Calculus 1 Quiz: Riemann Sum Approximations
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Riemann Sum ApproximationsQuestion 1 of 20

Let ff be a function defined by f(x)=2x2−8x+11f(x) = 2x^2 - 8x + 11. A midpoint Riemann sum M2M_2 with two equal subintervals is used to approximate ∫04f(x)dx\int_0^4 f(x) dx. What is the value of M2M_2?

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40
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Calculus 1 Quiz

Calculus 1 Quiz: Riemann Sum Approximations

Practice Riemann Sum Approximations 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 Sum Approximations, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

Let ff be a function defined by f(x)=2x2−8x+11f(x) = 2x^2 - 8x + 11. A midpoint Riemann sum M2M_2 with two equal subintervals is used to approximate ∫04f(x)dx\int_0^4 f(x) dx. What is the value of M2M_2?

  1. 20 (correct answer)
  2. 22
  3. 26
  4. 40
Explanation: The interval is [0,4][0,4] with n=2n=2 subintervals, so the width is Δx=(4−0)/2=2\Delta x = (4-0)/2 = 2. The subintervals are [0,2][0,2] and [2,4][2,4]. The midpoints of these subintervals are x1∗=1x_1^* = 1 and x2∗=3x_2^* = 3. We evaluate the function at these midpoints: f(1)=2(1)2−8(1)+11=5f(1) = 2(1)^2 - 8(1) + 11 = 5. f(3)=2(3)2−8(3)+11=18−24+11=5f(3) = 2(3)^2 - 8(3) + 11 = 18 - 24 + 11 = 5. The midpoint sum is M2=Δx[f(1)+f(3)]=2[5+5]=2[10]=20M_2 = \Delta x[f(1) + f(3)] = 2[5 + 5] = 2[10] = 20.

Question 2

A right Riemann sum with two equal subintervals is used to approximate ∫04(x3+k) dx\int_0^4 (x^3 + k) \,dx. If the approximation equals 164, what is the value of kk?

  1. 5 (correct answer)
  2. 27
  3. 37
  4. 46
Explanation: For the interval [0,4][0,4] with n=2n=2 subintervals, the width is Δx=(4−0)/2=2\Delta x = (4-0)/2 = 2. The subintervals are [0,2][0,2] and [2,4][2,4]. A right Riemann sum uses the right endpoints, which are x1=2x_1=2 and x2=4x_2=4. The function is f(x)=x3+kf(x) = x^3+k. The sum is R2=Δx[f(2)+f(4)]=2[(23+k)+(43+k)]=2[(8+k)+(64+k)]=2[72+2k]=144+4kR_2 = \Delta x [f(2) + f(4)] = 2[(2^3+k) + (4^3+k)] = 2[(8+k) + (64+k)] = 2[72+2k] = 144 + 4k. Setting this equal to 164 gives 144+4k=164144 + 4k = 164, which simplifies to 4k=204k = 20, so k=5k=5.

Question 3

The expression ∑k=092⋅(3+2k)2\sum_{k=0}^{9} 2 \cdot (3 + 2k)^2 is a left Riemann sum with 10 equal subintervals for which definite integral?

  1. ∫323x2 dx\int_3^{23} x^2 \,dx (correct answer)
  2. ∫3232x2 dx\int_3^{23} 2x^2 \,dx
  3. ∫321x2 dx\int_3^{21} x^2 \,dx
  4. ∫525x2 dx\int_5^{25} x^2 \,dx
Explanation: The sum has 10 terms (from k=0k=0 to k=9k=9), so n=10n=10. The term outside the function, 22, represents Δx\Delta x. Thus, the total width of the interval is b−a=n⋅Δx=10⋅2=20b-a = n \cdot \Delta x = 10 \cdot 2 = 20. The sum is a left Riemann sum, so the sample points are xk=a+kΔxx_k = a + k \Delta x. The expression inside the square is (3+2k)(3+2k), which represents the sample points being evaluated. Matching a+kΔxa + k \Delta x with 3+2k3+2k, we get a+k(2)=3+2ka+k(2) = 3+2k, which implies the starting point is a=3a=3. The endpoint is b=a+20=3+20=23b = a + 20 = 3 + 20 = 23. The function being evaluated is f(x)=x2f(x)=x^2. Therefore, the integral is ∫323x2 dx\int_3^{23} x^2 \,dx.

Question 4

Let Rn=∑k=1n2nf(1+2kn)R_n = \sum_{k=1}^n \frac{2}{n} f(1+\frac{2k}{n}) be a right Riemann sum for an integral of a function ff over an interval [a,b][a,b]. What are the values of aa and bb?

  1. a=0,b=2a=0, b=2
  2. a=1,b=2a=1, b=2
  3. a=1,b=3a=1, b=3 (correct answer)
  4. a=2,b=4a=2, b=4
Explanation: The general form for a right Riemann sum is ∑k=1nf(a+kΔx)Δx\sum_{k=1}^n f(a+k\Delta x)\Delta x, where Δx=(b−a)/n\Delta x = (b-a)/n. From the given sum, we can identify Δx=2/n\Delta x = 2/n. This means the length of the interval of integration is b−a=2b-a = 2. The sample points are xk=1+2knx_k = 1 + \frac{2k}{n}. Comparing this with the general form xk=a+kΔx=a+k2nx_k = a + k\Delta x = a + k\frac{2}{n}, we can see that the starting point is a=1a=1. Since the interval length is 2, the endpoint is b=a+2=1+2=3b = a + 2 = 1 + 2 = 3. Thus, the interval is [1,3][1,3].

Question 5

Let LnL_n be the left Riemann sum approximation for ∫abf(x) dx\int_a^b f(x) \,dx. Which of the following expressions represents the left Riemann sum approximation with nn subintervals for ∫a+2b+2f(x−2) dx\int_{a+2}^{b+2} f(x-2) \,dx?

  1. LnL_n (correct answer)
  2. Ln+2L_n + 2
  3. Ln−2L_n - 2
  4. 2Ln2L_n
Explanation: Let g(x)=f(x−2)g(x) = f(x-2). The integral ∫a+2b+2g(x) dx\int_{a+2}^{b+2} g(x) \,dx represents a horizontal shift of the function f(x)f(x) and the interval of integration by 2 units to the right. The area under the curve remains unchanged. By a u-substitution with u=x−2u=x-2, we find ∫a+2b+2f(x−2) dx=∫abf(u) du\int_{a+2}^{b+2} f(x-2) \,dx = \int_a^b f(u) \,du. Since the definite integrals are identical, their left Riemann sum approximations with the same number of subintervals will also be identical. The sample points are shifted, but the function values at those new points are the same as the original function values at the original points, and Δx\Delta x remains the same.

Question 6

Let ff be a function that is strictly decreasing and concave down on the interval [2,10][2, 10]. Let L4L_4, R4R_4, and T4T_4 be the left Riemann sum, right Riemann sum, and trapezoidal sum approximations of ∫210f(x) dx\int_2^{10} f(x) \,dx with 4 equal subintervals. Which of the following inequalities must be true?

  1. R4<T4<∫210f(x) dx<L4R_4 < T_4 < \int_2^{10} f(x) \,dx < L_4 (correct answer)
  2. L4<∫210f(x) dx<T4<R4L_4 < \int_2^{10} f(x) \,dx < T_4 < R_4
  3. R4<∫210f(x) dx<T4<L4R_4 < \int_2^{10} f(x) \,dx < T_4 < L_4
  4. L4<T4<∫210f(x) dx<R4L_4 < T_4 < \int_2^{10} f(x) \,dx < R_4
Explanation: For a strictly decreasing function, the left Riemann sum (L4L_4) uses the highest point in each subinterval, resulting in an overestimate. The right Riemann sum (R4R_4) uses the lowest point, resulting in an underestimate. For a function that is concave down, the secant lines used for the trapezoidal sum (T4T_4) lie below the curve, also resulting in an underestimate. Therefore, both R4R_4 and T4T_4 are underestimates, while L4L_4 is an overestimate. This means ∫210f(x) dx<L4\int_2^{10} f(x) \,dx < L_4. Furthermore, the trapezoidal sum is the average of the left and right sums (T4=L4+R42T_4 = \frac{L_4+R_4}{2}), so it must lie between them. Since ff is decreasing, R4<L4R_4 < L_4, which implies R4<T4<L4R_4 < T_4 < L_4. Combining these facts gives the full ordering: R4<T4<∫210f(x) dx<L4R_4 < T_4 < \int_2^{10} f(x) \,dx < L_4.

Question 7

The rate at which water flows into a reservoir is measured at various times. At time t=0t=0 hours, the rate is 100 cubic meters per hour. Two hours later (at t=2t=2), the rate is 150 m³/hr. One hour after that (at t=3t=3), the rate is 120 m³/hr. Finally, at t=5t=5 hours, the rate is 180 m³/hr.

Using a trapezoidal sum with subintervals given by the measurements, estimate the total volume of water that flowed into the reservoir during the 5-hour period.

  1. 685 m³ (correct answer)
  2. 590 m³
  3. 780 m³
  4. 1370 m³
Explanation: The total volume is the integral of the rate. We approximate this with a trapezoidal sum over the non-uniform intervals [0,2][0,2], [2,3][2,3], and [3,5][3,5]. The area of each trapezoid is 12Δt(r1+r2)\frac{1}{2} \Delta t (r_1 + r_2). Volume ≈12(2)(100+150)+12(1)(150+120)+12(2)(120+180)=1(250)+12(270)+1(300)=250+135+300=685\approx \frac{1}{2}(2)(100+150) + \frac{1}{2}(1)(150+120) + \frac{1}{2}(2)(120+180) = 1(250) + \frac{1}{2}(270) + 1(300) = 250 + 135 + 300 = 685 m³.

Question 8

Let f(x)=sin⁡(x2)f(x) = \sin(x^2). The integral I=∫0πsin⁡(x2) dxI = \int_0^{\sqrt{\pi}} \sin(x^2) \,dx is approximated using a right Riemann sum with nn subintervals, RnR_n. Which statement is the most accurate description of the error, En=∣I−Rn∣E_n = |I - R_n|?

  1. EnE_n is always zero if nn is large enough.
  2. E10E_{10} is guaranteed to be smaller than E5E_5.
  3. As n→∞n \to \infty, En→0E_n \to 0. (correct answer)
  4. The sign of the error I−RnI - R_n is always positive.
Explanation: The definite integral is defined as the limit of a Riemann sum as the number of subintervals nn approaches infinity. This means that as nn increases, the approximation RnR_n gets closer to the actual value of the integral II. Consequently, the absolute error En=∣I−Rn∣E_n = |I - R_n| approaches 0. Choice A is incorrect because the approximation is not exact for any finite nn. Choice B is not guaranteed; while the error generally decreases with larger nn, it's not a monotonic decrease for all functions. Choice D is incorrect because the function sin⁡(x2)\sin(x^2) is not monotonic on [0,π][0, \sqrt{\pi}], so the right Riemann sum will be an overestimate on some subintervals and an underestimate on others, making the sign of the total error indeterminate.

Question 9

Let f(x)={4if 0≤x<2x+2if 2≤x≤6f(x) = \begin{cases} 4 & \text{if } 0 \le x < 2 \\ x+2 & \text{if } 2 \le x \le 6 \end{cases}. Calculate the midpoint Riemann sum for ∫06f(x) dx\int_0^6 f(x) \,dx with 3 equal subintervals.

  1. 32 (correct answer)
  2. 28
  3. 36
  4. 30
Explanation: The interval [0,6][0,6] with n=3n=3 subintervals gives Δx=(6−0)/3=2\Delta x = (6-0)/3 = 2. The subintervals are [0,2][0,2], [2,4][2,4], and [4,6][4,6]. The midpoints are x1∗=1x_1^*=1, x2∗=3x_2^*=3, and x3∗=5x_3^*=5. We evaluate the function at these midpoints: f(1)=4f(1)=4 (using the first rule), f(3)=3+2=5f(3)=3+2=5 (using the second rule), and f(5)=5+2=7f(5)=5+2=7 (using the second rule). The midpoint sum is M3=Δx[f(1)+f(3)+f(5)]=2[4+5+7]=2[16]=32M_3 = \Delta x [f(1)+f(3)+f(5)] = 2[4+5+7] = 2[16] = 32.

Question 10

The sum S=14(1+1.25+1.5+1.75)S = \frac{1}{4} \left( \sqrt{1} + \sqrt{1.25} + \sqrt{1.5} + \sqrt{1.75} \right) is an approximation for ∫12x dx\int_1^2 \sqrt{x} \,dx. Which of the following is true?

  1. S is a right Riemann sum and it is an overestimate.
  2. S is a left Riemann sum and it is an overestimate.
  3. S is a right Riemann sum and it is an underestimate.
  4. S is a left Riemann sum and it is an underestimate. (correct answer)
Explanation: When you encounter a Riemann sum approximation problem, you need to identify two key aspects: whether it's a left or right sum, and whether the function's concavity makes it an over- or underestimate. To determine if this is a left or right Riemann sum, examine the interval [1,2][1,2] with width Δx=14\Delta x = \frac{1}{4}, giving us 4 subintervals. The x-values being used are 1, 1.25, 1.5, and 1.75. Since we're integrating from 1 to 2, these represent the left endpoints of each subinterval: [1,1.25][1, 1.25], [1.25,1.5][1.25, 1.5], [1.5,1.75][1.5, 1.75], and [1.75,2][1.75, 2]. This makes S a left Riemann sum. For f(x)=xf(x) = \sqrt{x}, we need to determine concavity. Taking derivatives: f′(x)=12xf'(x) = \frac{1}{2\sqrt{x}} and f′′(x)=−14x3/2<0f''(x) = -\frac{1}{4x^{3/2}} < 0 for all x>0x > 0. Since the second derivative is negative, the function is concave down on our interval. When a function is concave down, left Riemann sums underestimate the integral because the left endpoint values lie below the curve's average height over each subinterval. Answer D correctly identifies both aspects. Answer A incorrectly calls it a right sum. Answer B correctly identifies it as a left sum but wrongly claims it's an overestimate. Answer C incorrectly identifies it as a right sum, though it correctly states underestimate. Study tip: Remember that for concave down functions, left sums underestimate and right sums overestimate. For concave up functions, it's the opposite.

Question 11

A left Riemann sum with 10 equal subintervals is used to approximate I=∫05x2 dxI = \int_0^5 x^2 \,dx, and its value is L10L_{10}. What is the value of the left Riemann sum with 10 equal subintervals for the integral J=∫05(x+1)2 dxJ = \int_0^5 (x+1)^2 \,dx in terms of L10L_{10}?

  1. The value cannot be determined from L10L_{10}.
  2. L10+30L_{10} + 30
  3. L10+5L_{10} + 5
  4. L10+27.5L_{10} + 27.5 (correct answer)
Explanation: When you encounter Riemann sum problems involving transformed functions, look for ways to relate the new sum to the original by analyzing how the transformation affects each term. For the left Riemann sum L10L_{10} approximating ∫05x2 dx\int_0^5 x^2 \,dx, we have 10 subintervals of width Δx=0.5\Delta x = 0.5. The left endpoints are x0=0,x1=0.5,x2=1.0,…,x9=4.5x_0 = 0, x_1 = 0.5, x_2 = 1.0, \ldots, x_9 = 4.5, giving us L10=0.5∑i=09xi2L_{10} = 0.5 \sum_{i=0}^9 x_i^2. For ∫05(x+1)2 dx\int_0^5 (x+1)^2 \,dx, the corresponding left Riemann sum uses the same partition but evaluates (x+1)2(x+1)^2 at each left endpoint. This gives us 0.5∑i=09(xi+1)20.5 \sum_{i=0}^9 (x_i + 1)^2. Expanding (xi+1)2=xi2+2xi+1(x_i + 1)^2 = x_i^2 + 2x_i + 1, we get: 0.5∑i=09(xi2+2xi+1)=0.5∑i=09xi2+0.5∑i=092xi+0.5∑i=0910.5 \sum_{i=0}^9 (x_i^2 + 2x_i + 1) = 0.5 \sum_{i=0}^9 x_i^2 + 0.5 \sum_{i=0}^9 2x_i + 0.5 \sum_{i=0}^9 1 This equals L10+∑i=09xi+5L_{10} + \sum_{i=0}^9 x_i + 5. Since ∑i=09xi=0+0.5+1.0+⋯+4.5=22.5\sum_{i=0}^9 x_i = 0 + 0.5 + 1.0 + \cdots + 4.5 = 22.5, the answer is L10+22.5+5=L10+27.5L_{10} + 22.5 + 5 = L_{10} + 27.5. Choice (A) is wrong because we can systematically relate the sums through algebraic expansion. Choice (B) incorrectly calculates the additional terms. Choice (C) only accounts for the constant term contribution, missing the linear term 2xi2x_i. Strategy tip: When dealing with transformed functions in Riemann sums, expand the transformation algebraically and use linearity to separate the sum into manageable pieces you can relate back to the original.

Question 12

A left Riemann sum LnL_n and a right Riemann sum RnR_n are calculated for ∫abf(x)dx\int_a^b f(x)dx. If f(x)f(x) is a strictly increasing linear function, which expression represents the exact value of the integral?

  1. Tn=Ln+Rn2T_n = \frac{L_n + R_n}{2}
  2. MnM_n, the midpoint sum
  3. Both A and B are exact. (correct answer)
  4. Neither A nor B is exact.
Explanation: For any linear function, both the trapezoidal sum and the midpoint sum give the exact value of the definite integral for any number of subintervals nn. The top of each trapezoid (a secant line) will lie exactly on the linear function's graph. Similarly, the area of each midpoint rectangle is exactly equal to the area of the trapezoid formed by the interval on the x-axis, the vertical lines at the endpoints, and the tangent line at the midpoint. For a linear function, this tangent line is the function itself. Therefore, both approximations are exact. The expression in A is the formula for the trapezoidal sum, TnT_n.

Question 13

The average value of a continuous function ff on [a,b][a,b] is given by 1b−a∫abf(x)dx\frac{1}{b-a}\int_a^b f(x)dx. A right Riemann sum with nn equal subintervals, RnR_n, is used to approximate the integral. Which expression gives the corresponding approximation of the average value of ff?

  1. 1n(b−a)∑k=1nf(a+kb−an)\frac{1}{n(b-a)} \sum_{k=1}^n f(a+k\frac{b-a}{n})
  2. 1b−a∑k=1nf(a+kb−an)\frac{1}{b-a} \sum_{k=1}^n f(a+k\frac{b-a}{n})
  3. b−an2∑k=1nf(a+kb−an)\frac{b-a}{n^2} \sum_{k=1}^n f(a+k\frac{b-a}{n})
  4. 1n∑k=1nf(a+kb−an)\frac{1}{n} \sum_{k=1}^n f(a+k\frac{b-a}{n}) (correct answer)
Explanation: When working with Riemann sums and average values, remember that the average value formula involves dividing the integral by the interval length, but a Riemann sum already incorporates that interval length through its subinterval widths. Let's build this step by step. A right Riemann sum with n equal subintervals is: Rn=b−an∑k=1nf(a+kb−an)R_n = \frac{b-a}{n} \sum_{k=1}^n f\left(a+k\frac{b-a}{n}\right) The average value formula is 1b−a∫abf(x)dx\frac{1}{b-a}\int_a^b f(x)dx. When we approximate the integral with RnR_n, we get: Average value≈1b−a⋅b−an∑k=1nf(a+kb−an)=1n∑k=1nf(a+kb−an)\text{Average value} \approx \frac{1}{b-a} \cdot \frac{b-a}{n} \sum_{k=1}^n f\left(a+k\frac{b-a}{n}\right) = \frac{1}{n} \sum_{k=1}^n f\left(a+k\frac{b-a}{n}\right) This gives us answer D. Notice how the (b−a)(b-a) terms cancel out, leaving us with simply the arithmetic mean of the function values at the right endpoints. Choice A divides by an extra factor of (b−a)(b-a), which would underestimate the average. Choice B forgets to account for the width factor b−an\frac{b-a}{n} that's already built into the Riemann sum formula. Choice C incorrectly squares the denominator and moves (b−a)(b-a) to the numerator. Key insight: When approximating average values with Riemann sums, you're essentially taking the arithmetic mean of the sampled function values. The interval length factors cancel out naturally, leaving you with 1n\frac{1}{n} times the sum of function evaluations.

Question 14

Let SLS_L be the left Riemann sum and SRS_R be the right Riemann sum for ∫abf(x) dx\int_a^b f(x) \,dx with nn equal subintervals of width Δx\Delta x. Which expression is equivalent to SR−SLS_R - S_L?

  1. f(b)−f(a)f(b) - f(a)
  2. Δx(f(b)−f(a))\Delta x (f(b) - f(a)) (correct answer)
  3. 00
  4. f(b)−f(a)n\frac{f(b) - f(a)}{n}
Explanation: Let the subinterval endpoints be x0,x1,...,xnx_0, x_1, ..., x_n. SL=Δx[f(x0)+f(x1)+...+f(xn−1)]S_L = \Delta x [f(x_0) + f(x_1) + ... + f(x_{n-1})] and SR=Δx[f(x1)+f(x2)+...+f(xn)]S_R = \Delta x [f(x_1) + f(x_2) + ... + f(x_n)]. When we subtract SLS_L from SRS_R, most terms cancel out: SR−SL=Δx[(f(x1)+...+f(xn))−(f(x0)+...+f(xn−1))]=Δx[f(xn)−f(x0)]S_R - S_L = \Delta x [ (f(x_1) + ... + f(x_n)) - (f(x_0) + ... + f(x_{n-1})) ] = \Delta x [f(x_n) - f(x_0)]. Since xn=bx_n=b and x0=ax_0=a, this simplifies to Δx(f(b)−f(a))\Delta x (f(b) - f(a)).

Question 15

The integral ∫08g(x)dx\int_0^8 g(x) dx is approximated by a Riemann sum using the partition x0=0,x1=1,x2=4,x3=8x_0=0, x_1=1, x_2=4, x_3=8. The sample points used are the right endpoints of each subinterval. The values of the function are g(0)=5,g(1)=3,g(4)=6,g(8)=10g(0)=5, g(1)=3, g(4)=6, g(8)=10. What is the value of the approximation?

  1. 72
  2. 38
  3. 45
  4. 61 (correct answer)
Explanation: When you encounter Riemann sum problems, you're approximating the area under a curve by dividing it into rectangles. The key is identifying the width of each subinterval and the height (determined by your sample point). With the partition x0=0,x1=1,x2=4,x3=8x_0=0, x_1=1, x_2=4, x_3=8, you have three subintervals: [0,1][0,1], [1,4][1,4], and [4,8][4,8]. Their widths are Δx1=1−0=1\Delta x_1 = 1-0 = 1, Δx2=4−1=3\Delta x_2 = 4-1 = 3, and Δx3=8−4=4\Delta x_3 = 8-4 = 4. Since you're using right endpoints as sample points, the heights are g(1)=3g(1)=3, g(4)=6g(4)=6, and g(8)=10g(8)=10 respectively. The Riemann sum is: Area=Δx1⋅g(1)+Δx2⋅g(4)+Δx3⋅g(8)\text{Area} = \Delta x_1 \cdot g(1) + \Delta x_2 \cdot g(4) + \Delta x_3 \cdot g(8) =1⋅3+3⋅6+4⋅10=3+18+40=61= 1 \cdot 3 + 3 \cdot 6 + 4 \cdot 10 = 3 + 18 + 40 = 61 Answer choice A (72) likely results from incorrectly using all four function values without proper weighting. Choice B (38) might come from using left endpoints instead of right endpoints, giving 1⋅5+3⋅3+4⋅6=381 \cdot 5 + 3 \cdot 3 + 4 \cdot 6 = 38. Choice C (45) could result from using uniform widths of 8/3 instead of the actual partition widths. The correct answer is D (61). Study tip: Always organize Riemann sum problems in three steps: identify subinterval widths, determine heights using the specified sample points, then multiply and sum. Don't assume equal spacing—calculate each width from the given partition.

Question 16

Let f(x)=−x2f(x) = -x^2. Let AA be the area of the region bounded by the graph of f(x)f(x), the x-axis, and the lines x=0x=0 and x=3x=3. A midpoint Riemann sum with 3 subintervals (M3M_3) is used to approximate ∫03f(x) dx\int_0^3 f(x) \,dx. Which statement is true?

  1. M3M_3 approximates AA and M3>0M_3 > 0.
  2. M3M_3 approximates AA and M3<0M_3 < 0.
  3. M3M_3 approximates −A-A and M3<0M_3 < 0. (correct answer)
  4. M3M_3 approximates −A-A and M3>0M_3 > 0.
Explanation: The definite integral ∫03f(x) dx\int_0^3 f(x) \,dx calculates the signed area between the curve and the x-axis. Since f(x)=−x2f(x)=-x^2 is negative on (0,3](0,3], the integral's value will be negative. Area, AA, is by definition a positive quantity. The relationship is ∫03f(x) dx=−A\int_0^3 f(x) \,dx = -A. A Riemann sum, M3M_3, is an approximation of the definite integral. Therefore, M3M_3 approximates −A-A. Since Δx\Delta x is positive and all function values f(x)f(x) for the midpoints will be negative, the sum M3=Δx∑f(xk∗)M_3 = \Delta x \sum f(x_k^*) must be negative. Thus, M3<0M_3 < 0.

Question 17

The velocity of a particle is given by a differentiable function v(t)v(t), where v′(t)<0v'(t) < 0 for 0≤t≤80 \le t \le 8. Selected values of v(t)v(t) are v(0)=30v(0)=30, v(2)=25v(2)=25, v(4)=22v(4)=22, v(6)=18v(6)=18, and v(8)=10v(8)=10. Using a right Riemann sum with four equal subintervals based on the given values, what is the approximate distance traveled, and how does this approximation relate to the actual distance?

  1. 150, an underestimate of the actual distance. (correct answer)
  2. 150, an overestimate of the actual distance.
  3. 190, an underestimate of the actual distance.
  4. 190, an overestimate of the actual distance.
Explanation: The interval is [0,8][0, 8] with four subintervals, so the width of each subinterval is Δt=(8−0)/4=2\Delta t = (8-0)/4 = 2. A right Riemann sum uses the right endpoint of each subinterval: [0,2],[2,4],[4,6],[6,8][0,2], [2,4], [4,6], [6,8]. The approximation is Δt[v(2)+v(4)+v(6)+v(8)]=2[25+22+18+10]=2[75]=150\Delta t [v(2) + v(4) + v(6) + v(8)] = 2[25 + 22 + 18 + 10] = 2[75] = 150. Since v′(t)<0v'(t) < 0, the function v(t)v(t) is strictly decreasing. For a decreasing function, a right Riemann sum uses the minimum value in each subinterval, so it produces an underestimate of the actual integral (distance).

Question 18

Let F(x)=∫2xt3 dtF(x) = \int_2^x t^3 \,dt. Which of the following expressions represents a right Riemann sum approximation of F(6)F(6) with nn equal subintervals?

  1. ∑k=1n4n(2+4kn)3\sum_{k=1}^{n} \frac{4}{n} \left(2 + \frac{4k}{n}\right)^3 (correct answer)
  2. ∑k=1n6n(6kn)3\sum_{k=1}^{n} \frac{6}{n} \left(\frac{6k}{n}\right)^3
  3. ∑k=1n4n(4kn)3\sum_{k=1}^{n} \frac{4}{n} \left(\frac{4k}{n}\right)^3
  4. ∑k=1n6n(2+6kn)3\sum_{k=1}^{n} \frac{6}{n} \left(2 + \frac{6k}{n}\right)^3
Explanation: F(6)F(6) is the value of the definite integral ∫26t3 dt\int_2^6 t^3 \,dt. To find the right Riemann sum for this integral with nn subintervals, we first find Δt=(b−a)/n=(6−2)/n=4/n\Delta t = (b-a)/n = (6-2)/n = 4/n. The right endpoints are given by tk=a+kΔt=2+k(4/n)t_k = a + k \Delta t = 2 + k(4/n). The function is f(t)=t3f(t) = t^3. The Riemann sum is ∑k=1nf(tk)Δt=∑k=1n(2+4kn)3⋅4n\sum_{k=1}^n f(t_k) \Delta t = \sum_{k=1}^n (2 + \frac{4k}{n})^3 \cdot \frac{4}{n}, which matches choice A.

Question 19

Use a left Riemann sum to approximate ∫110f(x) dx\int_1^{10} f(x) \,dx based on the partition and values given: f(1)=10,f(3)=7,f(6)=11,f(10)=8f(1)=10, f(3)=7, f(6)=11, f(10)=8.

  1. 85 (correct answer)
  2. 79
  3. 84
  4. 82
Explanation: The partition is defined by the x-values 1,3,6,101, 3, 6, 10, which creates non-uniform subintervals: [1,3],[3,6],[6,10][1,3], [3,6], [6,10]. The widths are Δx1=3−1=2\Delta x_1 = 3-1=2, Δx2=6−3=3\Delta x_2 = 6-3=3, and Δx3=10−6=4\Delta x_3 = 10-6=4. A left Riemann sum uses the function value at the left endpoint of each interval. The sum is Δx1f(1)+Δx2f(3)+Δx3f(6)=2(10)+3(7)+4(11)=20+21+44=85\Delta x_1 f(1) + \Delta x_2 f(3) + \Delta x_3 f(6) = 2(10) + 3(7) + 4(11) = 20 + 21 + 44 = 85.

Question 20

If a trapezoidal sum with two equal subintervals is used to approximate ∫06x2dx\int_0^6 x^2 dx, what is the value of the approximation?

  1. 108
  2. 135
  3. 27
  4. 81 (correct answer)
Explanation: When you encounter a trapezoidal approximation problem, you're using the trapezoidal rule to estimate a definite integral by connecting function values with straight line segments instead of following the actual curve. For ∫06x2dx\int_0^6 x^2 dx with two equal subintervals, you first divide the interval [0,6] into two parts: [0,3] and [3,6]. Each subinterval has width Δx=3\Delta x = 3. The trapezoidal rule formula is: Δx2[f(x0)+2f(x1)+f(x2)]\frac{\Delta x}{2}[f(x_0) + 2f(x_1) + f(x_2)] You need the function values at the endpoints and midpoint:
  • f(0)=02=0f(0) = 0^2 = 0
  • f(3)=32=9f(3) = 3^2 = 9
  • f(6)=62=36f(6) = 6^2 = 36
Substituting into the formula: 32[0+2(9)+36]=32[54]=81\frac{3}{2}[0 + 2(9) + 36] = \frac{3}{2}[54] = 81 Choice A (108) likely comes from incorrectly using the rectangle rule with right endpoints: 3(9+36)=1353(9 + 36) = 135, then making an arithmetic error. Choice B (135) is exactly that rectangle rule calculation. Choice C (27) might result from using only one trapezoid incorrectly or miscalculating the area of a single trapezoid as 12⋅3⋅(9+9)=27\frac{1}{2} \cdot 3 \cdot (9+9) = 27. The correct answer is D (81). Study tip: Always identify your subinterval width first, then carefully apply the trapezoidal formula. Remember that interior points get coefficient 2, while endpoints get coefficient 1. Double-check your function evaluations—they're the most common source of errors.