Trapezoidal Rule - Business Calculus

1. A researcher wants to use the trapezoidal rule to estimate $\int_0^6 p(t) \, dt$ where $p(t)$ represents profit rate. The available data points are at $t = 0, 1.5, 3, 4.5, 6$. If the researcher instead uses only the data at $t = 0, 2, 4, 6$ (ignoring the other points), how does this affect the approximation?
   Answer: The approximation becomes less accurate because the subinterval width increases from 1.5 to 2, reducing the number of subintervals from 4 to 3 Originally, using points at t = 0, 1.5, 3, 4.5, 6 gives 4 subintervals of width 1.5 each. Using only t = 0, 2, 4, 6 gives 3 subintervals of width 2 each. Generally, smaller subinterval widths (more subintervals) give more accurate trapezoidal approximations. Choice A is wrong because the subinterval width does change. Choice B is wrong about the number of subintervals. Choice D is incorrect because even spacing doesn't necessarily improve accuracy when subinterval width increases.
2. Consider the integral $\int_0^4 \frac{120}{x+1} \, dx$. Let $T_2$ be the trapezoidal rule approximation with $n=2$ subintervals and $T_4$ be the approximation with $n=4$ subintervals. What is the value of $T_2 - T_4$?
   Answer: 22 Let $f(x) = \frac{120}{x+1}$. First, calculate $T_2$: $a=0, b=4, n=2$, so $\Delta x=2$. The points are 0, 2, 4. $T_2 = \frac{2}{2}[f(0) + 2f(2) + f(4)] = 1[\frac{120}{1} + 2(\frac{120}{3}) + \frac{120}{5}] = 120 + 2(40) + 24 = 224$ Next, calculate $T_4$: $a=0, b=4, n=4$, so $\Delta x=1$. The points are 0, 1, 2, 3, 4. $T_4 = \frac{1}{2}[f(0) + 2f(1) + 2f(2) + 2f(3) + f(4)]$ $T_4 = \frac{1}{2}[\frac{120}{1} + 2(\frac{120}{2}) + 2(\frac{120}{3}) + 2(\frac{120}{4}) + \frac{120}{5}]$ $T_4 = \frac{1}{2}[120 + 2(60) + 2(40) + 2(30) + 24] = \frac{1}{2}[120+120+80+60+24] = \frac{1}{2}[404] = 202$ Finally, find the difference: $T_2 - T_4 = 224 - 202 = 22$.
3. A company's revenue rate function $R'(t)$ is measured at hourly intervals over a 6-hour period. Using the trapezoidal rule to estimate total revenue $\int_0^6 R'(t) \, dt$, the calculation yields 420. However, due to a data collection error, one of the measured values was recorded as zero when it should have been positive. If this error caused the trapezoidal estimate to be 15% lower than it should be, what was the correct value of the mis-recorded data point?
   Answer: The missing value was 74 if it occurred at an interior point, or 148 if at an endpoint If the estimate is 15% too low, then 420 = 0.85 × (true estimate), so the true estimate is 420/0.85 ≈ 494.12. The difference is about 74. In the trapezoidal rule, interior points have weight 2 while endpoints have weight 1. If the missing value v occurred at an interior point, the correction to the estimate is (Δx/2) × 2v = Δx × v. If at an endpoint, the correction is (Δx/2) × v. With Δx = 1, we need: for interior point, 1 × v = 74, so v = 74; for endpoint, (1/2) × v = 74, so v = 148. Choice A gives wrong values. Choice C ignores position dependence. Choice D reverses the interior/endpoint relationship.
4. Consider applying the trapezoidal rule to estimate $\int_a^b f(x) \, dx$ using $n$ subintervals. If the function values at the endpoints are $f(a) = 8$ and $f(b) = 12$, and the sum of all interior function values is 45, what is the trapezoidal approximation in terms of $\Delta x$?
   Answer: $\frac{\Delta x}{2} \cdot (8 + 90 + 12)$ When you encounter trapezoidal rule problems, focus on the standard formula structure: the approximation equals $\frac{\Delta x}{2}$ times a sum that includes endpoint values once and interior values twice. The trapezoidal rule formula is: $T_n = \frac{\Delta x}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)]$ Here, $f(x_0) = f(a) = 8$ and $f(x_n) = f(b) = 12$ are the endpoint values. The sum of interior values is 45, but remember that each interior value gets multiplied by 2 in the formula, giving us $2 \times 45 = 90$. Therefore: $T_n = \frac{\Delta x}{2}[8 + 90 + 12] = \frac{\Delta x}{2}[110]$ This matches answer choice D: $\frac{\Delta x}{2} \cdot (8 + 90 + 12)$. A is wrong because it omits the $\frac{1}{2}$ factor entirely and doesn't account for doubling the interior values. B incorrectly doubles the endpoint values (giving 20 instead of 8 + 12 = 20) when only interior values should be doubled. C makes a similar error by doubling the endpoint sum to get 16, when the endpoints should remain as 8 + 12 = 20, not be doubled individually. Study tip: Always remember the trapezoidal rule pattern: $\frac{\Delta x}{2} \times$ [first endpoint + 2(sum of interior values) + last endpoint]. The factor of 2 applies only to interior points, never to the endpoints.
5. A company uses the trapezoidal rule with $n = 4$ subintervals to estimate $\int_0^8 f(x) \, dx$ where $f(x)$ represents the rate of production (in units per hour). If $f(0) = 12$, $f(2) = 18$, $f(4) = 15$, $f(6) = 21$, and $f(8) = 9$, what is the estimated total production over the 8-hour period?
   Answer: 144 units Using the trapezoidal rule with Δx = 2: T₄ = (Δx/2)[f(0) + 2f(2) + 2f(4) + 2f(6) + f(8)] = (2/2)[12 + 2(18) + 2(15) + 2(21) + 9] = 1[12 + 36 + 30 + 42 + 9] = 144. Choice A uses Δx = 1 instead of 2. Choice B omits the factor of 2 for interior points. Choice D incorrectly doubles the entire sum.
6. The trapezoidal rule approximation $T_n$ for $\int_a^b f(x) \, dx$ is derived by summing the areas of $n$ trapezoids whose top edges are line segments connecting points on the curve. If $f(x)$ is a linear function, such as $f(x) = mx+c$ with $m \neq 0$, what is the relationship between the trapezoidal rule approximation $T_n$ and the exact value of the integral for any number of subintervals $n \geq 1$?
   Answer: $T_n$ is always exactly equal to the exact value. The trapezoidal rule approximates a function on a subinterval with a straight line segment. If the function itself is a straight line (i.e., a linear function), then the approximating line segment is identical to the function's graph. Consequently, the area of each trapezoid is exactly equal to the area under the function on that subinterval. Since this is true for all subintervals, the sum of the trapezoid areas, $T_n$, is exactly equal to the true value of the integral for any choice of $n \geq 1$.
7. An analyst applies the trapezoidal rule to approximate $\int_1^5 g(x) \, dx$ using 4 equal subintervals. If the approximation yields 28.6 and the actual value of the integral is 30.2, what is the approximate percentage error in the trapezoidal rule estimate?
   Answer: 5.3% underestimate Percentage error = |approximation - actual|/|actual| × 100% = |28.6 - 30.2|/30.2 × 100% = 1.6/30.2 × 100% ≈ 5.3%. Since 28.6 < 30.2, this is an underestimate. Choice B uses the wrong base (28.6 instead of 30.2). Choices C and D confuse over/underestimate direction.
8. The marginal revenue function for a product is $MR(q) = 200 - 12\sqrt{q}$, where $q$ is the number of units sold. The total revenue from selling the first 100 units is calculated by $\int_0^{100} (200 - 12\sqrt{q}) \, dq$. An analyst uses the trapezoidal rule with $n=10$ subintervals ($T_{10}$) to estimate this total revenue. Which of the following statements is true about the estimate?
   Answer: The estimate $T_{10}$ is an overestimate of the actual total revenue. The trapezoidal rule provides an overestimate if the function is concave up ($f''(x) > 0$) and an underestimate if the function is concave down ($f''(x) < 0$). Let $f(q) = 200 - 12q^{1/2}$. The first derivative is $f'(q) = -6q^{-1/2}$. The second derivative is $f''(q) = (-6)(-\frac{1}{2})q^{-3/2} = 3q^{-3/2} = \frac{3}{\sqrt{q^3}}$. For the interval of integration $(0, 100]$, $q > 0$, so $f''(q)$ is always positive. This means the function $MR(q)$ is concave up on this interval. Therefore, the trapezoidal rule approximation $T_{10}$ will be an overestimate of the actual total revenue.
9. Use the trapezoidal rule with $n=2$ to approximate the definite integral $\int_0^4 (3x^2 + 1) \, dx$. What is the absolute error of this approximation?
   Answer: 8 This is a two-step problem. First, find the trapezoidal approximation $T_2$. Second, find the exact value of the integral. Finally, compute the absolute difference. 1. Approximation: Let $f(x)=3x^2+1$. With $a=0, b=4, n=2$, we have $\Delta x = \frac{4-0}{2}=2$. The points are $x_0=0, x_1=2, x_2=4$. The function values are $f(0)=1$, $f(2)=13$, $f(4)=49$. $T_2 = \frac{2}{2}[f(0) + 2f(2) + f(4)] = 1[1 + 2(13) + 49] = 1 + 26 + 49 = 76$ 2. Exact Value: $\int_0^4 (3x^2+1) \, dx = [x^3+x]_0^4 = (4^3+4)-(0^3+0) = 64+4=68$ 3. Absolute Error: $|\text{Exact} - \text{Approximation}| = |68 - 76| = |-8| = 8$
10. The trapezoidal rule with $n=4$ subintervals is used to approximate $\int_1^5 f(x) \, dx$. The approximation yields a value of 68. The following function values are known: $f(1)=10$, $f(2)=15$, $f(4)=25$, and $f(5)=30$. What is the value of $f(3)$?
    Answer: $8$ The formula for the trapezoidal rule is $T_n = \frac{\Delta x}{2} [f(x_0) + 2f(x_1) + \dots + 2f(x_{n-1}) + f(x_n)]$. Here, $a=1, b=5, n=4$, so $\Delta x = \frac{5-1}{4} = 1$. The partition points are $x_0=1, x_1=2, x_2=3, x_3=4, x_4=5$. We are given $T_4=68$. Substituting the known values into the formula gives: $68 = \frac{1}{2} [f(1) + 2f(2) + 2f(3) + 2f(4) + f(5)]$ $68 = \frac{1}{2} [10 + 2(15) + 2f(3) + 2(25) + 30]$ $136 = 10 + 30 + 2f(3) + 50 + 30$ $136 = 120 + 2f(3)$ $16 = 2f(3)$ $f(3) = 8$
11. The trapezoidal rule is applied to estimate $\int_2^{10} \sqrt{x} \, dx$ using $n = 4$ subintervals. Which of the following best describes the relationship between this approximation and the exact value of the integral?
    Answer: The approximation will overestimate the exact value since $\sqrt{x}$ is concave down on $[2,10]$ For f(x) = √x, f''(x) = -1/($4x^{3/2}$) < 0 for all x > 0, so √x is concave down on [2,10]. When a function is concave down, the trapezoidal rule overestimates the integral because the trapezoids lie above the curve. Choice A is wrong because the trapezoidal rule is exact only for linear functions. Choice C incorrectly states that √x is concave up. Choice D has the right concavity but wrong conclusion about over/underestimation.
12. A business analyst estimates $\int_0^8 r(x) \, dx$ using the trapezoidal rule with 4 subintervals and obtains 72. If the analyst doubles the number of subintervals to 8 (using the same function), the new approximation is 75. Based on these results, what can be concluded about the concavity of $r(x)$ on $[0,8]$?
    Answer: The function $r(x)$ is primarily concave up because the approximation is converging upward to the true value When you encounter problems comparing trapezoidal rule approximations with different numbers of subintervals, focus on how the error behaves and what it reveals about the function's concavity. The trapezoidal rule has a specific error pattern based on concavity. For concave up functions, the trapezoidal rule consistently underestimates the true integral value because the trapezoids lie below the curved function. For concave down functions, it overestimates because the trapezoids lie above the curve. Here, increasing from 4 to 8 subintervals improved the approximation from 72 to 75. Since more subintervals generally produce more accurate results, and the approximation increased, this suggests the true integral value is somewhere above 75. The pattern shows we're approaching the true value from below, which is characteristic of a concave up function where trapezoidal rule underestimates. Answer A incorrectly assumes that any increase in approximation indicates concave up behavior, but this reasoning is incomplete without considering whether we're approaching from above or below. Answer B makes the opposite error, suggesting concave down behavior when an increasing approximation actually points to concave up (since we're moving toward the true value from below). Answer C is overly cautious—the pattern of underestimation followed by improvement does provide meaningful information about concavity. Answer D correctly identifies both the concave up nature and the key insight: the approximation is converging upward toward the true value, indicating systematic underestimation typical of concave up functions. Remember: trapezoidal rule underestimates for concave up functions and overestimates for concave down functions.
13. Which step contains the student's first mistake?
    Answer: Step 4 Let's check each step. Step 1 correctly calculates $\Delta x = (b-a)/n = (9-1)/4 = 2$. Step 2 correctly lists the partition points starting at $a=1$ and adding $\Delta x$ successively. Step 3 correctly identifies the function values at these points. Step 4 sets up the sum. The trapezoidal rule formula is $T_n = \frac{\Delta x}{2}[f(x_0) + 2f(x_1) + \dots + f(x_n)]$. The student used $\Delta x$ as the multiplier, not $\frac{\Delta x}{2}$. The multiplier should be $\frac{2}{2}=1$, not 2. Therefore, the first mistake is in Step 4.
14. Which expression represents the approximation of $\int_1^3 \frac{1}{x^2+1} \, dx$ using the trapezoidal rule with $n=4$ subintervals?
    Answer: $0.25 \left[ \frac{1}{1^2+1} + \frac{2}{1.5^2+1} + \frac{2}{2^2+1} + \frac{2}{2.5^2+1} + \frac{1}{3^2+1} \right]$ For this integral, $a=1, b=3, n=4$, and $f(x) = \frac{1}{x^2+1}$. The width of each subinterval is $\Delta x = \frac{3-1}{4} = 0.5$. The partition points are $x_0=1, x_1=1.5, x_2=2, x_3=2.5, x_4=3$. The trapezoidal rule formula is $T_n = \frac{\Delta x}{2} [f(x_0) + 2f(x_1) + \dots + f(x_n)]$. Here, the pre-factor is $\frac{\Delta x}{2} = \frac{0.5}{2} = 0.25$. The expression is: $T_4 = 0.25 [f(1) + 2f(1.5) + 2f(2) + 2f(2.5) + f(3)]$ Substituting the function gives: $T_4 = 0.25 \left[ \frac{1}{1^2+1} + \frac{2}{1.5^2+1} + \frac{2}{2^2+1} + \frac{2}{2.5^2+1} + \frac{1}{3^2+1} \right]$
15. A company's profit flow is modeled by the function $P(t) = 150e^{-0.1t}$, where $t$ is in years and $P(t)$ is in thousands of dollars per year. Use the trapezoidal rule with $n=3$ to write an expression that estimates the total profit over the first 6 years, which is given by $\int_0^6 150e^{-0.1t} \, dt$.
    Answer: $150(1 + 2e^{-0.2} + 2e^{-0.4} + e^{-0.6})$ When you encounter a definite integral representing accumulation over time (like total profit from a profit rate), numerical integration methods like the trapezoidal rule become essential tools when exact integration is complex or unnecessary. The trapezoidal rule with $n=3$ subintervals over $[0,6]$ creates intervals of width $\Delta t = \frac{6-0}{3} = 2$. This gives you evaluation points at $t = 0, 2, 4, 6$. The trapezoidal rule formula is: $\int_a^b f(t)\,dt \approx \frac{\Delta t}{2}[f(t_0) + 2f(t_1) + 2f(t_2) + f(t_3)]$ Let's calculate the function values: $P(0) = 150e^0 = 150$, $P(2) = 150e^{-0.2}$, $P(4) = 150e^{-0.4}$, and $P(6) = 150e^{-0.6}$. Applying the formula: $\frac{2}{2}[150 + 2(150e^{-0.2}) + 2(150e^{-0.4}) + 150e^{-0.6}]$ Factoring out 150: $150[1 + 2e^{-0.2} + 2e^{-0.4} + e^{-0.6}]$ This matches answer choice D. Answer A omits the crucial factor of 2 for interior points and the endpoint values. Answer B incorrectly uses $\Delta t = 4$ instead of 2, doubling the coefficient to 300. Answer C forgets the factor of 2 for interior function values, which is essential to the trapezoidal rule. Remember: the trapezoidal rule always weights interior points by 2 and endpoints by 1, with the entire sum multiplied by $\frac{\Delta t}{2}$.
16. The total variable cost to produce $X$ units of a product is given by the integral of the marginal cost function, $\int_0^X C'(q) \, dq$. Let $T_n$ be the trapezoidal rule approximation of this integral using $n$ subintervals. Assuming $C'(q)$ is a continuous function, which statement best describes the behavior of $T_n$ as $n \to \infty$?
    Answer: $T_n$ approaches the exact total variable cost to produce $X$ units. This question tests your understanding of numerical integration and the Fundamental Theorem of Calculus in a business context. When you see the trapezoidal rule approximating an integral, focus on what happens as the approximation becomes more refined. The integral $\int_0^X C'(q) \, dq$ represents the exact total variable cost to produce $X$ units. By the Fundamental Theorem of Calculus, this equals $C(X) - C(0)$, where $C(0) = 0$ for variable costs, giving us the total variable cost $C(X)$. The trapezoidal rule $T_n$ approximates this integral by dividing the interval $[0,X]$ into $n$ subintervals and using trapezoids to estimate the area under the curve $C'(q)$. Since $C'(q)$ is continuous, as $n \to \infty$, the trapezoidal approximation converges to the exact value of the integral. Therefore, $T_n$ approaches the exact total variable cost to produce $X$ units, making B correct. Choice A is wrong because average variable cost would be $\frac{1}{X}\int_0^X C'(q) \, dq$, not the integral itself. Choice C is incorrect because the trapezoidal rule converges to a finite value for continuous functions—it doesn't diverge to infinity. Choice D confuses the integral with its derivative; $C'(X)$ is the marginal cost at point $X$, not the total variable cost from 0 to $X$. Study tip: Remember that numerical integration methods like the trapezoidal rule approximate definite integrals, and as you use more subintervals, you get closer to the exact integral value—not to derivatives or averages.

Start this quiz