Calculus 1 Quiz: Accumulations Of Change

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Accumulations Of ChangeQuestion 1 of 20

Let $g(x) = \int_1^x \frac{t^2}{t^2+1} dt$ . For what values of $x$ is the graph of $g(x)$ concave up?

For all real

x

For

x > 0

For

x < 0

For

x > 1

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Calculus 1 Quiz

Calculus 1 Quiz: Accumulations Of Change

Practice Accumulations Of Change in Calculus 1 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 1.

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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.

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Question 1

Let $g(x) = \int_1^x \frac{t^2}{t^2+1} dt$ . For what values of $x$ is the graph of $g(x)$ concave up?

For all real $x$
For $x > 0$ (correct answer)
For $x < 0$
For $x > 1$

Explanation: The concavity of

g(x)

is determined by the sign of its second derivative,

g''(x)

. By the Fundamental Theorem of Calculus, Part 1,

g'(x) = \frac{x^2}{x^2+1}

. Using the quotient rule to find the second derivative:

g''(x) = \frac{(2x)(x^2+1) - x^2(2x)}{(x^2+1)^2} = \frac{2x}{(x^2+1)^2}

. The graph of

g(x)

is concave up when

g''(x) > 0

. Since the denominator is always positive, we need the numerator

2x > 0

, which means

x > 0

Question 2

A function $f(x)$ has derivative $f'(x) = \cos(\pi x)$ . If $f(1) = 5$ , what is the value of $f(2)$ ?

1
4
5 (correct answer)
6

Explanation: According to the Fundamental Theorem of Calculus, the net change of a function is the integral of its derivative:

f(2) - f(1) = \int_1^2 f'(x) dx

. We can find

f(2)

using

f(2) = f(1) + \int_1^2 \cos(\pi x) dx

. The integral is

\int_1^2 \cos(\pi x) dx = [\frac{1}{\pi}\sin(\pi x)]_1^2 = \frac{1}{\pi}\sin(2\pi) - \frac{1}{\pi}\sin(\pi) = \frac{1}{\pi}(0) - \frac{1}{\pi}(0) = 0

. Therefore,

f(2) = f(1) + 0 = 5 + 0 = 5

Question 3

If $F(x) = \int_{\sin(x)}^{1} \sqrt{1+t^2} dt$ , what is $F'(x)$ ?

$\cos(x)\sqrt{1+\sin^2(x)}$
$-\cos(x)\sqrt{1+\sin^2(x)}$ (correct answer)
$\sqrt{1+\sin^2(x)}$
$-\sqrt{1+\sin^2(x)}$

Explanation: First, flip the limits of integration to get the variable in the upper limit:

F(x) = - \int_{1}^{\sin(x)} \sqrt{1+t^2} dt

. By the Fundamental Theorem of Calculus, Part 1, combined with the Chain Rule, we have

F'(x) = -\left(\sqrt{1+(\sin(x))^2} \cdot \frac{d}{dx}(\sin(x))\right)

. This simplifies to

F'(x) = -\sqrt{1+\sin^2(x)} \cdot \cos(x)

Question 4

The amount of snowfall is recorded during a storm. The rate of snowfall is $s(t) = 4t$ inches per hour for $t \ge 0$ .

If there were 3 inches of snow on the ground before the storm, at what time $t$ does the total depth of snow reach 19 inches?

$t = 2\sqrt{2}$ hours (correct answer)
$t = 4$ hours
$t = \sqrt{19/2}$ hours
$t = 19/4$ hours

Explanation: The total depth of snow at time

T

is the initial depth plus the accumulated snowfall:

D(T) = 3 + \int_0^T s(t) dt

. We want to find

T

when

D(T)=19

. So,

19 = 3 + \int_0^T 4t dt

. This means the accumulated snowfall must be

16 = \int_0^T 4t dt

. Evaluating the integral gives

\int_0^T 4t dt = [2t^2]_0^T = 2T^2

. Setting this equal to 16, we have

2T^2 = 16

, so

T^2 = 8

, and

T = \sqrt{8} = 2\sqrt{2}

hours.

Question 5

Over a 10-hour period, the rate of revenue a company earns is $R(t) = 20 - 2t$ dollars per hour, and the rate at which it incurs costs is $C(t) = 2 + 0.5t$ dollars per hour. What is the total profit accumulated during this 10-hour period, from $t=0$ to $t=10$ ?

$55 (correct answer)
$75
$100
$125

Explanation: Profit is revenue minus cost. The rate of change of profit is

P'(t) = R(t) - C(t)

. So,

P'(t) = (20 - 2t) - (2 + 0.5t) = 18 - 2.5t

. The total accumulated profit over the 10-hour period is the definite integral of the profit rate from

t=0

t=10

\int_0^{10} (18 - 2.5t) dt

. The antiderivative is

18t - \frac{2.5t^2}{2} = 18t - 1.25t^2

. Evaluating the definite integral:

[18t - 1.25t^2]_0^{10} = (18(10) - 1.25(10^2)) - (0) = 180 - 1.25(100) = 180 - 125 = 55

. The total profit is $55.

Question 6

Snow begins to fall at midnight ( $t=0$ ). From $t=0$ to $t=6$ hours, the rate of snowfall is a constant 0.5 inches per hour. From $t=6$ to $t=10$ , the rate increases linearly from 0.5 inches per hour to 2.5 inches per hour. What is the total accumulation of snow at $t=10$ hours?

3.0 inches
6.0 inches
9.0 inches (correct answer)
12.0 inches

Explanation: The total accumulation is the integral of the rate of snowfall. We can calculate this by finding the area under the rate graph. For the first interval,

0 \le t \le 6

, the accumulation is the area of a rectangle:

6 \text{ hours} \times 0.5 \text{ in/hr} = 3

inches. For the second interval,

6 \le t \le 10

, the rate graph is a trapezoid with base

10-6=4

hours and heights 0.5 and 2.5 in/hr. The area is

\frac{1}{2}(0.5+2.5)(4) = \frac{1}{2}(3)(4) = 6

inches. The total accumulation is the sum of the amounts from both intervals:

3 + 6 = 9

inches.

Question 7

A company's bank account balance is affected by deposits and withdrawals. The rate of deposits is given by $D(t) = 3t^2 - 12t + 15$ and the rate of withdrawals is given by $W(t) = 15 - t$ , both in thousands of dollars per month for $0 \le t \le 6$ . At what time $t$ on the interval $[0, 6]$ is the amount of money in the account at an absolute minimum?

$t = 0$
$t = 2$
$t = 11/3$ (correct answer)
$t = 6$

Explanation: To find the minimum amount, we need to find where the rate of change of the amount is zero. Let

M(t)

be the amount of money. The net rate of change is

M'(t) = D(t) - W(t) = (3t^2 - 12t + 15) - (15 - t) = 3t^2 - 11t

. We set

M'(t)=0

to find critical points:

t(3t - 11) = 0

, which gives

t=0

and

t=11/3

. To find the absolute minimum on

[0, 6]

, we check the critical points in the interval and the endpoints.

M'(t)

is negative for

t \in (0, 11/3)

and positive for

t \in (11/3, 6)

. This means the amount

M(t)

decreases until

t=11/3

and then increases. Therefore, the absolute minimum occurs at

t=11/3

Question 8

A population of bacteria in a lab culture grows at a rate modeled by $r(t) = 100e^{0.1t}$ bacteria per hour. At time $t=2$ hours, the population is 5000. Which of the following expressions represents the population $P(t)$ for any time $t > 2$ ?

$5000 + \int_0^t 100e^{0.1x} dx$
$\int_2^t 100e^{0.1x} dx$
$5000 + 100e^{0.1t}$
$5000 + \int_2^t 100e^{0.1x} dx$ (correct answer)

Explanation: The population

P(t)

at a time

t

is the initial population at a known time, say

t=a

, plus the accumulated growth from

a

t

. The accumulated growth is the integral of the rate function. Here, the known population is 5000 at

t=2

. So,

P(t) = P(2) + \int_2^t r(x) dx

. Substituting the given values, we get

P(t) = 5000 + \int_2^t 100e^{0.1x} dx

. Note that a dummy variable like

x

must be used for integration when the upper limit is

t

Question 9

The rate of change of air temperature on a certain day is modeled by $T'(t) = -2 \cos(\frac{\pi t}{12})$ degrees Celsius per hour, where $t$ is the number of hours after midnight. At 6 AM ( $t=6$ ), the temperature is $15^\circ$ C. What is the temperature at 6 PM ( $t=18$ )?

$(15 - \frac{48}{\pi}) ^\circ C$
$(15 + \frac{48}{\pi}) ^\circ C$ (correct answer)
$15^\circ C$
$\frac{48}{\pi} ^\circ C$

Explanation: When you encounter a rate of change problem like this, you're dealing with the Fundamental Theorem of Calculus. Since you're given

T'(t)

and need to find the actual temperature at a different time, you must integrate the rate function and use the initial condition. To find the temperature at

t = 18

given the temperature at

t = 6

, you need:

T(18) = T(6) + \int_6^{18} T'(t) \, dt

First, integrate

T'(t) = -2\cos(\frac{\pi t}{12})

\int T'(t) \, dt = -2 \cdot \frac{12}{\pi} \sin(\frac{\pi t}{12}) = -\frac{24}{\pi}\sin(\frac{\pi t}{12})

Now evaluate the definite integral:

\int_6^{18} T'(t) \, dt = \left[-\frac{24}{\pi}\sin(\frac{\pi t}{12})\right]_6^{18}

t = 18

\sin(\frac{18\pi}{12}) = \sin(\frac{3\pi}{2}) = -1

t = 6

\sin(\frac{6\pi}{12}) = \sin(\frac{\pi}{2}) = 1

So:

-\frac{24}{\pi}(-1) - (-\frac{24}{\pi})(1) = \frac{24}{\pi} + \frac{24}{\pi} = \frac{48}{\pi}

Therefore:

T(18) = 15 + \frac{48}{\pi}

, which is answer B. Answer A gives the negative change, suggesting integration error or wrong limits. Answer C implies zero net change, ignoring the actual integration. Answer D omits the initial temperature entirely, giving only the change in temperature. Study tip: Always remember that to go from rate to actual value, you integrate and add your initial condition. The definite integral gives you the change in temperature, not the final temperature itself.

Question 10

A particle moves along a line with velocity $v(t) = t^2 - 8t + 12$ for $t \ge 0$ . What is the total distance traveled by the particle during the time interval $0 \le t \le 6$ ?

0
$32/3$
36
$64/3$ (correct answer)

Explanation: Total distance traveled is the integral of the speed, which is the absolute value of velocity:

\int_0^6 |v(t)| dt

. First, find where

v(t) = 0

t^2 - 8t + 12 = (t-2)(t-6) = 0

, so

t=2

and

t=6

. On the interval

[0, 2]

v(t) > 0

. On

[2, 6]

v(t) < 0

. The integral must be split:

\int_0^2 (t^2 - 8t + 12) dt + \int_2^6 -(t^2 - 8t + 12) dt

. The antiderivative is

F(t) = \frac{t^3}{3} - 4t^2 + 12t

\int_0^2 v(t) dt = F(2) - F(0) = (\frac{8}{3} - 16 + 24) - 0 = 32/3

\int_2^6 v(t) dt = F(6) - F(2) = (\frac{216}{3} - 144 + 72) - 32/3 = 0 - 32/3 = -32/3

. The total distance is

|32/3| + |-32/3| = 64/3

Question 11

A quantity $Q(t)$ changes over time, with its initial value being $Q(0)$ . The rate of change of the quantity is given by the function $R(t)$ . If it is known that $\int_0^{10} R(t) dt < 0$ , which of the following statements must be true?

$Q(10)$ must be negative.
$Q(10) < Q(0)$ (correct answer)
$R(t) < 0$ for all $t$ in the interval $[0, 10]$ .
The function $Q(t)$ is decreasing on the interval $[0, 10]$ .

Explanation: The Fundamental Theorem of Calculus states that the net change in a quantity

Q(t)

from

t=a

t=b

is the definite integral of its rate of change,

R(t)

. Thus,

Q(10) - Q(0) = \int_0^{10} R(t) dt

. Given that

\int_0^{10} R(t) dt < 0

, it follows directly that

Q(10) - Q(0) < 0

, which simplifies to

Q(10) < Q(0)

. Choice A is incorrect because

Q(10)

could be positive if

Q(0)

is large enough. Choice C is a sufficient but not necessary condition; the integral can be negative even if

R(t)

is positive for some parts of the interval. Choice D is incorrect because

Q(t)

is decreasing only when its derivative,

R(t)

, is negative. We only know the net accumulation is negative, not that the rate is always negative.

Question 12

The rate at which water flows into a reservoir is modeled by the function $r(t) = 1000 e^{-t^2/50}$ gallons per hour, where $t$ is the number of hours since monitoring began. At $t=0$ , the reservoir contained 50,000 gallons of water. Which of the following expressions represents the total volume of water in the reservoir, in gallons, after 10 hours?

$50000 + \int_0^{10} 1000 e^{-t^2/50} dt$ (correct answer)
$\int_0^{10} 1000 e^{-t^2/50} dt$
$50000 + r(10)$
$r(10) - r(0)$

Explanation: The total amount of a quantity at a future time is its initial amount plus the net change. The net change is the definite integral of the rate of change. The initial volume is 50,000 gallons. The amount of water added over the first 10 hours is given by the integral of the rate function

r(t)

from

t=0

t=10

. Therefore, the total volume

V(10)

is given by the initial volume plus the accumulated volume:

V(10) = 50000 + \int_0^{10} 1000 e^{-t^2/50} dt

Question 13

The number of people entering a concert venue is modeled by the function $E(t) = -t^2 + 20t + 100$ people per hour. The number of people leaving the venue is modeled by $L(t) = 30t + 20$ people per hour. The venue opens at $t=0$ with 50 people already inside. Which of the following expressions gives the number of people in the venue at $t=4$ hours?

$50 + \int_0^4 (-t^2 - 10t + 80) dt$ (correct answer)
$\int_0^4 (-t^2 - 10t + 80) dt$
$50 + \int_0^4 (-t^2 + 50t + 120) dt$
$50 + (-4^2 - 10(4) + 80)$

Explanation: The total number of people is the initial amount plus the net accumulation. The net rate of change of people in the venue is the rate of people entering minus the rate of people leaving:

N(t) = E(t) - L(t) = (-t^2 + 20t + 100) - (30t + 20) = -t^2 - 10t + 80

. The total change in the number of people from

t=0

t=4

is the integral of this net rate,

\int_0^4 N(t) dt

. The number of people at

t=4

is the initial number (50) plus this net change. So, the expression is

50 + \int_0^4 (-t^2 - 10t + 80) dt

Question 14

A particle moves along the x-axis with velocity given by $v(t) = 3t^2 - 12t + 9$ for $t \ge 0$ . At time $t=0$ , the particle is at position $x(0) = -4$ . What is the position of the particle the first time it changes direction?

$4$
$0$ (correct answer)
$-4$
$8$

Explanation: The particle changes direction when its velocity

v(t)

changes sign. First, find when

v(t) = 0

3t^2 - 12t + 9 = 0 \implies 3(t^2 - 4t + 3) = 0 \implies 3(t-1)(t-3) = 0

. The first time the velocity is zero is at

t=1

. Since

v(t)

is a parabola opening upwards, the velocity changes from positive to negative at

t=1

, so the particle changes direction. The position of the particle at time

t=1

is given by the initial position plus the accumulated displacement:

x(1) = x(0) + \int_0^1 v(t) dt

. We have

x(0)=-4

. The integral is

\int_0^1 (3t^2 - 12t + 9) dt = [t^3 - 6t^2 + 9t]_0^1 = (1^3 - 6(1)^2 + 9(1)) - 0 = 1 - 6 + 9 = 4

. Therefore, the position is

x(1) = -4 + 4 = 0

Question 15

Sand is being added to a pile at a rate of $A(t) = 8\cos(0.2t)$ cubic feet per hour. At the same time, sand is being removed by erosion at a rate of $E(t) = 2t$ cubic feet per hour. The pile contains 50 cubic feet of sand at time $t=0$ . Which expression gives the total amount of sand in the pile at time $t=4$ hours?

$\int_0^4 (8\cos(0.2t) - 2t) dt$
$50 + \int_0^4 (8\cos(0.2t) - 2t) dt$ (correct answer)
$50 + \int_0^4 (8\cos(0.2t) + 2t) dt$
$50 + 4(8\cos(0.8) - 8)$

Explanation: The net rate of change of the amount of sand is the rate in minus the rate out, which is

A(t) - E(t) = 8\cos(0.2t) - 2t

. To find the total amount of sand at

t=4

, we need to add the net change in sand over the interval

[0, 4]

to the initial amount. The net change is the integral of the net rate of change from

t=0

t=4

, which is

\int_0^4 (8\cos(0.2t) - 2t) dt

. The initial amount is 50 cubic feet. Therefore, the total amount at

t=4

50 + \int_0^4 (8\cos(0.2t) - 2t) dt

Question 16

Let $F(x) = \int_2^{e^x} \frac{\ln t}{t} dt$ . For which value of $x$ does $F(x)$ have a local extremum?

$x = \ln 2$
$x = 1$
$x = 0$ (correct answer)
$x = e$

Explanation: To find local extrema of

F(x)

, we need to find its critical points by taking the derivative

F'(x)

and setting it to zero. We use the Fundamental Theorem of Calculus, Part 1, combined with the Chain Rule. Let

f(t) = \frac{\ln t}{t}

. Then

F(x) = \int_2^{u(x)} f(t) dt

where

u(x) = e^x

. The derivative is

F'(x) = f(u(x)) \cdot u'(x) = f(e^x) \cdot \frac{d}{dx}(e^x)

F'(x) = \frac{\ln(e^x)}{e^x} \cdot e^x = \frac{x}{e^x} \cdot e^x = x

. To find critical points, we set

F'(x) = 0

, which gives

x=0

. Since

F'(x)

changes sign from negative to positive at

x=0

, this corresponds to a local minimum, which is a type of local extremum.

Question 17

The rate at which a machine produces widgets is given by a differentiable function $R(t)$ , where $R$ is measured in widgets per minute and $t$ is measured in minutes from the start of a shift. What is the best interpretation of the expression $\frac{1}{60} \int_0^{60} R(t) dt$ ?

The total number of widgets produced during the first hour of the shift.
The rate of widget production at the end of the first hour of the shift.
The average rate of widget production, in widgets per minute, during the first hour of the shift. (correct answer)
The change in the rate of widget production between the start of the shift and the end of the first hour.

Explanation: The expression

\frac{1}{b-a} \int_a^b f(x) dx

gives the average value of the function

f(x)

on the interval

[a, b]

. In this context,

f(x)

is the rate function

R(t)

, and the interval is

[0, 60]

minutes, which is the first hour of the shift. Therefore,

\frac{1}{60-0} \int_0^{60} R(t) dt

represents the average value of the rate of production. The units of

R(t)

are widgets per minute, so the average value will also have units of widgets per minute. Thus, the expression is the average rate of widget production during the first hour.

Question 18

The temperature of a chemical reaction, $C(t)$ , in degrees Celsius, is initially $C(0) = 25$ . For $t \ge 0$ , the temperature changes at a rate of $C'(t) = 6t - 2t^2$ degrees per hour. What is the maximum temperature of the reaction?

$9^\circ$ C
$3^\circ$ C
$25^\circ$ C
$34^\circ$ C (correct answer)

Explanation: To find the maximum temperature, we must find when the rate of change,

C'(t)

, changes from positive to negative. First, find the critical points by setting

C'(t)=0

6t - 2t^2 = 2t(3-t) = 0

. This gives

t=0

and

t=3

. For

t

between 0 and 3,

C'(t)

is positive. For

t>3

C'(t)

is negative. Thus, a maximum temperature occurs at

t=3

. The maximum temperature is the initial temperature plus the accumulated change from

t=0

t=3

C(3) = C(0) + \int_0^3 (6t - 2t^2) dt

C(3) = 25 + [3t^2 - \frac{2}{3}t^3]_0^3

C(3) = 25 + ((3(3^2) - \frac{2}{3}(3^3)) - 0) = 25 + (27 - 18) = 25 + 9 = 34

. The maximum temperature is

34^\circ

Question 19

The internal temperature of a potato in an oven is given by a function $H(t)$ , with $H(0) = 20^\circ$ C. The temperature changes at a rate of $H'(t) = 15e^{-0.05t}$ degrees Celsius per minute for $t \ge 0$ . The potato is considered cooked when its internal temperature reaches $90^\circ$ C. Which of the following statements is true?

The potato's temperature approaches a maximum of $320^\circ$ C as time increases. (correct answer)
The potato's temperature approaches a maximum of $300^\circ$ C as time increases.
The potato reaches $90^\circ$ C at $t = 20\ln(\frac{7}{6})$ minutes.
The potato's temperature will never reach $90^\circ$ C because the rate of temperature change is always decreasing.

Explanation: This problem tests your understanding of how to find a function when given its derivative and initial condition, plus analyzing the long-term behavior of exponential functions. To find the temperature function

H(t)

, you need to integrate the rate of change:

H(t) = \int 15e^{-0.05t} dt = \frac{15}{-0.05}e^{-0.05t} + C = -300e^{-0.05t} + C

. Using the initial condition

H(0) = 20

20 = -300e^{0} + C = -300 + C

, so

C = 320

. Therefore,

H(t) = 320 - 300e^{-0.05t}

. As

t \to \infty

, the term

e^{-0.05t} \to 0

, so

H(t) \to 320

. The maximum temperature the potato approaches is

320°C

, making answer A correct. Answer B (

300°C

) likely comes from forgetting to add the constant of integration or misapplying the initial condition. Answer C attempts to solve

320 - 300e^{-0.05t} = 90

, which gives

e^{-0.05t} = \frac{23}{30}

, not the expression shown. Answer D reflects a common misconception—while

H'(t)

decreases over time, the temperature still increases toward its asymptotic limit. A decreasing rate doesn't mean the function stops growing; it just approaches its maximum more slowly. Study tip: When you have

f'(x)

and an initial condition, always integrate to find

f(x)

, then use the initial condition to solve for the constant. Pay attention to long-term behavior by examining what happens as the variable approaches infinity.

Question 20

Water is pumped into a tank at a rate of $R(t) = 20e^{-0.1t}$ liters per minute, and water leaks out at a constant rate of $L(t) = 4$ liters per minute. At time $t=0$ , the tank contains 100 liters of water.

Which of the following expressions represents the total amount of water, in liters, in the tank at time $t=5$ minutes?

$\int_0^5 (20e^{-0.1t} - 4) dt$
$100 + (20e^{-0.5} - 4)$
$100 + \int_0^5 20e^{-0.1t} dt$
$100 + \int_0^5 (20e^{-0.1t} - 4) dt$ (correct answer)

Explanation: The total amount of water at time

t=5

is the initial amount plus the net change. The initial amount is 100 liters. The net rate of change is the rate in minus the rate out,

R(t) - L(t) = 20e^{-0.1t} - 4

. The net change over the first 5 minutes is the integral of this net rate from 0 to 5. Therefore, the total amount is

100 + \int_0^5 (20e^{-0.1t} - 4) dt