Riemann Sums - Business Calculus

Let $C'(t)$ be the rate of growth of a company's profit, in millions of dollars per year. The function $C'(t)$ is known to be strictly increasing and concave up on the interval $[0, 5]$ . Let $I = \int_0^5 C'(t) dt$ be the actual total increase in profit over 5 years. Let $L_n$ , $R_n$ , and $T_n$ be the left Riemann sum, right Riemann sum, and trapezoidal sum approximations of $I$ with $n$ subintervals, respectively. Which of the following inequalities correctly orders these values?
Answer: $L_n < I < T_n < R_n$ 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 $C'(t)$ is strictly increasing and concave up on $[0,5]$ , 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 $L_n < I < R_n$ . 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 $T_n < I$ . However, since the trapezoidal sum averages the left and right endpoint values, we have $L_n < T_n < R_n$ . Looking at the choices: Choice A incorrectly places $I < R_n$ at the end rather than $T_n < R_n$ . Choice B wrongly suggests $T_n > R_n$ , which contradicts the fact that trapezoidal sums for concave up functions underestimate. Choice C incorrectly claims $T_n < L_n$ , but the trapezoidal sum must lie between the left and right sums. Choice D correctly captures the complete ordering: $L_n < I < T_n < R_n$ . Study tip: Remember that concavity determines whether trapezoidal sums over- or under-estimate: concave up means trapezoidal underestimates, concave down means it overestimates.
The marginal cost function for producing $x$ widgets is $C'(x) = 15 + 0.02x^2$ 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?
Answer: The left Riemann sum, because the marginal cost function is increasing on the interval $C'(x) = 15 + 0.02x^2$ is increasing on $[10,20]$ since $C''(x) = 0.04x > 0$ for all positive $x$ . 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.
The marginal cost function for producing a certain electronic component, $C'(q)$ , is known to be a continuous and strictly decreasing function for $q > 0$ . Let $A$ be the actual cost of increasing production from $q=50$ to $q=100$ . Let $R_5$ be the right Riemann sum approximation of $\int_{50}^{100} C'(q) dq$ with 5 subintervals of equal width. Which of the following statements correctly compares $R_5$ to $A$ ?
Answer: $R_5$ is an underestimate of $A$ . The actual cost $A$ is given by the definite integral $A = \int_{50}^{100} C'(q) dq$ . 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.
The rate of oil consumption by a factory is modeled by $f(t) = 50 + 10\sin(\pi t/12)$ gallons per hour, where $t$ is hours after midnight. If a left Riemann sum with 3 subintervals is used to estimate oil consumption from $t = 0$ to $t = 6$ hours, which expression represents this approximation?
Answer: $2[f(0) + f(2) + f(4)]$ For a left Riemann sum from $t = 0$ to $t = 6$ with 3 subintervals: $\Delta t = \frac{6-0}{3} = 2$ . The left endpoints of the subintervals $[0,2], [2,4], [4,6]$ are $t = 0, 2, 4$ . The approximation is $\Delta t \cdot [f(0) + f(2) + f(4)] = 2[f(0) + f(2) + f(4)]$ . Choice B uses right endpoints, choice C has wrong $\Delta t$ , and choice D has wrong $\Delta t$ .
The rate of visitors, $v(t)$ , entering a theme park is measured in hundreds of people per hour, where $t$ is hours after opening at 9 a.m. Park management observes that the rate of entry is always increasing between 9 a.m. ( $t=0$ ) and 1 p.m. ( $t=4$ ). A left Riemann sum, $L_4 = \sum_{i=0}^{3} v(t_i) \Delta t$ , is calculated with $\Delta t=1$ to estimate the total number of visitors who entered during this period. Which statement best describes this estimate?
Answer: The estimate is an underestimate of the total visitors because the rate function is increasing. The Riemann sum $\sum v(t_i) \Delta t$ approximates the definite integral $\int_0^4 v(t) \,dt$ , which represents the total accumulation of visitors between $t=0$ and $t=4$ . The question specifies that a left Riemann sum ( $L_4$ ) is used and that the rate function $v(t)$ 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.
The marginal revenue from selling $x$ units of a product is $R'(x)$ dollars per unit. The total revenue from selling units 21 through 40 is approximated using a left Riemann sum with $n=4$ subintervals. What is the economic interpretation of the second term in this sum?
Answer: The approximate revenue generated by selling units 26 through 30. The integral to be approximated is $\int_{20}^{40} R'(x) dx$ . The number of subintervals is $n=4$ . The width of each subinterval is $\Delta x = (40-20)/4 = 5$ . The subintervals are [20, 25], [25, 30], [30, 35], and [35, 40]. A left Riemann sum uses the left endpoints: $x=20, 25, 30, 35$ . The sum is $L_4 = R'(20)\Delta x + R'(25)\Delta x + R'(30)\Delta x + R'(35)\Delta x$ . The second term is $R'(25)\Delta x = R'(25) \cdot 5$ . $R'(25)$ is the marginal revenue (rate of change of revenue) at 25 units. Multiplying this rate by a quantity of units ( $\Delta x=5$ ) gives an approximation of the total revenue generated by that block of units. The second term uses the rate at $x=25$ (the left endpoint) to approximate the revenue over the second interval, which is [25, 30]. This interval represents the units from 26 through 30.
The rate of return on an investment, $r(t)$ , is a continuous function that is known to be both decreasing and concave up over the interval $0 \le t \le 5$ years. Four estimates are calculated for the total return $\int_0^5 r(t) \,dt$ using $n=10$ subintervals: Left Riemann Sum ( $L$ ), Right Riemann Sum ( $R$ ), Midpoint Riemann Sum ( $M$ ), and Trapezoidal Rule ( $T$ ). Which of the following correctly orders these four estimates from smallest to largest?
Answer: $R < M < T < L$ When you encounter Riemann sum comparison problems, the key is analyzing how the function's shape affects each approximation method. Since $r(t)$ is decreasing and concave up, you can predict the relative accuracy of each estimate. For a decreasing function, the Left Riemann Sum ( $L$ ) uses the higher left endpoint of each subinterval, creating an overestimate. The Right Riemann Sum ( $R$ ) uses the lower right endpoint, creating an underestimate. So $R < \text{true value} < L$ . 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 ( $M$ ) underestimates the integral because the function values at midpoints lie below the curve's average over each subinterval. The Trapezoidal Rule ( $T$ ) 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 $R < M < T < L$ . Choice A incorrectly places $M$ as the smallest estimate. Choice B reverses the relationship between $L$ and $R$ for a decreasing function. Choice C incorrectly suggests $L < R$ , which contradicts the decreasing property. Choice D correctly captures all relationships: $R < M < T < L$ . Study tip: Remember the pattern "RMTL" for decreasing, concave up functions. The decreasing property determines that $R < L$ , while concavity determines where $M$ and $T$ fall between them.
The velocity of a particle is $v(t) = t^2 - 4t + 5$ ft/sec. When comparing left and right Riemann sums with 4 equal subintervals to approximate displacement from $t = 0$ to $t = 4$ seconds, which statement is true?
Answer: The right sum overestimates because the velocity function is concave up throughout the interval $v(t) = t^2 - 4t + 5$ has $v'(t) = 2t - 4$ , so it decreases on $[0,2]$ and increases on $[2,4]$ . However, $v''(t) = 2 > 0$ , 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.
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: $\sum_{k=1}^{60} 500(1.02)^{k/20} \cdot \frac{1}{20}$ Which of the following definite integrals does this sum approximate?
Answer: $\int_{0}^{3} 500(1.02)^{x} dx$ Let's analyze the components of the Riemann sum. The term $\frac{1}{20}$ represents $\Delta x$ . The summation is from $k=1$ to $n=60$ . The total width of the interval is $b-a = n \cdot \Delta x = 60 \cdot \frac{1}{20} = 3$ . The sample points are determined by the term inside the function, which is $k/20$ . This corresponds to right endpoints $x_k = a + k \Delta x$ . If we assume the interval starts at $a=0$ , then $x_k = 0 + k \frac{1}{20} = k/20$ . The interval is therefore $[0, 3]$ . The function being evaluated is $f(x_k) = 500(1.02)^{x_k}$ , which means the function is $f(x) = 500(1.02)^x$ . Combining these, the sum approximates the definite integral $\int_{0}^{3} 500(1.02)^{x} dx$ .
The marginal cost $C'(q)$ for producing $q$ items is a continuous function that is strictly increasing for $q \ge 0$ . Let $\Delta C$ be the true change in cost from $q=10$ to $q=50$ . Let $L$ be the left Riemann sum approximation and $R$ be the right Riemann sum approximation for $\int_{10}^{50} C'(q) \,dq$ , both using $n=10$ subintervals of equal width. Which of the following correctly orders these three values?
Answer: $L < \Delta C < R$ For a strictly increasing function, a left Riemann sum ( $L$ ) uses the minimum value on each subinterval, resulting in an underestimate of the true integral value ( $\Delta C$ ). Thus, $L < \Delta C$ . A right Riemann sum ( $R$ ) uses the maximum value on each subinterval, resulting in an overestimate of the true integral value. Thus, $\Delta C < R$ . Combining these two inequalities gives the correct order: $L < \Delta C < R$ .
The rate of appreciation of an investment is given by a differentiable function $V'(t)$ , where $t$ is time in years. It is known that the rate of appreciation is slowing down, meaning the graph of $V'(t)$ is concave down for $t \ge 0$ . An approximation for the total appreciation over the first 5 years, $\int_0^5 V'(t) dt$ , is calculated using the Trapezoidal Rule with $n=10$ subintervals. How does this approximation, $T_{10}$ , compare to the actual appreciation, $A$ ?
Answer: $T_{10}$ is an underestimate of $A$ . 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 $V'(t)$ 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 $V'(t)$ , making $T_{10}$ an underestimate of the true integral $A = \int_0^5 V'(t) dt$ . Choice A incorrectly claims $T_{10}$ overestimates $A$ . This would only be true if $V'(t)$ were concave up, where the curve would lie above the connecting line segments. Choice B suggests the relationship depends on whether $V'(t)$ 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 $V'(t)$ 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.
A chemical spill is causing a pond to be contaminated. The rate at which the pollutant enters the pond is given by $P(t) = 15e^{0.1t}$ liters per hour, where $t$ 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 $t=1$ to $t=2$ ), using a midpoint Riemann sum with 4 subintervals?
Answer: 0.25(15 $e^{0.1(1.125)}$ + 15 $e^{0.1(1.375)}$ + 15 $e^{0.1(1.625)}$ + 15 $e^{0.1(1.875)}$ ) The goal is to approximate $\int_{1}^{2} P(t) dt$ with a midpoint sum using $n=4$ . First, calculate the width of each subinterval: $\Delta t = \frac{2-1}{4} = 0.25$ . 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: $m_1 = (1+1.25)/2 = 1.125$ $m_2 = (1.25+1.5)/2 = 1.375$ $m_3 = (1.5+1.75)/2 = 1.625$ $m_4 = (1.75+2)/2 = 1.875$ The midpoint Riemann sum is $M_4 = \Delta t [P(m_1) + P(m_2) + P(m_3) + P(m_4)]$ . Substituting the values gives: $M_4 = 0.25[15e^{0.1(1.125)} + 15e^{0.1(1.375)} + 15e^{0.1(1.625)} + 15e^{0.1(1.875)}]$ . This matches choice B.
The marginal profit function $P'(x)$ for a product is positive and strictly increasing for $x \ge 0$ . 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?
Answer: A left Riemann sum with 20 subintervals. The total profit is the integral of the marginal profit function, $\int_{100}^{200} P'(x) dx$ . 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 ( $n=20$ ) 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.
A storage tank is leaking oil at a rate of $r(t) = 15e^{-0.1t}$ liters per hour, where $t$ 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.
Answer: 82.5 liters The interval is $[0, 8]$ and there are $n=4$ subintervals, so the width of each subinterval is $\Delta t = (8-0)/4 = 2$ . The subintervals are $[0, 2]$ , $[2, 4]$ , $[4, 6]$ , and $[6, 8]$ . The midpoints of these subintervals are $t_1=1$ , $t_2=3$ , $t_3=5$ , and $t_4=7$ . The midpoint Riemann sum is calculated as $M_4 = \Delta t [r(1) + r(3) + r(5) + r(7)]$ . First, evaluate the function at the midpoints: $r(1) = 15e^{-0.1} \approx 13.57$ , $r(3) = 15e^{-0.3} \approx 11.11$ , $r(5) = 15e^{-0.5} \approx 9.10$ , $r(7) = 15e^{-0.7} \approx 7.45$ . The sum is $M_4 = 2(13.57 + 11.11 + 9.10 + 7.45) = 2(41.23) = 82.46$ . The closest answer is 82.5 liters.
The expression $S = \sum_{k=1}^{40} \left(1 + \frac{2k}{40}\right)^3 \cdot \frac{1}{20}$ is a right Riemann sum approximation for $\int_a^b f(x) \,dx$ . Which of the following are the correct function $f(x)$ and interval $[a, b]$ ?
Answer: $f(x) = x^3$ on the interval $[1, 3]$ The general form of a right Riemann sum is $\sum_{k=1}^{n} f(a+k\Delta x) \Delta x$ , where $\Delta x = (b-a)/n$ . In the given sum, $n=40$ . Let's identify $\Delta x = 1/20$ . The term inside the function is $1 + \frac{2k}{40} = 1 + \frac{k}{20}$ . This matches the form $a+k\Delta x$ with $a=1$ and $\Delta x = 1/20$ . The function being applied is cubing the term, so $f(x)=x^3$ . The interval width is $b-a = n \Delta x = 40 \cdot (1/20) = 2$ . Since $a=1$ , the interval is $[1, 1+2] = [1, 3]$ . Therefore, the sum approximates $\int_1^3 x^3 \,dx$ .
A water tank is being filled at a variable rate described by $r(t) = 8 + 2\cos(\pi t/6)$ gallons per minute, where $t$ is time in minutes. The difference between the right Riemann sum and left Riemann sum approximations (right minus left) using 6 subintervals over $[0, 12]$ is:
Answer: $0$ gallons With 6 subintervals on $[0,12]$ , $\Delta t = 2$ . Left endpoints: $0, 2, 4, 6, 8, 10$ . Right endpoints: $2, 4, 6, 8, 10, 12$ . The difference is $2[r(12) - r(0)] = 2[(8 + 2\cos(2\pi)) - (8 + 2\cos(0))] = 2[(8 + 2) - (8 + 2)] = 2[10 - 10] = 0$ . Since $\cos(2\pi) = \cos(0) = 1$ , the difference is zero. Choices A, B, and D represent common computational errors.

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