Business Calculus Quiz: Riemann Sums
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Riemann SumsQuestion 1 of 16

The marginal cost function for producing a certain electronic component, C(q)C'(q), is known to be a continuous and strictly decreasing function for q>0q > 0. Let AA be the actual cost of increasing production from q=50q=50 to q=100q=100. Let R5R_5 be the right Riemann sum approximation of 50100C(q)dq\int_{50}^{100} C'(q) dq with 5 subintervals of equal width. Which of the following statements correctly compares R5R_5 to AA?

R5R_5 is an overestimate of AA.
R5R_5 is an underestimate of AA.
R5R_5 is exactly equal to AA.
The relationship cannot be determined without knowing the function C(q)C'(q).
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Business Calculus Quiz

Business Calculus Quiz: Riemann Sums

Practice Riemann Sums in Business Calculus 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, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

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.

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Question 1

The marginal cost function for producing a certain electronic component, C(q)C'(q), is known to be a continuous and strictly decreasing function for q>0q > 0. Let AA be the actual cost of increasing production from q=50q=50 to q=100q=100. Let R5R_5 be the right Riemann sum approximation of 50100C(q)dq\int_{50}^{100} C'(q) dq with 5 subintervals of equal width. Which of the following statements correctly compares R5R_5 to AA?

  1. R5R_5 is an overestimate of AA.
  2. R5R_5 is an underestimate of AA. (correct answer)
  3. R5R_5 is exactly equal to AA.
  4. The relationship cannot be determined without knowing the function C(q)C'(q).
Explanation: The actual cost AA is given by the definite integral A=50100C(q)dqA = \int_{50}^{100} C'(q) dq. For a function that is strictly decreasing on an interval, a right Riemann sum will always be an underestimate of the definite integral over that interval. This is because the height of each rectangle is determined by the function's value at the right endpoint, which is the lowest value in that subinterval. Therefore, all the rectangles lie entirely below the curve, making the sum of their areas less than the area under the curve.

Question 2

The marginal cost function for producing xx widgets is C(x)=15+0.02x2C'(x) = 15 + 0.02x^2 dollars per widget. If the current production level changes from 10 to 20 widgets, and you must choose between a left Riemann sum with 5 subintervals or a right Riemann sum with 5 subintervals to estimate the change in total cost, which approach gives a more conservative (lower) estimate?

  1. The left Riemann sum, because the marginal cost function is decreasing on the interval
  2. The left Riemann sum, because the marginal cost function is increasing on the interval (correct answer)
  3. The right Riemann sum, because it uses endpoint values that minimize the sum
  4. Both methods give identical estimates since the function is linear in this region
Explanation: C(x)=15+0.02x2C'(x) = 15 + 0.02x^2 is increasing on [10,20][10,20] since C(x)=0.04x>0C''(x) = 0.04x > 0 for all positive xx. For increasing functions, left Riemann sums underestimate the integral (giving lower values), while right sums overestimate. So the left sum gives the more conservative estimate. Choice A has the wrong direction for the function. Choice C is incorrect about which method minimizes. Choice D is wrong because the function is quadratic, not linear.

Question 3

The marginal profit function P(x)P'(x) for a product is positive and strictly increasing for x0x \ge 0. To estimate the total profit gained from selling the 101st unit through the 200th unit, which of the following Riemann sum approximations would provide the most accurate, guaranteed underestimate?

  1. A left Riemann sum with 10 subintervals.
  2. A right Riemann sum with 10 subintervals.
  3. A left Riemann sum with 20 subintervals. (correct answer)
  4. A right Riemann sum with 20 subintervals.
Explanation: The total profit is the integral of the marginal profit function, 100200P(x)dx\int_{100}^{200} P'(x) dx. For a strictly increasing function, a left Riemann sum provides an underestimate, as the height of each rectangle is determined by the lowest point in its subinterval. A right Riemann sum provides an overestimate. Therefore, to get a guaranteed underestimate, we must use a left Riemann sum. Between the two left Riemann sum options (10 vs. 20 subintervals), the approximation with more subintervals (n=20n=20) will be more accurate and closer to the true value of the integral. Since it is an underestimate, a more accurate underestimate will be a larger value, making it the 'best' or 'tightest' lower bound among the choices.

Question 4

The rate of appreciation of an investment is given by a differentiable function V(t)V'(t), where tt is time in years. It is known that the rate of appreciation is slowing down, meaning the graph of V(t)V'(t) is concave down for t0t \ge 0. An approximation for the total appreciation over the first 5 years, 05V(t)dt\int_0^5 V'(t) dt, is calculated using the Trapezoidal Rule with n=10n=10 subintervals. How does this approximation, T10T_{10}, compare to the actual appreciation, AA?

  1. T10T_{10} is an overestimate of AA.
  2. The relationship depends on whether V(t)V'(t) is increasing or decreasing.
  3. T10T_{10} is exactly equal to AA.
  4. T10T_{10} is an underestimate of AA. (correct answer)
Explanation: This question tests your understanding of how the Trapezoidal Rule behaves when approximating integrals of functions with different concavity patterns. When you see a numerical integration problem that mentions concavity, immediately think about whether the approximation will be an overestimate or underestimate. The Trapezoidal Rule connects function values with straight line segments, then calculates the area under these segments. Since V(t)V'(t) is concave down (the problem states "the rate of appreciation is slowing down"), the actual curve lies below the straight line segments connecting any two points on the curve. This means the trapezoidal approximation captures more area than what's actually under the curve V(t)V'(t), making T10T_{10} an underestimate of the true integral A=05V(t)dtA = \int_0^5 V'(t) dt. Choice A incorrectly claims T10T_{10} overestimates AA. This would only be true if V(t)V'(t) were concave up, where the curve would lie above the connecting line segments. Choice B suggests the relationship depends on whether V(t)V'(t) is increasing or decreasing, but this misses the point entirely. The key factor is concavity, not the sign of the derivative. Choice C claims exact equality, which would only occur if V(t)V'(t) were linear (neither concave up nor concave down). Study tip: Remember the concavity rule for Trapezoidal Rule approximations: concave down functions are underestimated, concave up functions are overestimated. The "curve position relative to connecting lines" mental image will serve you well on integration approximation problems.

Question 5

The rate of oil consumption by a factory is modeled by f(t)=50+10sin(πt/12)f(t) = 50 + 10\sin(\pi t/12) gallons per hour, where tt is hours after midnight. If a left Riemann sum with 3 subintervals is used to estimate oil consumption from t=0t = 0 to t=6t = 6 hours, which expression represents this approximation?

  1. 2[f(0)+f(2)+f(4)]2[f(0) + f(2) + f(4)] (correct answer)
  2. 2[f(2)+f(4)+f(6)]2[f(2) + f(4) + f(6)]
  3. 6[f(0)+f(3)+f(6)]6[f(0) + f(3) + f(6)]
  4. 3[f(0)+f(2)+f(4)]3[f(0) + f(2) + f(4)]
Explanation: For a left Riemann sum from t=0t = 0 to t=6t = 6 with 3 subintervals: Δt=603=2\Delta t = \frac{6-0}{3} = 2. The left endpoints of the subintervals [0,2],[2,4],[4,6][0,2], [2,4], [4,6] are t=0,2,4t = 0, 2, 4. The approximation is Δt[f(0)+f(2)+f(4)]=2[f(0)+f(2)+f(4)]\Delta t \cdot [f(0) + f(2) + f(4)] = 2[f(0) + f(2) + f(4)]. Choice B uses right endpoints, choice C has wrong Δt\Delta t, and choice D has wrong Δt\Delta t.

Question 6

A chemical spill is causing a pond to be contaminated. The rate at which the pollutant enters the pond is given by P(t)=15e0.1tP(t) = 15e^{0.1t} liters per hour, where tt is the number of hours since the spill began. What is the approximate total volume of pollutant that enters the pond during the second hour (from t=1t=1 to t=2t=2), using a midpoint Riemann sum with 4 subintervals?

  1. 4(15e0.1(1.125)e^{0.1(1.125)} + 15e0.1(1.375)e^{0.1(1.375)} + 15e0.1(1.625)e^{0.1(1.625)} + 15e0.1(1.875)e^{0.1(1.875)})
  2. 0.25(15e0.1(1.125)e^{0.1(1.125)} + 15e0.1(1.375)e^{0.1(1.375)} + 15e0.1(1.625)e^{0.1(1.625)} + 15e0.1(1.875)e^{0.1(1.875)}) (correct answer)
  3. 0.25(15e0.1(1.0)e^{0.1(1.0)} + 15e0.1(1.25)e^{0.1(1.25)} + 15e0.1(1.5)e^{0.1(1.5)} + 15e0.1(1.75)e^{0.1(1.75)})
  4. 0.25(15e0.1(1.25)e^{0.1(1.25)} + 15e0.1(1.5)e^{0.1(1.5)} + 15e0.1(1.75)e^{0.1(1.75)} + 15e0.1(2.0)e^{0.1(2.0)})
Explanation: The goal is to approximate 12P(t)dt\int_{1}^{2} P(t) dt with a midpoint sum using n=4n=4. First, calculate the width of each subinterval: Δt=214=0.25\Delta t = \frac{2-1}{4} = 0.25. The subintervals are [1, 1.25], [1.25, 1.5], [1.5, 1.75], and [1.75, 2]. Next, find the midpoint of each subinterval: m1=(1+1.25)/2=1.125m_1 = (1+1.25)/2 = 1.125 m2=(1.25+1.5)/2=1.375m_2 = (1.25+1.5)/2 = 1.375 m3=(1.5+1.75)/2=1.625m_3 = (1.5+1.75)/2 = 1.625 m4=(1.75+2)/2=1.875m_4 = (1.75+2)/2 = 1.875 The midpoint Riemann sum is M4=Δt[P(m1)+P(m2)+P(m3)+P(m4)]M_4 = \Delta t [P(m_1) + P(m_2) + P(m_3) + P(m_4)]. Substituting the values gives: M4=0.25[15e0.1(1.125)+15e0.1(1.375)+15e0.1(1.625)+15e0.1(1.875)]M_4 = 0.25[15e^{0.1(1.125)} + 15e^{0.1(1.375)} + 15e^{0.1(1.625)} + 15e^{0.1(1.875)}]. This matches choice B.

Question 7

The marginal revenue from selling xx units of a product is R(x)R'(x) dollars per unit. The total revenue from selling units 21 through 40 is approximated using a left Riemann sum with n=4n=4 subintervals. What is the economic interpretation of the second term in this sum?

  1. The approximate revenue generated by selling the 25th unit.
  2. The approximate revenue generated by selling units 21 through 25.
  3. The approximate revenue generated by selling units 26 through 30. (correct answer)
  4. The total revenue from selling the first 30 units.
Explanation: The integral to be approximated is 2040R(x)dx\int_{20}^{40} R'(x) dx. The number of subintervals is n=4n=4. The width of each subinterval is Δx=(4020)/4=5\Delta x = (40-20)/4 = 5. The subintervals are [20, 25], [25, 30], [30, 35], and [35, 40]. A left Riemann sum uses the left endpoints: x=20,25,30,35x=20, 25, 30, 35. The sum is L4=R(20)Δx+R(25)Δx+R(30)Δx+R(35)ΔxL_4 = R'(20)\Delta x + R'(25)\Delta x + R'(30)\Delta x + R'(35)\Delta x. The second term is R(25)Δx=R(25)5R'(25)\Delta x = R'(25) \cdot 5. R(25)R'(25) is the marginal revenue (rate of change of revenue) at 25 units. Multiplying this rate by a quantity of units (Δx=5\Delta x=5) gives an approximation of the total revenue generated by that block of units. The second term uses the rate at x=25x=25 (the left endpoint) to approximate the revenue over the second interval, which is [25, 30]. This interval represents the units from 26 through 30.

Question 8

Let C(t)C'(t) be the rate of growth of a company's profit, in millions of dollars per year. The function C(t)C'(t) is known to be strictly increasing and concave up on the interval [0,5][0, 5]. Let I=05C(t)dtI = \int_0^5 C'(t) dt be the actual total increase in profit over 5 years. Let LnL_n, RnR_n, and TnT_n be the left Riemann sum, right Riemann sum, and trapezoidal sum approximations of II with nn subintervals, respectively. Which of the following inequalities correctly orders these values?

  1. Ln<Tn<I<RnL_n < T_n < I < R_n
  2. Ln<I<Rn<TnL_n < I < R_n < T_n
  3. Tn<Ln<I<RnT_n < L_n < I < R_n
  4. Ln<I<Tn<RnL_n < I < T_n < R_n (correct answer)
Explanation: When you encounter a problem about Riemann sum approximations and their relationship to the actual integral value, you need to analyze how the function's properties affect each approximation method. Since C(t)C'(t) is strictly increasing and concave up on [0,5][0,5], this tells us everything we need to know about the ordering. For an increasing function, left Riemann sums always underestimate the integral (they use the smaller function value at the left endpoint of each subinterval), while right Riemann sums overestimate it (they use the larger value at the right endpoint). This immediately gives us Ln<I<RnL_n < I < R_n. The key insight is understanding where the trapezoidal sum fits. For a concave up function, the trapezoidal approximation (which connects endpoints with straight lines) lies entirely below the actual curve, since straight lines lie below concave up curves. This means Tn<IT_n < I. However, since the trapezoidal sum averages the left and right endpoint values, we have Ln<Tn<RnL_n < T_n < R_n. Looking at the choices: Choice A incorrectly places I<RnI < R_n at the end rather than Tn<RnT_n < R_n. Choice B wrongly suggests Tn>RnT_n > R_n, which contradicts the fact that trapezoidal sums for concave up functions underestimate. Choice C incorrectly claims Tn<LnT_n < L_n, but the trapezoidal sum must lie between the left and right sums. Choice D correctly captures the complete ordering: Ln<I<Tn<RnL_n < I < T_n < R_n. Study tip: Remember that concavity determines whether trapezoidal sums over- or under-estimate: concave up means trapezoidal underestimates, concave down means it overestimates.

Question 9

A financial analyst models the total accumulated value of an asset over 3 years using a definite integral. The following right Riemann sum is used to approximate this value: k=160500(1.02)k/20120\sum_{k=1}^{60} 500(1.02)^{k/20} \cdot \frac{1}{20} Which of the following definite integrals does this sum approximate?

  1. 03500(1.02)xdx\int_{0}^{3} 500(1.02)^{x} dx (correct answer)
  2. 060500(1.02)x/20dx\int_{0}^{60} 500(1.02)^{x/20} dx
  3. 03500(1.02)t/20dt\int_{0}^{3} 500(1.02)^{t/20} dt
  4. 03500(1.02)xdx20\int_{0}^{3} 500(1.02)^{x} \frac{dx}{20}
Explanation: Let's analyze the components of the Riemann sum. The term 120\frac{1}{20} represents Δx\Delta x. The summation is from k=1k=1 to n=60n=60. The total width of the interval is ba=nΔx=60120=3b-a = n \cdot \Delta x = 60 \cdot \frac{1}{20} = 3. The sample points are determined by the term inside the function, which is k/20k/20. This corresponds to right endpoints xk=a+kΔxx_k = a + k \Delta x. If we assume the interval starts at a=0a=0, then xk=0+k120=k/20x_k = 0 + k \frac{1}{20} = k/20. The interval is therefore [0,3][0, 3]. The function being evaluated is f(xk)=500(1.02)xkf(x_k) = 500(1.02)^{x_k}, which means the function is f(x)=500(1.02)xf(x) = 500(1.02)^x. Combining these, the sum approximates the definite integral 03500(1.02)xdx\int_{0}^{3} 500(1.02)^{x} dx.

Question 10

The marginal cost C(q)C'(q) for producing qq items is a continuous function that is strictly increasing for q0q \ge 0. Let ΔC\Delta C be the true change in cost from q=10q=10 to q=50q=50. Let LL be the left Riemann sum approximation and RR be the right Riemann sum approximation for 1050C(q)dq\int_{10}^{50} C'(q) \,dq, both using n=10n=10 subintervals of equal width. Which of the following correctly orders these three values?

  1. R<ΔC<LR < \Delta C < L
  2. ΔC<L<R\Delta C < L < R
  3. L<ΔC<RL < \Delta C < R (correct answer)
  4. L<R<ΔCL < R < \Delta C
Explanation: For a strictly increasing function, a left Riemann sum (LL) uses the minimum value on each subinterval, resulting in an underestimate of the true integral value (ΔC\Delta C). Thus, L<ΔCL < \Delta C. A right Riemann sum (RR) uses the maximum value on each subinterval, resulting in an overestimate of the true integral value. Thus, ΔC<R\Delta C < R. Combining these two inequalities gives the correct order: L<ΔC<RL < \Delta C < R.

Question 11

A storage tank is leaking oil at a rate of r(t)=15e0.1tr(t) = 15e^{-0.1t} liters per hour, where tt is the number of hours since the leak began. Use a midpoint Riemann sum with four subintervals of equal width to approximate the total amount of oil that leaks out during the first 8 hours.

  1. 91.1 liters
  2. 82.5 liters (correct answer)
  3. 74.6 liters
  4. 41.2 liters
Explanation: The interval is [0,8][0, 8] and there are n=4n=4 subintervals, so the width of each subinterval is Δt=(80)/4=2\Delta t = (8-0)/4 = 2. The subintervals are [0,2][0, 2], [2,4][2, 4], [4,6][4, 6], and [6,8][6, 8]. The midpoints of these subintervals are t1=1t_1=1, t2=3t_2=3, t3=5t_3=5, and t4=7t_4=7. The midpoint Riemann sum is calculated as M4=Δt[r(1)+r(3)+r(5)+r(7)]M_4 = \Delta t [r(1) + r(3) + r(5) + r(7)]. First, evaluate the function at the midpoints: r(1)=15e0.113.57r(1) = 15e^{-0.1} \approx 13.57, r(3)=15e0.311.11r(3) = 15e^{-0.3} \approx 11.11, r(5)=15e0.59.10r(5) = 15e^{-0.5} \approx 9.10, r(7)=15e0.77.45r(7) = 15e^{-0.7} \approx 7.45. The sum is M4=2(13.57+11.11+9.10+7.45)=2(41.23)=82.46M_4 = 2(13.57 + 11.11 + 9.10 + 7.45) = 2(41.23) = 82.46. The closest answer is 82.5 liters.

Question 12

The expression S=k=140(1+2k40)3120S = \sum_{k=1}^{40} \left(1 + \frac{2k}{40}\right)^3 \cdot \frac{1}{20} is a right Riemann sum approximation for abf(x)dx\int_a^b f(x) \,dx. Which of the following are the correct function f(x)f(x) and interval [a,b][a, b]?

  1. f(x)=x3f(x) = x^3 on the interval [1,3][1, 3] (correct answer)
  2. f(x)=x3f(x) = x^3 on the interval [0,2][0, 2]
  3. f(x)=(1+x)3f(x) = (1+x)^3 on the interval [0,1][0, 1]
  4. f(x)=x3f(x) = x^3 on the interval [1,2][1, 2]
Explanation: The general form of a right Riemann sum is k=1nf(a+kΔx)Δx\sum_{k=1}^{n} f(a+k\Delta x) \Delta x, where Δx=(ba)/n\Delta x = (b-a)/n. In the given sum, n=40n=40. Let's identify Δx=1/20\Delta x = 1/20. The term inside the function is 1+2k40=1+k201 + \frac{2k}{40} = 1 + \frac{k}{20}. This matches the form a+kΔxa+k\Delta x with a=1a=1 and Δx=1/20\Delta x = 1/20. The function being applied is cubing the term, so f(x)=x3f(x)=x^3. The interval width is ba=nΔx=40(1/20)=2b-a = n \Delta x = 40 \cdot (1/20) = 2. Since a=1a=1, the interval is [1,1+2]=[1,3][1, 1+2] = [1, 3]. Therefore, the sum approximates 13x3dx\int_1^3 x^3 \,dx.

Question 13

The rate of return on an investment, r(t)r(t), is a continuous function that is known to be both decreasing and concave up over the interval 0t50 \le t \le 5 years. Four estimates are calculated for the total return 05r(t)dt\int_0^5 r(t) \,dt using n=10n=10 subintervals: Left Riemann Sum (LL), Right Riemann Sum (RR), Midpoint Riemann Sum (MM), and Trapezoidal Rule (TT). Which of the following correctly orders these four estimates from smallest to largest?

  1. M<T<L<RM < T < L < R
  2. R<T<M<LR < T < M < L
  3. L<T<M<RL < T < M < R
  4. R<M<T<LR < M < T < L (correct answer)
Explanation: When you encounter Riemann sum comparison problems, the key is analyzing how the function's shape affects each approximation method. Since r(t)r(t) is decreasing and concave up, you can predict the relative accuracy of each estimate. For a decreasing function, the Left Riemann Sum (LL) uses the higher left endpoint of each subinterval, creating an overestimate. The Right Riemann Sum (RR) uses the lower right endpoint, creating an underestimate. So R<true value<LR < \text{true value} < L. The concave up property is crucial for the remaining methods. When a function is concave up, it curves upward like a smile. This means the Midpoint Rule (MM) underestimates the integral because the function values at midpoints lie below the curve's average over each subinterval. The Trapezoidal Rule (TT) also underestimates because straight-line segments connecting endpoints lie below the concave up curve. However, the Midpoint Rule is generally more accurate than the Trapezoidal Rule for concave functions, so R<M<T<LR < M < T < L. Choice A incorrectly places MM as the smallest estimate. Choice B reverses the relationship between LL and RR for a decreasing function. Choice C incorrectly suggests L<RL < R, which contradicts the decreasing property. Choice D correctly captures all relationships: R<M<T<LR < M < T < L. Study tip: Remember the pattern "RMTL" for decreasing, concave up functions. The decreasing property determines that R<LR < L, while concavity determines where MM and TT fall between them.

Question 14

The rate of visitors, v(t)v(t), entering a theme park is measured in hundreds of people per hour, where tt is hours after opening at 9 a.m. Park management observes that the rate of entry is always increasing between 9 a.m. (t=0t=0) and 1 p.m. (t=4t=4). A left Riemann sum, L4=i=03v(ti)ΔtL_4 = \sum_{i=0}^{3} v(t_i) \Delta t, is calculated with Δt=1\Delta t=1 to estimate the total number of visitors who entered during this period. Which statement best describes this estimate?

  1. The estimate is an overestimate of the total visitors because the rate function is increasing.
  2. The estimate is an underestimate of the total visitors because the rate function is increasing. (correct answer)
  3. The estimate represents the instantaneous rate of entry at 1 p.m. and is an overestimate.
  4. The estimate represents the average number of visitors per hour and is an underestimate.
Explanation: The Riemann sum v(ti)Δt\sum v(t_i) \Delta t approximates the definite integral 04v(t)dt\int_0^4 v(t) \,dt, which represents the total accumulation of visitors between t=0t=0 and t=4t=4. The question specifies that a left Riemann sum (L4L_4) is used and that the rate function v(t)v(t) is increasing on the interval. For an increasing function, the left endpoint of any subinterval is the minimum value on that subinterval. Therefore, a left Riemann sum will be an underestimate of the true total number of visitors.

Question 15

A water tank is being filled at a variable rate described by r(t)=8+2cos(πt/6)r(t) = 8 + 2\cos(\pi t/6) gallons per minute, where tt is time in minutes. The difference between the right Riemann sum and left Riemann sum approximations (right minus left) using 6 subintervals over [0,12][0, 12] is:

  1. 44 gallons
  2. 88 gallons
  3. 00 gallons (correct answer)
  4. 4-4 gallons
Explanation: With 6 subintervals on [0,12][0,12], Δt=2\Delta t = 2. Left endpoints: 0,2,4,6,8,100, 2, 4, 6, 8, 10. Right endpoints: 2,4,6,8,10,122, 4, 6, 8, 10, 12. The difference is 2[r(12)r(0)]=2[(8+2cos(2π))(8+2cos(0))]=2[(8+2)(8+2)]=2[1010]=02[r(12) - r(0)] = 2[(8 + 2\cos(2\pi)) - (8 + 2\cos(0))] = 2[(8 + 2) - (8 + 2)] = 2[10 - 10] = 0. Since cos(2π)=cos(0)=1\cos(2\pi) = \cos(0) = 1, the difference is zero. Choices A, B, and D represent common computational errors.

Question 16

The velocity of a particle is v(t)=t24t+5v(t) = t^2 - 4t + 5 ft/sec. When comparing left and right Riemann sums with 4 equal subintervals to approximate displacement from t=0t = 0 to t=4t = 4 seconds, which statement is true?

  1. The left sum overestimates because the velocity function is always increasing on the given interval
  2. The right sum overestimates because the velocity function is concave up throughout the interval (correct answer)
  3. The left sum underestimates because the velocity function decreases then increases on the interval
  4. Both sums give the same approximation because the function has a minimum at the midpoint
Explanation: v(t)=t24t+5v(t) = t^2 - 4t + 5 has v(t)=2t4v'(t) = 2t - 4, so it decreases on [0,2][0,2] and increases on [2,4][2,4]. However, v(t)=2>0v''(t) = 2 > 0, so the function is concave up throughout. For concave up functions, right Riemann sums overestimate the integral. Choice A is wrong because velocity isn't always increasing. Choice C is wrong about the direction of error. Choice D is incorrect because the sums differ.