Calculus 2 Quiz: Accumulations Of Change

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Accumulations Of ChangeQuestion 1 of 20

Water flows into a cylindrical tank at a rate of $r(t) = 20 + 10\sin(\pi t/6)$ gallons per minute, where $t$ is time in minutes. If the tank initially contains 50 gallons, what is the average rate of change in the tank's volume over the first 12 minutes?

Exactly 20 gallons per minute since the sine term averages to zero

Approximately 18.33 gallons per minute accounting for initial conditions

Exactly 24.17 gallons per minute including the initial volume effect

Approximately 22.5 gallons per minute due to the oscillating inflow rate

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Calculus 2 Quiz

Calculus 2 Quiz: Accumulations Of Change

Practice Accumulations Of Change in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Accumulations Of Change, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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

Exactly 20 gallons per minute since the sine term averages to zero (correct answer)
Approximately 18.33 gallons per minute accounting for initial conditions
Exactly 24.17 gallons per minute including the initial volume effect
Approximately 22.5 gallons per minute due to the oscillating inflow rate

Explanation: The volume at time

t

V(t) = 50 + \int_0^t (20 + 10\sin(\pi s/6))ds = 50 + 20t - \frac{60}{\pi}\cos(\pi t/6) + \frac{60}{\pi}

. At

t = 12

V(12) = 50 + 240 - \frac{60}{\pi}\cos(2\pi) + \frac{60}{\pi} = 50 + 240 = 290

. The average rate of change is

\frac{V(12) - V(0)}{12 - 0} = \frac{290 - 50}{12} = 20

gallons per minute. The sine term integrates to zero over a complete period, leaving only the constant term.

Question 2

Let $f(x)$ be a differentiable function. The expression $\frac{1}{h} \int_c^{c+h} f'(t) dt$ represents the...

average value of $f(t)$ on the interval $[c, c+h]$ .
average rate of change of $f(t)$ on the interval $[c, c+h]$ . (correct answer)
instantaneous rate of change of $f(t)$ at $t=c$ .
net change of $f'(t)$ on the interval $[c, c+h]$ .

Explanation: By the Net Change Theorem,

\int_c^{c+h} f'(t) dt = f(c+h) - f(c)

. Therefore, the expression is

\frac{f(c+h) - f(c)}{h}

. This is the definition of the average rate of change of the function

f(t)

over the interval

[c, c+h]

. It is also the average value of the rate function

f'(t)

over that interval.

Question 3

The rate of temperature change in a chemical reaction follows $\frac{dT}{dt} = \frac{3t}{t^2 + 1}$ degrees per minute. If the reaction starts at 25°C when $t = 0$ , during which one-minute interval does the temperature increase most rapidly?

From $t = 0$ to $t = 1$ minute, when the rate function achieves its highest values (correct answer)
From $t = 0.5$ to $t = 1.5$ minutes, centered around the peak rate
From $t = 1$ to $t = 2$ minutes, accounting for sustained moderate rates
From $t = 2$ to $t = 3$ minutes, during the gradual decline phase

Explanation: To find when temperature increases most rapidly over one-minute intervals, we compare

\int_0^1 \frac{3t}{t^2 + 1}dt

\int_1^2 \frac{3t}{t^2 + 1}dt

, etc. For the first integral:

\int_0^1 \frac{3t}{t^2 + 1}dt = \frac{3}{2}\ln(t^2 + 1)|_0^1 = \frac{3}{2}\ln(2) ≈ 1.04°C

. Since

\frac{dT}{dt}

has its maximum at

t = 1

and decreases afterward, the interval

[0,1]

captures the period of highest rates and yields the largest temperature increase.

Question 4

A particle moves along the x-axis with a velocity given by $v(t) = 3t^2 - 12$ for $t \ge 0$ . What is the total distance traveled by the particle during the time interval $0 \le t \le 3$ ?

23 units (correct answer)
9 units
-9 units
7 units

Explanation: Total distance traveled requires integrating the absolute value of velocity,

\int_0^3 |v(t)| dt

. First, find where

v(t) = 0

3t^2 - 12 = 0 \Rightarrow t^2 = 4 \Rightarrow t = 2

(since

t \ge 0

). The velocity is negative on

[0, 2)

and positive on

(2, 3]

. The integral must be split:

\int_0^3 |3t^2 - 12| dt = \int_0^2 -(3t^2 - 12) dt + \int_2^3 (3t^2 - 12) dt = [12t - t^3]_0^2 + [t^3 - 12t]_2^3 = ((24 - 8) - 0) + ((27 - 36) - (8 - 24)) = 16 + (-9 - (-16)) = 16 + 7 = 23

Question 5

A population of insects increases at a rate of $g(t) = 100 + 6t$ insects per day and decreases due to natural causes at a rate of $d(t) = 40 + 2t$ insects per day. If the initial population at $t=0$ is 500, what is the insect population at $t=10$ days?

800
1300 (correct answer)
1100
600

Explanation: The net rate of change of the population is

P'(t) = g(t) - d(t) = (100 + 6t) - (40 + 2t) = 60 + 4t

. The change in population over 10 days is the integral of the net rate:

\Delta P = \int_0^{10} (60 + 4t) dt = [60t + 2t^2]_0^{10} = (600 + 2(100)) - 0 = 800

. The final population is the initial population plus the change:

P(10) = P(0) + \Delta P = 500 + 800 = 1300

Question 6

The rate of flow of oil through a pipeline is given by $R(t) = 100 \sin(\frac{\pi t}{12})$ barrels per hour, where $t$ is the number of hours from the start of the day (t=0). What is the total flow of oil during the first 6 hours of the day (from t=0 to t=6)?

$\frac{1200}{\pi}$ barrels (correct answer)
$\frac{600}{\pi}$ barrels
100 barrels
$1200\pi$ barrels

Explanation: Total flow is the definite integral of the rate of flow. We need to calculate

\int_0^6 100 \sin(\frac{\pi t}{12}) dt

. The antiderivative is

100 \cdot \frac{-12}{\pi} \cos(\frac{\pi t}{12}) = -\frac{1200}{\pi} \cos(\frac{\pi t}{12})

. Evaluating the definite integral:

[-\frac{1200}{\pi} \cos(\frac{\pi t}{12})]_0^6 = -\frac{1200}{\pi} [\cos(\frac{6\pi}{12}) - \cos(0)] = -\frac{1200}{\pi} [\cos(\frac{\pi}{2}) - 1] = -\frac{1200}{\pi} [0 - 1] = \frac{1200}{\pi}

barrels.

Question 7

A force of $F(x) = \frac{10}{(x+1)^2}$ Newtons is applied to an object to move it along the x-axis from $x=0$ to $x=4$ meters. The work done in moving an object from $x=a$ to $x=b$ with force $F(x)$ is $W = \int_a^b F(x) dx$ . It is known that 1 Joule = 1 Newton-meter. What is the work done?

8 Joules (correct answer)
-2 Joules
12 Joules
10 Joules

Explanation: The work done is the integral of the force function over the given interval.

W = \int_0^4 \frac{10}{(x+1)^2} dx

. Let

u = x+1

, so

du = dx

. The new bounds are

u(0)=1

and

u(4)=5

. The integral becomes

\int_1^5 \frac{10}{u^2} du = 10 \int_1^5 u^{-2} du = 10 [-u^{-1}]_1^5 = -10[\frac{1}{5} - \frac{1}{1}] = -10[-\frac{4}{5}] = 8

Joules.

Question 8

A particle's position at time $t=1$ is $p(1)=5$ . The velocity of the particle is given by $v(t) = \frac{1}{t^2}$ for $t > 0$ . What is the position of the particle at $t=e$ ?

$6 - e^{-1}$ (correct answer)
$5 - e^{-1}$
$4 + e^{-1}$
$1 - e^{-1}$

Explanation: The final position is the initial position plus the net change (displacement). The displacement from

t=1

t=e

\int_1^e v(t) dt = \int_1^e \frac{1}{t^2} dt = [-t^{-1}]_1^e = [-\frac{1}{t}]_1^e = -\frac{1}{e} - (-\frac{1}{1}) = 1 - \frac{1}{e}

. The position at

t=e

p(e) = p(1) + \text{displacement} = 5 + (1 - e^{-1}) = 6 - e^{-1}

Question 9

The acceleration of a particle is given by $a(t) = 6t - 2$ . The particle's velocity at $t=1$ is $v(1)=4$ , and its position at $t=1$ is $s(1)=10$ . What is the position of the particle at $t=2$ ?

7
11
17 (correct answer)
15

Explanation: When you encounter a problem involving acceleration, velocity, and position, you're working with the fundamental relationship between these three quantities through differentiation and integration. Since acceleration is the derivative of velocity, and velocity is the derivative of position, you'll need to integrate twice to go from acceleration to position. Starting with

a(t) = 6t - 2

, integrate to find velocity:

v(t) = \int (6t - 2) dt = 3t^2 - 2t + C_1

. Using the initial condition

v(1) = 4

4 = 3(1)^2 - 2(1) + C_1 = 1 + C_1

, so

C_1 = 3

. Therefore,

v(t) = 3t^2 - 2t + 3

. Next, integrate velocity to find position:

s(t) = \int (3t^2 - 2t + 3) dt = t^3 - t^2 + 3t + C_2

. Using

s(1) = 10

10 = (1)^3 - (1)^2 + 3(1) + C_2 = 3 + C_2

, so

C_2 = 7

. Thus,

s(t) = t^3 - t^2 + 3t + 7

. At

t = 2

s(2) = (2)^3 - (2)^2 + 3(2) + 7 = 8 - 4 + 6 + 7 = 17

. The answer is C. Choice A (7) represents just the constant term from the position function. Choice B (11) is what you'd get if you made an error with one of the integration constants. Choice D (15) results from miscalculating the final substitution, likely

8 + 6 + 1

instead of the correct arithmetic. Always work systematically through both integrations and carefully apply initial conditions to find each constant—this prevents computational errors in multi-step integration problems.

Question 10

A factory's production rate is $p(t) = 120 - 20\sqrt{t}$ units per hour, where $t$ is in hours. The factory runs for a 9-hour shift. What is the average number of units produced per hour during the shift?

90 units/hour
60 units/hour
720 units/hour
80 units/hour (correct answer)

Explanation: When you encounter a problem asking for the "average rate" over an interval given a rate function, you need to use the formula for average value of a function:

\text{Average} = \frac{1}{b-a}\int_a^b f(t) \, dt

Here, you need to find the average production rate over the 9-hour shift, so you'll calculate:

\text{Average rate} = \frac{1}{9-0}\int_0^9 (120 - 20\sqrt{t}) \, dt

First, evaluate the integral:

\int_0^9 (120 - 20\sqrt{t}) \, dt = \int_0^9 (120 - 20t^{1/2}) \, dt = \left[120t - 20 \cdot \frac{t^{3/2}}{3/2}\right]_0^9

This simplifies to:

\left[120t - \frac{40t^{3/2}}{3}\right]_0^9 = 120(9) - \frac{40(9)^{3/2}}{3} = 1080 - \frac{40(27)}{3} = 1080 - 360 = 720

Therefore, the average rate is

\frac{720}{9} = 80

units per hour. Choice A (90 units/hour) likely comes from evaluating the rate function at a single point, like

t = 0

, rather than finding the true average. Choice B (60 units/hour) might result from incorrectly evaluating the rate at the endpoint (

t = 9

) instead of averaging over the interval. Choice C (720 units/hour) is the total production during the shift, not the average rate—this confuses total output with rate. Remember: when finding average rates, always use the average value formula. Don't confuse total production with average production rate, and avoid evaluating the function at just one point.

Question 11

The velocity of a particle is $v(t) = 2^t \ln(2)$ meters per second. If the particle is at the origin at $t=0$ , what is its position at $t=3$ ?

9 meters
8 meters
$8\ln(2)$ meters
7 meters (correct answer)

Explanation: When you encounter a velocity function and need to find position, you're working with the fundamental relationship between velocity and position: velocity is the derivative of position with respect to time. To find position from velocity, you need to integrate. Given

v(t) = 2^t \ln(2)

, you need to find the position function

s(t)

by integrating the velocity function. The antiderivative of

2^t \ln(2)

2^t + C

, where

C

is the constant of integration. You can verify this because the derivative of

2^t

is indeed

2^t \ln(2)

. To find

C

, use the initial condition that the particle is at the origin when

t = 0

. Setting

s(0) = 0

s(0) = 2^0 + C = 1 + C = 0

, so

C = -1

. Therefore,

s(t) = 2^t - 1

. At

t = 3

s(3) = 2^3 - 1 = 8 - 1 = 7

meters, confirming answer D. Answer A (9 meters) might result from calculating

2^3 + 1

instead of

2^3 - 1

, forgetting the correct sign of the constant. Answer B (8 meters) comes from ignoring the initial condition entirely and using

s(t) = 2^t

. Answer C (

8\ln(2)

meters) appears if you incorrectly integrate

2^t \ln(2)

\frac{(2^t)^2 \ln(2)}{2} = 4^t \ln(2)/2

, then evaluate at

t = 3

. Remember: always use initial conditions to determine the constant of integration, and double-check your antiderivative by differentiating it back to the original function.

Question 12

The current flowing into a capacitor is given by $I(t) = 3 \cos(t)$ amperes. The charge $Q(t)$ on the capacitor is related to the current by $I(t) = Q'(t)$ . If the initial charge at $t=0$ is 2 Coulombs, what is the charge at $t=\pi/2$ seconds?

3 Coulombs
5 Coulombs (correct answer)
2 Coulombs
-1 Coulombs

Explanation: The change in charge is the integral of the current. The charge at time

t=\pi/2

is given by

Q(\pi/2) = Q(0) + \int_0^{\pi/2} I(t) dt

. We have

Q(0) = 2

. The integral is

\int_0^{\pi/2} 3 \cos(t) dt = [3 \sin(t)]_0^{\pi/2} = 3 \sin(\pi/2) - 3 \sin(0) = 3(1) - 3(0) = 3

. So, the final charge is

Q(\pi/2) = 2 + 3 = 5

Coulombs.

Question 13

A vat initially contains 200 kg of a salt solution. A process removes salt at a rate of $R(t) = 3t^2$ kg/hr for the first 2 hours. After 2 hours, a different process begins removing salt at a rate of $R(t) = 12$ kg/hr. How much salt remains in the vat after 4 hours?

152 kg
176 kg
160 kg
168 kg (correct answer)

Explanation: This is a piecewise rate problem where you need to calculate cumulative change over different time intervals with different removal rates. Since salt is being removed at different rates during different time periods, you must integrate the rate function over each interval separately. For the first 2 hours,

R(t) = 3t^2

kg/hr, so the salt removed is

\int_0^2 3t^2 \, dt = [t^3]_0^2 = 8

kg. From hours 2 to 4,

R(t) = 12

kg/hr (constant rate), so the salt removed is

\int_2^4 12 \, dt = 12(4-2) = 24

kg. Total salt removed = 8 + 24 = 32 kg. Starting with 200 kg, the remaining salt is 200 - 32 = 168 kg. Answer A (152 kg) represents removing 48 kg total, which you'd get if you incorrectly calculated the first integral as

[3t^3/3]_0^2 = 24

instead of

[t^3]_0^2 = 8

. Answer B (176 kg) corresponds to removing only 24 kg total, suggesting you ignored the variable rate period entirely and only calculated the constant rate removal. Answer C (160 kg) represents removing 40 kg, which occurs if you miscalculate the second period as 32 kg instead of 24 kg. When solving rate problems with piecewise functions, always break the problem into separate time intervals and integrate each piece over its specific domain. Don't forget that the total change is the sum of changes from each interval.

Question 14

Snow is falling at a rate of $s(t) = 2e^{-t/3}$ inches per hour. Concurrently, snow is melting at a rate of $m(t) = \frac{1}{2}$ inch per hour. If there are 3 inches of snow on the ground at $t=0$ , which integral represents the depth of snow on the ground at $t=6$ hours?

$\int_0^6 (2e^{-t/3} - \frac{1}{2}) dt$
$3 + \int_0^6 (2e^{-t/3} - \frac{1}{2}) dt$ (correct answer)
$3 + \int_0^6 (2e^{-t/3} + \frac{1}{2}) dt$
$\int_0^6 2e^{-t/3} dt - 3$

Explanation: The amount of snow at a given time is the initial amount plus the net accumulation. The net rate of change of snow depth is the rate of snowfall minus the rate of melting,

s(t) - m(t)

. The total accumulated change over 6 hours is the integral of this net rate from 0 to 6. To find the final amount, this net change must be added to the initial amount of 3 inches. Thus, the expression is

3 + \int_0^6 (s(t) - m(t)) dt = 3 + \int_0^6 (2e^{-t/3} - \frac{1}{2}) dt

Question 15

Water flows into a conical tank at a rate of $r(t) = t^2$ cubic feet per minute. The tank is initially empty. How much more water flows into the tank during the second minute (from $t=1$ to $t=2$ ) than during the first minute (from $t=0$ to $t=1$ )?

$\frac{7}{3}$ cubic feet
$\frac{8}{3}$ cubic feet
2 cubic feet (correct answer)
3 cubic feet

Explanation: Let V1 be the volume that flows in during the first minute and V2 be the volume during the second minute.

V1 = \int_0^1 t^2 dt = [\frac{t^3}{3}]_0^1 = \frac{1}{3}

V2 = \int_1^2 t^2 dt = [\frac{t^3}{3}]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}

. The question asks for the difference,

V2 - V1 = \frac{7}{3} - \frac{1}{3} = \frac{6}{3} = 2

cubic feet. The calculation

\frac{7}{3}

is a distractor representing only the amount in the second minute.

Question 16

An object is heated and cooled such that its rate of temperature change is $f(t) = 4\cos(t)$ degrees Celsius per hour for $0 \le t \le \pi$ . What is the total magnitude of temperature change over this interval? Note: This is not asking for net change.

8 degrees (correct answer)
0 degrees
4 degrees
$4\pi$ degrees

Explanation: The total magnitude of change is the integral of the absolute value of the rate,

\int_0^\pi |4\cos(t)| dt

. On the interval

[0, \pi/2]

\cos(t)

is positive. On

[\pi/2, \pi]

\cos(t)

is negative. So we split the integral:

\int_0^{\pi/2} 4\cos(t) dt + \int_{\pi/2}^\pi -4\cos(t) dt = [4\sin(t)]_0^{\pi/2} + [-4\sin(t)]_{\pi/2}^\pi = (4\sin(\pi/2) - 4\sin(0)) + (-4\sin(\pi) - (-4\sin(\pi/2))) = (4-0) + (0 - (-4)) = 4+4=8

degrees.

Question 17

The population density $x$ miles from the center of a city is given by $D(x) = 10000e^{-0.5x}$ people per square mile. Which integral represents the total population living within a 5-mile radius of the city center?

$\int_0^5 10000e^{-0.5x} dx$
$\int_0^5 2\pi x (10000e^{-0.5x}) dx$ (correct answer)
$2\pi \int_0^5 10000e^{-0.5x} dx$
$\pi \int_0^5 (10000e^{-0.5x})^2 dx$

Explanation: To find the total population, we must integrate the population density over the area. We can think of the area as a series of thin concentric rings. The area of a ring at radius

x

with a small thickness

dx

is approximately its circumference times its thickness, which is

dA = 2\pi x dx

. The number of people in this ring is the density at that radius times the area of the ring:

D(x) \cdot dA = 10000e^{-0.5x} (2\pi x dx)

. To find the total population within a 5-mile radius, we accumulate (integrate) this quantity from the center (

x=0

) to the edge (

x=5

). This gives the integral

\int_0^5 2\pi x (10000e^{-0.5x}) dx

Question 18

A reservoir's water level changes according to the net inflow rate $R(t) = 4\cos(\frac{\pi t}{12}) - 1$ feet per hour, where $t$ is hours after midnight. If the water level is 15 feet at midnight, at what time during the first 24 hours is the water level at its minimum?

At 6:00 AM, when the cosine function equals zero and rate becomes most negative
At 12:00 PM, corresponding to the minimum of the rate function over one period
At 6:00 PM, when the accumulated negative change reaches its maximum extent (correct answer)
At 12:00 AM (end of day), completing the full cycle of water level changes

Explanation: Water level:

h(t) = 15 + \int_0^t (4\cos(\frac{\pi s}{12}) - 1)ds = 15 + \frac{48}{\pi}\sin(\frac{\pi t}{12}) - t

. Setting

\frac{dh}{dt} = 4\cos(\frac{\pi t}{12}) - 1 = 0

gives

\cos(\frac{\pi t}{12}) = \frac{1}{4}

. This yields

t = \frac{12}{\pi}\arccos(\frac{1}{4}) ≈ 5.1

hours and

t ≈ 18.9

hours. Since

\frac{d^2h}{dt^2} = -\frac{\pi}{3}\sin(\frac{\pi t}{12}) > 0

t ≈ 18.9

, this corresponds to a minimum. Therefore, the minimum occurs at approximately 6:54 PM, closest to 6:00 PM.

Question 19

Oil leaks from a tank at a rate of $L(t) = 5e^{-0.2t}$ gallons per minute, where $t$ is minutes since the leak began. Simultaneously, oil is pumped back into the tank at a constant rate of 2 gallons per minute starting at $t = 5$ minutes. What is the net change in tank volume from $t = 0$ to $t = 15$ minutes?

Approximately $-15.2$ gallons, representing net loss due to early uncompensated leakage
Approximately $-16.8$ gallons, accounting for exponential decay and delayed pump activation (correct answer)
Approximately $-12.3$ gallons, considering the partial compensation from pumping operations
Approximately $-21.1$ gallons, reflecting maximum loss scenario with pump inefficiency

Explanation: Net rate =

-5e^{-0.2t}

for

0 ≤ t < 5

and

-5e^{-0.2t} + 2

for

5 ≤ t ≤ 15

. Net change =

\int_0^5 (-5e^{-0.2t})dt + \int_5^{15} (-5e^{-0.2t} + 2)dt

. The first integral:

[\frac{5}{0.2}e^{-0.2t}]_0^5 = 25(e^{-1} - 1) ≈ -15.8

. The second integral:

[\frac{5}{0.2}e^{-0.2t} + 2t]_5^{15} = 25(e^{-3} - e^{-1}) + 20 ≈ -1.0

. Total net change ≈

-15.8 + (-1.0) = -16.8

gallons.

Question 20

Let $G(x) = \int_2^x f(t) dt$ . If $f(t)$ is always positive, which of the following statements about $G(x)$ must be true?

$G(x)$ is always positive.
$G(x)$ is always increasing. (correct answer)
$G(x)$ is always concave up.
$G(x)$ has a local minimum at $x=2$ .

Explanation: By the Second Fundamental Theorem of Calculus,

G'(x) = f(x)

. Since we are given that

f(t)

is always positive, it means that

G'(x)

is always positive. A function whose derivative is always positive is always increasing.

G(x)

is not always positive; for example, if

x<2

G(x) = \int_2^x f(t) dt = -\int_x^2 f(t) dt

would be negative. Concavity depends on

G''(x) = f'(x)

, about which we have no information.

G(x)

has a critical point at

x=2

since

G'(2)=f(2)

, but we don't know if it's a minimum; we only know the function is increasing through that point.