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AP Calculus BC Quiz

AP Calculus BC Quiz: Exploring Accumulations Of Change

Practice Exploring Accumulations Of Change in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

A runner’s speed is v(t)=2+cos⁡tv(t)=2+\cos tv(t)=2+cost m/s for 0≤t≤2π0\le t\le2\pi0≤t≤2π; what distance is traveled?

Select an answer to continue

What this quiz covers

This quiz focuses on Exploring Accumulations Of Change, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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.

All questions

Question 1

A runner’s speed is v(t)=2+cos⁡tv(t)=2+\cos tv(t)=2+cost m/s for 0≤t≤2π0\le t\le2\pi0≤t≤2π; what distance is traveled?

  1. 000 meters
  2. 2π2\pi2π meters
  3. 4π4\pi4π meters (correct answer)
  4. 6π6\pi6π meters
  5. 222 meters

Explanation: This problem assesses the skill of exploring accumulations of change by calculating the total distance traveled from a speed function. The speed is v(t)=2+cos⁡tv(t) = 2 + \cos tv(t)=2+cost meters per second over 000 to 2π2\pi2π seconds. Since speed is always positive, the distance is the integral of v(t)v(t)v(t) from 000 to 2π2\pi2π, accumulating the rate of distance over time. This gives [2t+sin⁡t][2t + \sin t][2t+sint] from 000 to 2π2\pi2π = 4π+0=4π4\pi + 0 = 4\pi4π+0=4π meters. A tempting distractor is 2π2\pi2π meters, from integrating only the constant 2 and ignoring cos⁡t\cos tcost, but the full function must be included. In general, to find the net accumulation of a quantity from its rate of change, compute the definite integral of the rate function over the specified interval.

Question 2

A leak changes oil amount at rate A′(t)=−5tA'(t)=-5\sqrt{t}A′(t)=−5t​ liters/hour for 0≤t≤40\le t\le 40≤t≤4; what is net change in oil?

  1. −803-\frac{80}{3}−380​ liters (correct answer)
  2. 803\frac{80}{3}380​ liters
  3. −20-20−20 liters/hour
  4. −40-40−40 liters
  5. 403\frac{40}{3}340​ liters

Explanation: This problem uses accumulation reasoning to find net change in oil from the leak rate. The net change equals ∫A'(t)dt = ∫(-5√t)dt from t = 0 to t = 4. Computing: ∫(-5t^(1/2))dt = -5(2t^(3/2)/3) = -10t^(3/2)/3 evaluated from 0 to 4 gives -10(8)/3 - 0 = -80/3 liters. Choice C incorrectly gives a rate (liters/hour) rather than the accumulated volume change. When integrating power functions with fractional exponents, apply the power rule carefully and track negative signs.

Question 3

For 0≤t≤30\le t\le 30≤t≤3, a population changes at rate P′(t)=2e−tP'(t)=2e^{-t}P′(t)=2e−t thousand/day; what is total population increase?

  1. 2e−32e^{-3}2e−3 thousand
  2. 2(1−e−3)2(1-e^{-3})2(1−e−3) thousand (correct answer)
  3. 6e−36e^{-3}6e−3 thousand/day
  4. 2(1+e−3)2(1+e^{-3})2(1+e−3) thousand
  5. 666 thousand

Explanation: This problem requires accumulation reasoning to find total population increase from the rate of change. The total increase equals ∫P'(t)dt = ∫2e^(-t)dt from t = 0 to t = 3. Computing the integral: ∫2e^(-t)dt = -2e^(-t) evaluated from 0 to 3 gives -2e^(-3) - (-2e^0) = -2e^(-3) + 2 = 2(1 - e^(-3)) thousand. Choice C incorrectly gives a rate (thousand/day) rather than the accumulated population change. To find total change from an exponential rate, integrate carefully and evaluate at the bounds.

Question 4

For 2≤t≤52 \le t \le 52≤t≤5, temperature changes at rate T′(t)=1t−2T'(t) = \frac{1}{t} - 2T′(t)=t1​−2 °C/min; what is net temperature change?

  1. ln⁡ ⁣(52)−6\ln\!\left(\frac{5}{2}\right) - 6ln(25​)−6 °C (correct answer)
  2. ln⁡ ⁣(52)+6\ln\!\left(\frac{5}{2}\right) + 6ln(25​)+6 °C
  3. ln⁡ ⁣(52)\ln\!\left(\frac{5}{2}\right)ln(25​) °C/min
  4. −6-6−6 °C/min
  5. 6−ln⁡ ⁣(52)6 - \ln\!\left(\frac{5}{2}\right)6−ln(25​) °C

Explanation: This problem requires accumulation reasoning to find net temperature change from the rate. The net change equals ∫T′(t) dt=∫(1t−2)dt\int T'(t) \, dt = \int \left( \frac{1}{t} - 2 \right) dt∫T′(t)dt=∫(t1​−2)dt from t = 2 to t = 5. Computing: ∫(1t−2)dt=ln⁡∣t∣−2t\int \left( \frac{1}{t} - 2 \right) dt = \ln |t| - 2t∫(t1​−2)dt=ln∣t∣−2t evaluated from 2 to 5 gives (ln⁡5−10)−(ln⁡2−4)=ln⁡5−ln⁡2−6=ln⁡(52)−6(\ln 5 - 10) - (\ln 2 - 4) = \ln 5 - \ln 2 - 6 = \ln\left( \frac{5}{2} \right) - 6(ln5−10)−(ln2−4)=ln5−ln2−6=ln(25​)−6 °C. Choice C incorrectly gives units of rate (°C/min) rather than accumulated temperature change. To find net change, integrate each term separately and combine the results at the bounds.

Question 5

Water flows with rate r(t)=6t+1r(t)=\frac{6}{t+1}r(t)=t+16​ L/min for 0≤t≤30\le t\le30≤t≤3; how many liters enter total?

  1. ln⁡4\ln 4ln4
  2. 6ln⁡46\ln 46ln4 (correct answer)
  3. 18ln⁡418\ln 418ln4
  4. 64\frac{6}{4}46​
  5. 184\frac{18}{4}418​

Explanation: This problem tests the skill of exploring accumulations of change by using integration to find the total water volume from a flow rate. The total liters entering is the definite integral of r(t) = 6/(t+1) from t=0 to t=3, accumulating the flow over the interval. The antiderivative 6 ln|t+1| evaluated gives 6(ln4 - ln1) = 6 ln4 liters. This reflects the decreasing flow rate still yielding a positive accumulation. A tempting distractor like ln4 might arise from omitting the coefficient 6, but it fails as the integral requires multiplying by that factor. In general, to find the net accumulation over an interval, compute the definite integral of the rate function from the lower to upper limit.

Question 6

The rate of change of profit is R′(t)=t2−4tR'(t)=t^2-4tR′(t)=t2−4t thousand dollars/month for 0≤t≤40 \le t \le 40≤t≤4; find net change in profit.

  1. −323-\frac{32}{3}−332​ (correct answer)
  2. −8-8−8
  3. 000
  4. 888
  5. 323\frac{32}{3}332​

Explanation: This problem tests the skill of exploring accumulations of change by using integration to find the net change in profit from a rate of change. The net change is the definite integral of R′(t)=t2−4tR'(t) = t^2 - 4tR′(t)=t2−4t from t=0 to t=4, accumulating the monthly profit variations. The antiderivative 13t3−2t2\frac{1}{3}t^3 - 2t^231​t3−2t2 evaluated gives (643−32)−0=−323\left( \frac{64}{3} - 32 \right) - 0 = -\frac{32}{3}(364​−32)−0=−332​ thousand dollars. This negative accumulation results from the rate being negative initially and not fully compensated later. A tempting distractor like 0 might stem from finding where the rate is zero instead of integrating, but it fails as the integral captures the overall deficit. In general, to find the net accumulation over an interval, compute the definite integral of the rate function from the lower to upper limit.

Question 7

A runner’s velocity is v(t)=sin⁡tv(t)=\sin tv(t)=sint m/s for 0≤t≤π0\le t\le\pi0≤t≤π; what is total distance traveled?​​

  1. 000 meters
  2. 222 meters (correct answer)
  3. π\piπ meters
  4. 111 meter
  5. −2-2−2 meters

Explanation: This problem asks for total distance, not displacement, which requires accumulation reasoning with absolute values. Since v(t) = sin(t) ≥ 0 for 0≤t≤π, the runner always moves forward, so distance = ∫₀ᵖ|sin(t)|dt = ∫₀ᵖsin(t)dt = [-cos(t)]₀ᵖ = -cos(π)-(-cos(0)) = -(-1)-(-1) = 2 meters. Choice A (0 meters) would be the displacement if velocity changed sign symmetrically, but sin(t) stays positive on [0,π]. For distance traveled, integrate the absolute value of velocity; for displacement, integrate velocity directly.

Question 8

For 0≤t≤60\le t\le60≤t≤6, a machine’s production rate is p(t)={2t,0≤t≤312−2t,3≤t≤6p(t)=\begin{cases}2t,&0\le t\le3\\12-2t,&3\le t\le6\end{cases}p(t)={2t,12−2t,​0≤t≤33≤t≤6​; how many items are produced?

  1. 181818 items (correct answer)
  2. 363636 items
  3. 999 items
  4. 121212 items
  5. 666 items

Explanation: This problem evaluates the skill of exploring accumulations of change by determining total production from a piecewise rate function. The production rate is 2t for 0 to 3 and 12 - 2t for 3 to 6. Total items produced is the integral over each piece, accumulating the rate over time. This gives [t²] from 0 to 3 = 9, plus [12t - t²] from 3 to 6 = 36 - 27 = 9, totaling 18 items. A tempting distractor is 36 items, from doubling the first integral without computing the second properly, but both pieces must be integrated separately. In general, to find the net accumulation of a quantity from its rate of change, compute the definite integral of the rate function over the specified interval.

Question 9

A battery’s charge changes at Q′(t)=−3t+1Q'(t)=-\frac{3}{t+1}Q′(t)=−t+13​ coulombs/min for 0≤t≤30\le t\le30≤t≤3; what is Q(3)−Q(0)Q(3)-Q(0)Q(3)−Q(0)?

  1. −3ln⁡4-3\ln 4−3ln4 coulombs (correct answer)
  2. 3ln⁡43\ln 43ln4 coulombs
  3. −94-\dfrac{9}{4}−49​ coulombs
  4. −9-9−9 coulombs
  5. ln⁡4\ln 4ln4 coulombs

Explanation: This problem assesses the skill of exploring accumulations of change by finding the net change in charge from a rate function. The rate is Q'(t) = -3/(t+1) coulombs per minute over 0 to 3 minutes. The net change Q(3) - Q(0) is the integral of Q'(t) from 0 to 3, accumulating the rate over time. This gives -3 [ln(t+1)] from 0 to 3 = -3(ln4 - ln1) = -3 ln4 coulombs. A tempting distractor is 3 ln4, from omitting the negative sign in the rate, but the sign indicates discharge. In general, to find the net accumulation of a quantity from its rate of change, compute the definite integral of the rate function over the specified interval.

Question 10

A function satisfies f′(x)=x1+x2f'(x)=\frac{x}{1+x^2}f′(x)=1+x2x​ for 0≤x≤20\le x\le20≤x≤2; what is f(2)−f(0)f(2)-f(0)f(2)−f(0)?

  1. ln⁡5\ln 5ln5
  2. 25\dfrac{2}{5}52​
  3. 12ln⁡5\dfrac{1}{2}\ln 521​ln5 (correct answer)
  4. ln⁡2\ln 2ln2
  5. 222

Explanation: This problem assesses the skill of exploring accumulations of change by finding the net change in a function from its derivative. The derivative is f′(x)=x1+x2f'(x) = \frac{x}{1 + x^2}f′(x)=1+x2x​ over 0 to 2. The change f(2)−f(0)f(2) - f(0)f(2)−f(0) is the integral of f′(x)f'(x)f′(x) from 0 to 2, accumulating the rate over the interval. Using substitution u=1+x2u = 1 + x^2u=1+x2, this yields 12[ln⁡u]\frac{1}{2} [\ln u]21​[lnu] from 1 to 5 = 12ln⁡5\frac{1}{2} \ln 521​ln5. A tempting distractor is ln⁡5\ln 5ln5, from forgetting the 1/2 factor in the substitution, but the du=2x dxdu = 2x \, dxdu=2xdx requires it. In general, to find the net accumulation of a quantity from its rate of change, compute the definite integral of the rate function over the specified interval.

Question 11

A cyclist’s velocity is v(t)=sin⁡tv(t)=\sin tv(t)=sint m/s for 0≤t≤π0\le t\le\pi0≤t≤π; what is the cyclist’s displacement?

  1. 000
  2. 111
  3. 222 (correct answer)
  4. π\piπ
  5. 2π2\pi2π

Explanation: This problem tests the skill of exploring accumulations of change by using integration to find the displacement from a velocity function. The cyclist's displacement is the definite integral of v(t)=sin⁡tv(t) = \sin tv(t)=sint from t=0t=0t=0 to t=πt=\pit=π, which accumulates the position changes over the interval. The antiderivative −cos⁡t-\cos t−cost evaluated at the limits gives −cos⁡(π)−(−cos⁡(0))=−(−1)−(−1)=1+1=2-\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2−cos(π)−(−cos(0))=−(−1)−(−1)=1+1=2 meters. This positive accumulation reflects the velocity being non-negative throughout the interval, leading to forward movement. A tempting distractor like 0 might come from confusing displacement with total distance or misapplying the fundamental theorem, but it fails as the integral clearly yields 2. In general, to find the net accumulation over an interval, compute the definite integral of the rate function from the lower to upper limit.

Question 12

A battery’s charge changes at rate Q′(t)=6t+1Q'(t)=\frac{6}{t+1}Q′(t)=t+16​ coulombs/hour for 0≤t≤20\le t\le 20≤t≤2; find net charge change.

  1. 121212 coulombs
  2. 6ln⁡36\ln 36ln3 coulombs (correct answer)
  3. 6ln⁡26\ln 26ln2 coulombs/hour
  4. 3ln⁡33\ln 33ln3 coulombs
  5. 6ln⁡36\ln 36ln3 coulombs/hour

Explanation: This problem requires accumulation reasoning to find net charge change from the charging rate. The net change equals ∫Q'(t)dt = ∫6/(t+1)dt from t = 0 to t = 2. Computing: ∫6/(t+1)dt = 6ln|t+1| evaluated from 0 to 2 gives 6ln(3) - 6ln(1) = 6ln(3) - 0 = 6ln(3) coulombs. Choice E incorrectly includes units of rate (coulombs/hour) rather than accumulated charge (coulombs). To find accumulated change from a rational function rate, integrate using logarithms and evaluate at the bounds.

Question 13

A reservoir’s outflow rate is q(t)=2+e−tq(t)=2+e^{-t}q(t)=2+e−t m3^33/hr for 0≤t≤20\le t\le20≤t≤2; how much water leaves in total?

  1. 2+e−22+e^{-2}2+e−2
  2. 4+e−24+e^{-2}4+e−2
  3. 4+(1−e−2)4+(1-e^{-2})4+(1−e−2) (correct answer)
  4. 2(1−e−2)2(1-e^{-2})2(1−e−2)
  5. 4−(1−e−2)4-(1-e^{-2})4−(1−e−2)

Explanation: This problem tests the skill of exploring accumulations of change by using integration to find the total water outflow from a rate. The total water leaving is the definite integral of q(t) = 2 + e^{-t} from t=0 to t=2, accumulating the volume over the interval. The antiderivative 2t - e^{-t} evaluated gives (4 - e^{-2}) - (0 - 1) = 4 - e^{-2} + 1 = 5 - e^{-2}, or equivalently 4 + (1 - e^{-2}) m³. This accumulation reflects the decreasing but positive outflow rate. A tempting distractor like 4 + e^{-2} might result from signing the exponential integral incorrectly, but it fails as the proper antiderivative yields the subtraction. In general, to find the net accumulation over an interval, compute the definite integral of the rate function from the lower to upper limit.

Question 14

A population changes at rate P′(t)=3sin⁡tP'(t)=3\sin tP′(t)=3sint thousand/year for 0≤t≤π0\le t\le\pi0≤t≤π; what is the total population change?

  1. 000 thousand
  2. 666 thousand (correct answer)
  3. 3π3\pi3π thousand
  4. −6-6−6 thousand
  5. 222 thousand

Explanation: This problem evaluates the skill of exploring accumulations of change by determining the total population change from a rate function. The rate of change is P'(t) = 3 sin t thousand per year over 0 to π years. The total change is the definite integral of P'(t) from 0 to π, accumulating the rate over the interval. This yields 3 [-cos t] from 0 to π = 3(1 + 1) = 6 thousand. A tempting distractor is 3π thousand, from multiplying the maximum rate by the interval length instead of integrating, but integration accounts for the varying rate. In general, to find the net accumulation of a quantity from its rate of change, compute the definite integral of the rate function over the specified interval.

Question 15

Water flows into a tank at rate r(t)=6−2tr(t)=6-2tr(t)=6−2t liters/min for 0≤t≤40\le t\le40≤t≤4; what is the net change in volume?

  1. 888 liters (correct answer)
  2. 444 liters
  3. 161616 liters
  4. 242424 liters
  5. 000 liters

Explanation: This problem assesses the skill of exploring accumulations of change by calculating the net change in volume from a given flow rate. The flow rate is r(t) = 6 - 2t liters per minute over the interval from t=0 to t=4. The net change in volume is the definite integral of r(t) from 0 to 4, which accumulates the rate over time. Computing this integral yields [6t - t²] from 0 to 4 = 24 - 16 = 8 liters. A tempting distractor is 24 liters, which comes from integrating only the constant term 6 and ignoring the -2t term, but the full rate function must be integrated. In general, to find the net accumulation of a quantity from its rate of change, compute the definite integral of the rate function over the specified interval.

Question 16

The rate of change of profit is P′(t)=10t+2−3P'(t)=\frac{10}{t+2}-3P′(t)=t+210​−3 thousand dollars/week for 0≤t≤40\le t\le 40≤t≤4; find net profit change.

  1. 10ln⁡3−1210\ln 3-1210ln3−12 thousand dollars (correct answer)
  2. 10ln⁡6−1210\ln 6-1210ln6−12 thousand dollars
  3. 10ln⁡310\ln 310ln3 thousand dollars
  4. 106−3\frac{10}{6}-3610​−3 thousand dollars/week
  5. 12−10ln⁡312-10\ln 312−10ln3 thousand dollars

Explanation: This problem uses accumulation reasoning to find net profit change from a rate function. The net change equals the integral of P'(t) = 10/(t+2) - 3 from t = 0 to t = 4. Computing ∫[10/(t+2) - 3]dt = 10ln|t+2| - 3t, and evaluating from 0 to 4 gives [10ln(6) - 12] - [10ln(2) - 0] = 10ln(6) - 12 - 10ln(2) = 10ln(6/2) - 12 = 10ln(3) - 12 thousand dollars. The answer 10ln(6) - 12 (choice B) forgets to subtract the initial value 10ln(2) when evaluating the definite integral. When computing definite integrals, always evaluate at both limits and subtract: F(b) - F(a).

Question 17

A lake’s depth changes at h′(t)=t−2h'(t)=t-2h′(t)=t−2 cm/day for 0≤t≤50\le t\le50≤t≤5; what is the net change in depth?

  1. 52\dfrac{5}{2}25​ cm (correct answer)
  2. 000 cm
  3. 555 cm
  4. −52-\dfrac{5}{2}−25​ cm
  5. 333 cm/day

Explanation: This problem tests the skill of exploring accumulations of change by calculating the net change in depth from a rate function. The rate is h'(t) = t - 2 cm per day over 0 to 5 days. The net change is the integral of h'(t) from 0 to 5, accumulating the rate over the interval. This yields [t²/2 - 2t] from 0 to 5 = 12.5 - 10 = 2.5 cm, or 5/2 cm. A tempting distractor is 0 cm, from evaluating h'(5) = 3 instead of integrating, but integration captures the full accumulation. In general, to find the net accumulation of a quantity from its rate of change, compute the definite integral of the rate function over the specified interval.

Question 18

An object’s temperature changes at rate T′(t)=−3cos⁡tT'(t)=-3\cos tT′(t)=−3cost °C/hr for 0≤t≤π20 \le t \le \frac{\pi}{2}0≤t≤2π​; find T(π2)−T(0)T(\frac{\pi}{2})-T(0)T(2π​)−T(0).

  1. −3-3−3 (correct answer)
  2. - rac{3\pi}{2}
  3. 000
  4. 333
  5. 3π2\frac{3\pi}{2}23π​

Explanation: This problem tests the skill of exploring accumulations of change by using integration to find the net temperature change from a rate. The change T(π/2)−T(0)T(\pi/2) - T(0)T(π/2)−T(0) is the definite integral of T′(t)=−3cos⁡tT'(t) = -3 \cos tT′(t)=−3cost from t=0t=0t=0 to t=π/2t=\pi/2t=π/2, accumulating temperature shifts. The antiderivative −3sin⁡t-3 \sin t−3sint evaluated gives −3(1−0)=−3-3(1 - 0) = -3−3(1−0)=−3 °C. This negative accumulation reflects the consistently negative rate causing cooling. A tempting distractor like 0 might arise from ignoring the negative sign or misintegrating, but it fails as the evaluation confirms the decrease. In general, to find the net accumulation over an interval, compute the definite integral of the rate function from the lower to upper limit.

Question 19

A population changes at rate P′(t)=100e−0.2tP'(t)=100e^{-0.2t}P′(t)=100e−0.2t people/day for 0≤t≤100\le t\le100≤t≤10; what is total increase?

  1. 100e−2100e^{-2}100e−2
  2. 100(1−e−2)100(1-e^{-2})100(1−e−2)
  3. 500(1−e−2)500(1-e^{-2})500(1−e−2) (correct answer)
  4. 500e−2500e^{-2}500e−2
  5. 1000(1−e−2)1000(1-e^{-2})1000(1−e−2)

Explanation: This problem tests the skill of exploring accumulations of change by using integration to find the total population increase from a growth rate. The total increase is the definite integral of P'(t) = 100e^{-0.2t} from t=0 to t=10, accumulating the daily additions over the interval. The antiderivative -500e^{-0.2t} evaluated gives -500e^{-2} - (-500e^0) = 500(1 - e^{-2}) people. This accumulation shows the decelerating growth rate leading to a finite total increase. A tempting distractor like 100(1 - e^{-2}) might result from forgetting the factor in the antiderivative, but it fails because the correct scalar is 500. In general, to find the net accumulation over an interval, compute the definite integral of the rate function from the lower to upper limit.

Question 20

A car’s velocity is v(t)=t−2v(t)=t-2v(t)=t−2 m/s for 0≤t≤50 \le t \le 50≤t≤5; what is net displacement over the interval?

  1. −10-10−10
  2. 2.52.52.5 (correct answer)
  3. 555
  4. 101010
  5. 151515

Explanation: This problem tests the skill of exploring accumulations of change by using integration to find the net displacement from a velocity function. The net displacement is the definite integral of v(t)=t−2v(t) = t - 2v(t)=t−2 from t=0t=0t=0 to t=5t=5t=5, accumulating position changes over the interval. The antiderivative 12t2−2t\frac{1}{2}t^2 - 2t21​t2−2t evaluated gives (252−10)−0=2.5(\frac{25}{2} - 10) - 0 = 2.5(225​−10)−0=2.5 meters. This positive net results despite initial backward movement, as later forward motion dominates. A tempting distractor like −10-10−10 might come from evaluating at a point instead of integrating, but it fails because the full integral accounts for the entire interval. In general, to find the net accumulation over an interval, compute the definite integral of the rate function from the lower to upper limit.