Accumulation Functions, Definite Intervals, Applied Contexts - AP Calculus BC

A population of insects is growing at a rate of $g(t) = 100 + 8t$ insects per day. If the initial population at $t=0$ is 500 insects, which of the following gives the total population at time $t=10$ days?
Answer: $500 + \int_0^{10} (100 + 8t) dt$ The total population at a given time is the initial population plus the net change in population. The net change is the integral of the growth rate function over the time interval. Therefore, the population at $t=10$ is the initial population of 500 plus the accumulated growth from $t=0$ to $t=10$ , which is given by $500 + \int_0^{10} (100 + 8t) dt$ .
The rate of growth of a forest's biomass is modeled by $B'(t) = k\cos(0.2t)$ tons per year, where $t$ is time in years. Which of the following expressions represents the total change in the forest's biomass from year $t=5$ to year $t=10$ ?
Answer: $\int_5^{10} k\cos(0.2t) dt$ To find the total change in biomass over a time interval, one must integrate the rate of change of biomass, $B'(t)$ , over that interval. The time interval is from $t=5$ to $t=10$ . Therefore, the correct expression for the total change is the definite integral $\int_5^{10} k\cos(0.2t) dt$ .
A lake's water level changes at rate $h'(t)$ meters/day for $0\le t\le 12$ . What does $\int_0^{12} h'(t)\,dt$ represent?
Answer: The net change in water level from day 0 to day 12, in meters This question requires interpreting a definite integral of a derivative as net change in an environmental context. Since h'(t) represents the rate of water level change in meters per day, the definite integral ∫₀¹² h'(t)dt accumulates all the instantaneous rates of change over the 12-day interval. By the Fundamental Theorem of Calculus, this equals h(12) - h(0), which is the net change in water level from day 0 to day 12, measured in meters. Choice A (water level at t=12) would only be correct if we knew h(0) and could add it to the integral result. Always check units: (meters/day) × (days) = meters, confirming a change in water level.
A piece of metal is heated. Its temperature changes at a rate of $T'(t) = 2t + 5$ degrees Celsius per minute. If the temperature of the metal is 350 degrees Celsius after 10 minutes, what was the initial temperature of the metal?
Answer: 200 The temperature at time $t=10$ is given by $T(10) = T(0) + \int_0^{10} T'(t) dt$ . We are given $T(10) = 350$ . The integral is $\int_0^{10} (2t + 5) dt = [t^2 + 5t]_0^{10} = 100 + 50 = 150$ . So, $350 = T(0) + 150$ . Solving for the initial temperature, $T(0) = 350 - 150 = 200$ degrees Celsius.
Suppose the amount of snow on the slope is the same at $t=24$ as it was at $t=0$ . Which of the following must be true?
Answer: $\int_0^{24} (S(t) - M(t)) dt = 0$ 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.
Fuel is consumed at rate $c(t)$ gallons/hour; what does $\int_{1}^{5} c(t)\,dt$ represent during the trip?
Answer: The total fuel consumed from $t=1$ to $t=5$ , in gallons This question tests the skill of interpreting accumulation functions using definite integrals in applied contexts. The definite integral ∫ from 1 to 5 of c(t) dt represents the total fuel consumed during the trip over the interval from t=1 to t=5 hours. Since c(t) is the consumption rate in gallons per hour, integrating over time yields the accumulated usage, in gallons. This measures the net amount of fuel used in that timeframe. A tempting distractor is choice A, the average fuel consumption rate, but that requires dividing the integral by the 4-hour interval to obtain gallons per hour. To verify such interpretations, perform a units check: multiplying rate units (gal/hr) by time (hr) gives gal, confirming total consumed instead of a rate.
Electric charge flows with current $I(t)$ coulombs/second for $0\le t\le 3$ . What does $\int_0^3 I(t)\,dt$ represent?
Answer: The total charge transferred from $t=0$ to $t=3$ , in coulombs This question requires interpreting a definite integral as accumulation in an electrical context. Since I(t) represents electric current in coulombs per second (which is the definition of current), the definite integral ∫₀³ I(t)dt accumulates all the instantaneous current values over the 3-second interval. This accumulation gives the total electric charge transferred from t=0 to t=3, measured in coulombs. Choice E (net change in current) would require integrating the derivative of current, not current itself. For unit verification: (coulombs/second) × (seconds) = coulombs, confirming total charge transferred.
A bicyclist's acceleration is $a(t)$ m/s $^2$ ; what does $\int_{2}^{7} a(t)\,dt$ represent?
Answer: The net change in velocity from $t=2$ to $t=7$ , in meters per second This question tests the skill of interpreting accumulation functions using definite integrals in applied contexts. The definite integral ∫ from 2 to 7 of a(t) dt represents the net change in the bicyclist's velocity over the interval from t=2 to t=7 seconds. Since a(t) is acceleration in meters per second squared, integrating over time yields the change in velocity, in meters per second. This is v(7) - v(2) by the Fundamental Theorem of Calculus. A tempting distractor is choice C, the displacement, but that would require integrating velocity, not acceleration. To verify such interpretations, perform a units check: multiplying acceleration units (m/s²) by time (s) gives m/s, matching velocity change rather than position or acceleration units.
The amount of water in a reservoir is changing at a rate of $R(t) = 1000 - 40t$ gallons per day for $0 \le t \le 40$ . The reservoir contains 50,000 gallons of water at time $t=0$ . At what time $t$ is the amount of water in the reservoir at its maximum?
Answer: $t=25$ Let $W(t)$ be the amount of water. The rate of change is $W'(t) = R(t)$ . To find the maximum amount, we find critical points by setting $R(t)=0$ . $1000 - 40t = 0$ gives $t=25$ . Since $R(t) > 0$ for $t<25$ and $R(t) < 0$ for $t>25$ , the amount of water is increasing up to $t=25$ and then decreasing. Thus, the absolute maximum occurs at $t=25$ days.
Rain falls at rate $f(t)$ inches/hour for $6\le t\le 10$ . What does $\int_6^{10} f(t)\,dt$ represent?
Answer: The total rainfall accumulated from $t=6$ to $t=10$ , in inches This problem tests your ability to interpret definite integrals as accumulation in a weather context. Since f(t) represents the rainfall rate in inches per hour, the definite integral ∫₆¹⁰ f(t)dt accumulates all the instantaneous rainfall rates over the 4-hour interval from t=6 to t=10. This gives the total rainfall accumulated during this time period, measured in inches. Choice B (average rainfall rate) is incorrect because it still has rate units (inches/hour), while the integral converts rate to total accumulation. Always verify units: (inches/hour) × (hours) = inches, confirming total rainfall amount.
A runner's speed is $s(t)$ meters/second for $0 \le t \le 12$ seconds. What does $\int_4^{10} s(t)\,dt$ represent?
Answer: The distance the runner travels from $t=4$ to $t=10$ , in meters 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).
A medication enters the bloodstream at rate $m(t)$ mg/min for $0 \le t \le 120$ . What does $\int_{30}^{90} m(t)\,dt$ represent?
Answer: The total amount of medication entering from $t=30$ to $t=90$ , in milligrams This problem requires interpreting a definite integral of an entry rate as total accumulation in a pharmacological context. Since m(t) represents the rate at which medication enters the bloodstream in milligrams per minute, the definite integral $\int_{30}^{90} m(t)\,dt$ accumulates these rates over time, giving the total amount of medication that entered. The integral multiplies rate (mg/min) by time (minutes), yielding units of milligrams—representing the total medication absorbed from t=30 to t=90. Choice D incorrectly suggests a change in rate with units $\text{mg/min}^2$ , but the integral of a rate gives total quantity, not rate change. Remember: rate $\times$ time = total quantity, so $\text{mg/min} \times \text{min} = \text{mg}$ of medication.
The doors open at $t=0$ and there are no people in the venue initially. Which of the following integrals represents the number of people in the venue at time $t=4$ hours?
Answer: $\int_0^4 (E(t) - L(t)) dt$ Since the venue starts empty, the number of people inside at time $t=4$ is the net accumulation of people from $t=0$ to $t=4$ . The net rate of change of people is the rate of entry minus the rate of leaving, $E(t) - L(t)$ . The total accumulation is the definite integral of this net rate, which is $\int_0^4 (E(t) - L(t)) dt$ .
A website gains users at rate $G(t)$ users/day; what does $\int_{10}^{20} G(t)\,dt$ represent?
Answer: The total number of users gained from day 10 to day 20, in users This question tests the skill of interpreting accumulation functions using definite integrals in applied contexts. The definite integral ∫ from 10 to 20 of G(t) dt represents the total number of users gained by the website over the interval from day 10 to day 20. Since G(t) is the gain rate in users per day, integrating over time accumulates the net increase, in users. This measures the overall growth in user base during that period. A tempting distractor is choice B, the average gain rate, but that involves dividing the integral by the 10-day interval to get users per day. To verify such interpretations, perform a units check: multiplying rate units (users/day) by time (days) gives users, aligning with total gained rather than a rate.
The marginal profit for a company, in dollars per unit, is given by $P'(x) = 150 - 0.2x$ , where $x$ is the number of units sold. Which of the following expressions gives the total change in profit from selling the 101st unit through the 200th unit?
Answer: $\int_{100}^{200} (150 - 0.2x) dx$ 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 $\int_{100}^{200} (150 - 0.2x) dx$ .
The velocity of a particle moving along the y-axis is given by $v(t)$ , where $v(t)$ is measured in meters per second and $t$ is in seconds. What does the value of $\int_{a}^{b} v(t) dt$ represent?
Answer: The net displacement of the particle, in meters, from time $t=a$ to $t=b$ . The definite integral of a velocity function over a time interval gives the net change in position, which is known as displacement. The total distance would be the integral of the speed, $\int_a^b |v(t)| dt$ . The average velocity would require dividing by the length of the time interval. The change in velocity would be $v(b)-v(a)$ .
Water leaks from a pool at rate $L(t)$ gallons/hr; what does $\int_{0}^{10} L(t)\,dt$ represent?
Answer: The total amount of water leaked from $t=0$ to $t=10$ , in gallons This question tests the skill of interpreting accumulation functions using definite integrals in applied contexts. The definite integral ∫ from 0 to 10 of L(t) dt represents the total volume of water leaked from the pool over the interval from t=0 to t=10 hours. Since L(t) is the leak rate in gallons per hour, integrating over time yields the accumulated loss, in gallons. This measures the net amount of water that has escaped during that period. A tempting distractor is choice D, the average leak rate, but that would mean dividing the integral by 10 hours to get gallons per hour. To verify such interpretations, perform a units check: multiplying rate units (gal/hr) by time (hr) gives gal, confirming total volume rather than a rate.
How many gallons of fuel does the car consume during this one-hour interval?
Answer: $\frac{31}{3}$ First, express the fuel consumption rate as a function of time, $t$ , by substituting $v(t)$ into $F(v)$ . $R(t) = F(v(t)) = 2 + 0.01(50t)^2 = 2 + 25t^2$ . To find the total fuel consumed, integrate this rate from $t=0$ to $t=1$ : $\int_0^1 (2 + 25t^2) dt = [2t + \frac{25}{3}t^3]_0^1 = 2 + \frac{25}{3} = \frac{31}{3}$ gallons.
Data is added to a server at a rate modeled by $D(t) = 10 + 5\cos(\frac{\pi t}{12})$ gigabytes per hour, where $t$ is the number of hours past midnight. How many gigabytes of data are added to the server between 6 AM ( $t=6$ ) and noon ( $t=12$ )?
Answer: $60 - \frac{60}{\pi}$ The total data added is the integral of the rate from $t=6$ to $t=12$ . $\int_6^{12} (10 + 5\cos(\frac{\pi t}{12})) dt = [10t + 5\frac{12}{\pi}\sin(\frac{\pi t}{12})]_6^{12} = (120 + \frac{60}{\pi}\sin(\pi)) - (60 + \frac{60}{\pi}\sin(\frac{\pi}{2})) = 120 - (60 + \frac{60}{\pi}(1)) = 60 - \frac{60}{\pi}$ gigabytes.
A savings account balance changes at rate $B'(t)$ dollars/month for $0\le t\le 9$ . What does $\int_0^9 B'(t)\,dt$ represent?
Answer: The net change in account balance from month 0 to month 9, in dollars This question requires interpreting a definite integral of a derivative as net change in a financial context. Since B'(t) represents the rate of change of the account balance in dollars per month, the definite integral ∫₀⁹ B'(t)dt accumulates all the instantaneous rates of change over the 9-month interval. By the Fundamental Theorem of Calculus, this equals B(9) - B(0), which is the net change in account balance from month 0 to month 9, measured in dollars. Choice A (account balance at t=9) would require knowing the initial balance B(0) to add to this net change. To check units: (dollars/month) × (months) = dollars, confirming a change in balance amount.

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