AP Calculus BC Quiz: Accumulation Functions Definite Intervals Applied Contexts

What this quiz covers

This quiz focuses on Accumulation Functions Definite Intervals Applied Contexts, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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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.

Question 1

1. $\int_2^6 r'(t) dt$ (correct answer)
2. $r'(6) - r'(2)$
3. $\frac{1}{4} \int_2^6 r'(t) dt$
4. $r(6)$

Explanation: The Fundamental Theorem of Calculus states that the integral of a rate of change of a quantity gives the net change in that quantity. Here, $r'(t)$ is the rate of change of the radius. Therefore, the total increase (net change) in the radius from $t=2$ to $t=6$ is given by the definite integral $\int_2^6 r'(t) dt$.

Question 2

1. $500 - \int_0^4 (20t + 50) dt$ (correct answer)
2. $\int_0^4 (20t + 50) dt$
3. $500 - (20(4) + 50)$
4. $500 + \int_0^4 (20t + 50) dt$

Explanation: The initial number of orders is 500. The total number of orders processed from $t=0$ to $t=4$ is the integral of the rate of processing, $\int_0^4 P(t) dt$. The number of orders remaining is the initial amount minus the total amount processed. Thus, the expression is $500 - \int_0^4 (20t + 50) dt$.

Question 3

1. The rate of absorption equals the rate of elimination, $A(t) = E(t)$. (correct answer)
2. The net rate of change of the drug, $A(t) - E(t)$, is at its maximum value.
3. The total amount of drug absorbed equals the total amount of drug eliminated.
4. The derivative of the net rate, $A'(t) - E'(t)$, is equal to zero.

Explanation: Let $M(t)$ be the amount of the drug in the bloodstream. The rate of change of this amount is $M'(t) = A(t) - E(t)$. To find a maximum value for $M(t)$, we must find a critical point by setting its derivative to zero: $M'(t) = A(t) - E(t) = 0$. This occurs when $A(t) = E(t)$, where the rate of change switches from positive to negative.

Question 4

1. $\frac{\pi}{4}$
2. $5.5$
3. $5 + \ln(2)$
4. $5 + \frac{\pi}{4}$ (correct answer)

Explanation: The value of $Q(1)$ is the initial value $Q(0)$ plus the net change from $t=0$ to $t=1$. $Q(1) = Q(0) + \int_0^1 Q'(t) dt = 5 + \int_0^1 \frac{1}{t^2+1} dt = 5 + [\arctan(t)]_0^1 = 5 + (\arctan(1) - \arctan(0)) = 5 + (\frac{\pi}{4} - 0) = 5 + \frac{\pi}{4}$.

Question 5

1. 80
2. 150 (correct answer)
3. 240
4. 390

Explanation: The net rate of change of people in the park is $R(t) = E(t) - L(t) = (100 + 20t) - 80 = 20 + 20t$. The net change from $t=0$ to $t=3$ is the integral of the net rate: $\int_0^3 (20 + 20t) dt = [20t + 10t^2]_0^3 = (20(3) + 10(3^2)) - 0 = 60 + 90 = 150$ people.

Question 6

1. $50000 + \int_0^{100} 150\sqrt{t} dt$
2. $50000 - \int_0^{100} 150\sqrt{t} dt$ (correct answer)
3. $\int_0^{100} 150\sqrt{t} dt$
4. $50000 - 150\sqrt{100}$

Explanation: The amount of fuel remaining is the initial amount minus the total amount of fuel burned. The amount burned is the integral of the rate of consumption from $t=0$ to $t=100$. Thus, the remaining fuel is $50000 - \int_0^{100} 150\sqrt{t} dt$.

Question 7

1. $\int_0^{24} (S(t) - M(t)) dt = 0$ (correct answer)
2. $S(t) = M(t)$ for all $t$ in $[0, 24]$
3. The average rates of snowmaking and melting over the 24-hour period are equal
4. $S(24) = M(24)$

Explanation: If the amount of snow is the same at $t=24$ as at $t=0$, then the net change in snow is zero. The net change equals the integral of the net rate of change: $\int_0^{24} (S(t) - M(t)) dt = 0$. This doesn't require that $S(t) = M(t)$ at every instant, only that the total snow added equals the total snow melted over the entire period.

Question 8

1. $\frac{C(5) - C(2)}{5 - 2}$ (correct answer)
2. $C(5) - C(2)$
3. $\frac{1}{3} \int_2^5 C(t) dt$
4. $C'(5) - C'(2)$

Explanation: The question asks for the average rate of change of the number of cars. Since $C(t)$ is the total number of cars (an accumulation function), the rate is $C'(t)$. The average value of the rate $C'(t)$ over $[2, 5]$ is $\frac{1}{5-2}\int_2^5 C'(t)dt$. By the Fundamental Theorem of Calculus, this is equal to $\frac{C(5)-C(2)}{5-2}$. This is also the standard definition of average rate of change for the function $C(t)$.

Question 9

1. $6 + \frac{9}{\pi}$ (correct answer)
2. $6 - \frac{3}{\pi}$
3. $9 + \frac{6}{\pi}$
4. $1.5 + \sin(\frac{\pi}{3})$

Explanation: The amount of snow that fell from $t=2$ to $t=6$ is given by $S(6) - S(2) = \int_2^6 s(x) dx$. $\int_2^6 (1.5 + \sin(\frac{\pi x}{6})) dx = [1.5x - \frac{6}{\pi}\cos(\frac{\pi x}{6})]_2^6 = (1.5(6) - \frac{6}{\pi}\cos(\pi)) - (1.5(2) - \frac{6}{\pi}\cos(\frac{\pi}{3})) = (9 - \frac{6}{\pi}(-1)) - (3 - \frac{6}{\pi}(\frac{1}{2})) = (9 + \frac{6}{\pi}) - (3 - \frac{3}{\pi}) = 6 + \frac{9}{\pi}$ cm.

Question 10

1. $\int_{101}^{200} (150 - 0.2x) dx$
2. $P'(200) - P'(100)$
3. $\frac{1}{100} \int_{100}^{200} (150 - 0.2x) dx$
4. $\int_{100}^{200} (150 - 0.2x) dx$ (correct answer)

Explanation: The change in profit from selling the 101st unit through the 200th unit corresponds to the accumulation of profit from $x=100$ to $x=200$. The total change is the definite integral of the marginal profit function over this interval, which is $\int_{100}^{200} (150 - 0.2x) dx$.

Question 11

1. The total number of items produced from day 5 to day 12, in items (correct answer)
2. The production rate on day 12, in items per day
3. The average production rate from day 5 to day 12, in items per day
4. The change in production rate from day 5 to day 12, in items per day
5. The number of days needed to produce 12 items, in days

Explanation: This question tests the skill of interpreting definite integrals as accumulation in applied contexts. The definite integral ∫ from 5 to 12 of m(t) dt represents the total number of items produced over the 7-day interval, as m(t) is the production rate in items per day. Integrating the rate over time accumulates the net output of items from day 5 to day 12. This gives the overall count of items manufactured during that period. A tempting distractor is choice C, the average production rate, which fails because it requires dividing the integral by the number of days (7) to find the average, not just the integral itself. Always check units: (items/day) × days yields items, matching total produced, whereas average rate would be in items/day.

Question 12

1. The rainfall rate at $t=6$, in inches per hour
2. The total rainfall accumulated from $t=2$ to $t=6$, in inches (correct answer)
3. The average rainfall rate from $t=2$ to $t=6$, in inches per hour
4. The change in rainfall rate from $t=2$ to $t=6$, in inches per hour
5. The number of hours it rained, in hours

Explanation: This question tests the skill of interpreting definite integrals as accumulation in applied contexts. The definite integral ∫ from 2 to 6 of p(t) dt represents the total rainfall accumulated over the 4-hour interval, as p(t) is the rainfall rate in inches per hour. Integrating the rate over time accumulates the total depth of rain that fell from t=2 to t=6. This measures the overall inches of precipitation during that period. A tempting distractor is choice C, the average rainfall rate, which fails because it requires dividing the integral by the interval length (4 hours) to compute the average, not the integral alone. Always check units: (inches/hour) × hours yields inches, matching total rainfall, whereas average rate would remain in inches/hour.

Question 13

1. The concentration of pollutant in the lake at $t=10$, in kilograms per day
2. The total mass of pollutant added from $t=0$ to $t=10$, in kilograms (correct answer)
3. The average input rate over $0\le t\le 10$, in kilograms per day
4. The change in input rate between $t=0$ and $t=10$, in kilograms per day
5. The time when the input rate is largest, in days

Explanation: This question tests the skill of interpreting definite integrals as accumulation in applied contexts. The definite integral ∫ from 0 to 10 of c(t) dt represents the total mass of pollutant added to the lake over the 10-day interval, as c(t) is the input rate in kilograms per day. Integrating the rate over time accumulates the net kilograms of pollutant entering from t=0 to t=10. This measures the overall pollution load added during that period. A tempting distractor is choice C, the average input rate, which fails because it requires dividing the integral by the time interval (10 days) to compute the average, not the integral alone. Always check units: (kg/day) × days yields kg, matching total mass, whereas average rate would remain in kg/day.

Question 14

1. The average inflow rate over $0\le t\le 8$, in liters per minute
2. The total amount of water added to the tank from $t=0$ to $t=8$, in liters (correct answer)
3. The inflow rate at $t=8$, in liters per minute
4. The change in time from $0$ to $8$, in minutes
5. The total time required to add 8 liters, in minutes

Explanation: This problem tests your ability to interpret definite integrals as accumulation in applied contexts. Since r(t) represents the rate of water flow in liters per minute, the definite integral ∫₀⁸ r(t)dt accumulates all the instantaneous rates over the 8-minute interval. This gives the total amount of water that flows into the tank from t=0 to t=8, measured in liters. Choice A (average inflow rate) would require dividing the integral by 8, which is a common misconception when students confuse total accumulation with average rate. To verify units, remember that (liters/minute) × (minutes) = liters, confirming that the integral represents a total quantity of water, not a rate.

Question 15

1. The average filling rate of the tank from $t=2$ to $t=6$, in liters per minute
2. The total number of liters added to the tank from $t=2$ to $t=6$ (correct answer)
3. The instantaneous amount of water in the tank at $t=6$, in liters
4. The change in time required to add one liter between $t=2$ and $t=6$, in minutes per liter
5. The total filling rate accumulated from $t=2$ to $t=6$, in liters per minute

Explanation: This problem tests your ability to interpret definite integrals as accumulation in applied contexts. Since r(t) represents the filling rate in liters per minute, the definite integral ∫₂⁶ r(t)dt accumulates these rates over time, giving the total volume of water added to the tank. The integral multiplies rate (liters/minute) by time (minutes), yielding units of liters—representing the total amount added from t=2 to t=6. Choice A incorrectly suggests an average rate, but that would require dividing the integral by (6-2). Always verify units: when integrating a rate with respect to time, you get the total accumulated quantity, not a rate.

Question 16

1. The average power used from hour 3 to hour 7, in kilowatts
2. The total energy consumed from hour 3 to hour 7, in kilowatt-hours (correct answer)
3. The instantaneous energy consumption rate at hour 7, in kilowatt-hours
4. The change in power from hour 3 to hour 7, in kilowatts per hour
5. The total power used from hour 3 to hour 7, in kilowatts

Explanation: This problem tests your understanding of definite integrals as accumulation in energy contexts. Since P(t) represents power usage in kilowatts (energy per time), the definite integral ∫₃⁷ P(t)dt accumulates this power over time, giving the total energy consumed. The integral multiplies power (kilowatts) by time (hours), yielding units of kilowatt-hours—the standard unit for energy consumption from hour 3 to hour 7. Choice E incorrectly suggests "total power," but power is already a rate; we accumulate energy, not power itself. Always check units: rate × time = total quantity, so kilowatts × hours = kilowatt-hours of energy.

Question 17

1. The amount of rainfall accumulated from $t=0$ to $t=3$, in inches (correct answer)
2. The rainfall rate at $t=3$, in inches per hour
3. The average rainfall rate from $t=0$ to $t=3$, in inches
4. The total number of hours of rain from $t=0$ to $t=3$, in hours
5. The change in rainfall rate from $t=0$ to $t=3$, in inches per hour squared

Explanation: This problem tests your ability to interpret definite integrals as accumulation in meteorological contexts. Since R(t) represents the rainfall rate in inches per hour, the definite integral ∫₀³ R(t)dt accumulates these rates over time, giving the total rainfall accumulated. The integral multiplies rate (inches/hour) by time (hours), yielding units of inches—representing the total depth of rain that fell from t=0 to t=3. Choice C incorrectly suggests an average rate, but the integral gives a total amount, not a rate (average would require dividing by 3). Always verify units: rate × time = accumulated quantity, so inches/hour × hours = inches of rainfall.

Question 18

1. The volume of water in the reservoir at day 9, in cubic meters
2. The total volume of water that leaves from day 2 to day 9, in cubic meters (correct answer)
3. The average leaving rate from day 2 to day 9, in cubic meters
4. The net change in leaving rate from day 2 to day 9, in cubic meters per day
5. The time it takes for one cubic meter to leave between day 2 and day 9, in days per cubic meter

Explanation: This problem tests your ability to interpret definite integrals as accumulation in hydrological contexts. Since L(t) represents the rate at which water leaves in cubic meters per day, the definite integral ∫₂⁹ L(t)dt accumulates these rates over time, giving the total volume of water that left the reservoir. The integral multiplies rate (cubic meters/day) by time (days), yielding units of cubic meters—representing the total water loss from day 2 to day 9. Choice A incorrectly suggests the volume remaining in the reservoir, but the integral measures what left, not what remains. Always focus on what the rate describes: L(t) is a leaving rate, so its integral is total volume that left.

Question 19

1. The runner’s speed at $t=10$, in meters per second
2. The runner’s average speed from $t=4$ to $t=10$, in meters per second
3. The distance the runner travels from $t=4$ to $t=10$, in meters (correct answer)
4. The change in speed from $t=4$ to $t=10$, in meters per second per second
5. The time the runner spends traveling from $t=4$ to $t=10$, in seconds

Explanation: This problem requires interpreting a definite integral of speed as distance traveled in an applied context. Since s(t) represents the runner's speed in meters per second, the definite integral $\int_4^{10} s(t)\, dt$ accumulates these speeds over time, giving the total distance traveled. The integral multiplies speed (meters/second) by time (seconds), yielding units of meters—representing how far the runner traveled from t=4 to t=10. Choice B incorrectly suggests average speed, but that would require dividing the integral by $(10-4)$. Remember: speed integrated over time gives distance traveled (always positive), while velocity integrated gives displacement (can be negative).

Question 20

1. The number of fish in the population at day 20
2. The net change in the population from day 10 to day 20, in fish (correct answer)
3. The average rate of change of the population from day 10 to day 20, in fish per day
4. The time required for the population to change by one fish between days 10 and 20, in days per fish
5. The total rate of change from day 10 to day 20, in fish per day

Explanation: This problem requires interpreting a definite integral of a population rate as net change in an applied context. Since p(t) represents the rate of population change in fish per day, the definite integral ∫₁₀²⁰ p(t)dt accumulates these rates over time, giving the net change in population size. The integral multiplies rate of change (fish/day) by time (days), yielding units of fish—representing how much the population increased or decreased from day 10 to day 20. Choice A incorrectly suggests the actual population size at day 20, but the integral gives change, not absolute value. Remember: integrating a rate of change gives the net change, not the final value.