemann sum approximation and the true value of the integral depends on the function's monotonic behavior for left/right sums. We need to determine if C(t)C(t) is increasing or decreasing. The derivative is C′(t)=−(−0.1)e−0.1t=0.1e−0.1tC'(t) = -(-0.1)e^{-0.1t} = 0.1e^{-0.1t}. Since the exponential function is always positive, C′(t)>0C'(t) > 0 for all tt. This means C(t)C(t) is a strictly increasing function. For an increasing function, a right Riemann sum uses the maximum value of the function in each subinterval, resulting in an overestimate of the integral. Concavity determines over/underestimation for trapezoidal and midpoint rules, but not for left/right sums.

Question 19

Let LnL_n be the left Riemann sum, RnR_n be the right Riemann sum, and MnM_n be the midpoint Riemann sum for ∫02(x2−2x) dx\int_0^2 (x^2-2x) \,dx with nn equal subintervals. Which statement is true for any n≥2n \ge 2?

  1. LnL_n is an underestimate.
  2. RnR_n is an overestimate.
  3. Ln=RnL_n = R_n. (correct answer)
  4. The relationship between the sums and the integral cannot be determined.
Explanation: The function f(x)=x2−2x=x(x−2)f(x) = x^2-2x = x(x-2) is a parabola opening upwards with roots at x=0x=0 and x=2x=2. This function is symmetric about its axis of symmetry at x=0+22=1x = \frac{0+2}{2} = 1. Over the interval [0,2][0, 2], the function is decreasing on [0,1][0, 1] and increasing on [1,2][1, 2]. Because the function is not monotonic over the entire interval [0,2][0, 2], we cannot conclude that LnL_n or RnR_n is always an overestimate or underestimate. However, there is a special property for the difference: Rn−Ln=(f(b)−f(a))ΔxR_n - L_n = (f(b) - f(a))\Delta x. Here, a=0,b=2a=0, b=2. f(0)=0f(0)=0 and f(2)=22−2(2)=0f(2)=2^2-2(2)=0. Thus, Rn−Ln=(0−0)2n=0R_n - L_n = (0-0)\frac{2}{n} = 0, which means Ln=RnL_n = R_n for any nn.

Question 20

The velocity of a particle is given by v(t)v(t) for t≥0t \ge 0. The expression ∑k=110v(0.5k)⋅(0.5)\sum_{k=1}^{10} v(0.5k) \cdot (0.5) represents an approximation of the particle's displacement. Which of the following does this sum represent?

  1. A left Riemann sum on the interval [0,10][0, 10] with 20 subintervals.
  2. A right Riemann sum on the interval [0,5][0, 5] with 10 subintervals. (correct answer)
  3. A midpoint Riemann sum on the interval [0,5][0, 5] with 5 subintervals.
  4. A right Riemann sum on the interval [0,10][0, 10] with 10 subintervals.
Explanation: Let's analyze the given sum: ∑k=110v(0.5k)⋅(0.5)\sum_{k=1}^{10} v(0.5k) \cdot (0.5). This is a Riemann sum with n=10n=10 terms. The width of each subinterval is Δx=0.5\Delta x = 0.5. The summation index kk runs from 1 to 10. The sample points are tk=0.5kt_k = 0.5k. When k=1,t1=0.5k=1, t_1=0.5. When k=10,t10=5k=10, t_{10}=5. Since the index starts at 1 and the sample points are a+kΔxa+k\Delta x (with a=0a=0), this is a right Riemann sum. The interval ends at b=a+nΔx=0+10(0.5)=5b = a + n\Delta x = 0 + 10(0.5) = 5. So, the sum represents a right Riemann sum approximation for ∫05v(t) dt\int_0^5 v(t)\,dt with 10 subintervals.