Calculus 2 Quiz: Ftc And Accumulation Functions

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Ftc And Accumulation FunctionsQuestion 1 of 20

The function $f(t)$ is continuous and positive for all $t$ . Let $G(x) = \int_0^x f(t) dt$ . If the graph of $f(t)$ is increasing for $t>0$ , which of the following statements about the graph of $G(x)$ must be true for $x>0$ ?

G(x)

is increasing and concave up.

G(x)

is increasing and concave down.

G(x)

is decreasing and concave up.

G(x)

is decreasing and concave down.

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Calculus 2 Quiz

Calculus 2 Quiz: Ftc And Accumulation Functions

Practice Ftc And Accumulation Functions 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 Ftc And Accumulation Functions, 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

$G(x)$ is increasing and concave up. (correct answer)
$G(x)$ is increasing and concave down.
$G(x)$ is decreasing and concave up.
$G(x)$ is decreasing and concave down.

Explanation: By the Fundamental Theorem of Calculus Part 1,

G'(x) = f(x)

. Since we are given that

f(t)

is positive for all

t

G'(x) = f(x) > 0

for

x>0

. This means that

G(x)

is an increasing function. To determine concavity, we look at the second derivative,

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

. We are given that

f(t)

is an increasing function, which means its derivative,

f'(t)

, must be positive. Therefore,

G''(x) = f'(x) > 0

for

x>0

. This means that

G(x)

is concave up. Thus,

G(x)

is both increasing and concave up.

Question 2

Let $F(x) = \int_3^x (c-t)e^{-t^2} dt$ . If $F(x)$ has a local maximum at $x=5$ , what is the value of the constant $c$ ?

3
$(5-c)e^{-25}$
$e^{-25}$
5 (correct answer)

Explanation: This problem tests your understanding of the Fundamental Theorem of Calculus and how to find critical points of functions defined by integrals. When you see a function defined as an integral with a variable upper limit, you'll need to differentiate it to find where extrema occur. To find where

F(x)

has a local maximum, you need to find where

F'(x) = 0

. Using the Fundamental Theorem of Calculus, if

F(x) = \int_3^x (c-t)e^{-t^2} dt

, then

F'(x) = (c-x)e^{-x^2}

. Since

F(x)

has a local maximum at

x = 5

, we know

F'(5) = 0

. Substituting:

F'(5) = (c-5)e^{-25} = 0

. Since

e^{-25} > 0

(exponential functions are always positive), the only way this product equals zero is if

c - 5 = 0

, which means

c = 5

. Looking at the wrong answers: Choice (A) gives

c = 3

, which would make

F'(5) = -2e^{-25} \neq 0

. Choice (B) represents

(5-c)e^{-25}

, which is actually the negative of

F'(5)

and represents a common algebra error. Choice (C) gives

c = e^{-25}

, which would make

F'(5) = (e^{-25} - 5)e^{-25} \neq 0

since

e^{-25} \neq 5

. Remember: when finding critical points of integral functions, differentiate using the Fundamental Theorem of Calculus first, then set the derivative equal to zero. The exponential factor will never be zero, so focus on when the polynomial factor equals zero.

Question 3

Let $f$ be a function such that $f'(x) = \cos(x^2)$ . If $f(1) = 5$ , which of the following expressions represents $f(2)$ ?

$5 + \int_1^2 \cos(x^2) dx$ (correct answer)
$5 + \cos(4)$
$\int_1^2 (5+\cos(x^2)) dx$
$5 + \cos(4) - \cos(1)$

Explanation: This question is a direct application of the accumulation concept derived from the Fundamental Theorem of Calculus Part 2. The theorem states that

\int_a^b F'(x) dx = F(b) - F(a)

. We can rewrite this as

F(b) = F(a) + \int_a^b F'(x) dx

. In this problem, we are given

f'(x) = \cos(x^2)

and we want to find

f(2)

given

f(1)

. Using the formula with

F=f

a=1

, and

b=2

, we get

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

. Substituting the given values, we have

f(2) = 5 + \int_1^2 \cos(x^2) dx

Question 4

The velocity of a particle moving along the x-axis is given by $v(t) = 3t^2 - 12$ for $t \ge 0$ . The particle starts at position $x=5$ at time $t=0$ . What is the position of the particle at $t=3$ ?

-9
-4 (correct answer)
5
14

Explanation: The position function

x(t)

is the antiderivative of the velocity function

v(t)

. The position at time

t=3

can be found using the initial position and the net displacement from

t=0

t=3

. The displacement is given by the definite integral of the velocity function:

\int_0^3 v(t) dt

. The position is

x(3) = x(0) + \int_0^3 (3t^2 - 12) dt

. First, calculate the integral:

\int_0^3 (3t^2 - 12) dt = [t^3 - 12t]_0^3 = (3^3 - 12(3)) - (0) = 27 - 36 = -9

. This is the displacement. The final position is the initial position plus the displacement:

x(3) = 5 + (-9) = -4

Question 5

The rate at which a contaminant is leaking into a pond is given by $r(t) = 2 - \sqrt{t}$ gallons per hour, where $t$ is time in hours. The leak is stopped at $t=4$ hours. Let $A(t) = \int_0^t r(x) dx$ . Which statement correctly describes the amount of contaminant in the pond?

The amount of contaminant is at its maximum at $t=2$ , because the rate of leakage is highest before it begins to decrease.
The amount of contaminant is at its maximum at $t=4$ , because the rate of leakage is non-negative on the interval $[0, 4]$ . (correct answer)
The total amount of contaminant that has leaked into the pond after 4 hours is given by the rate $r(4)$ .
The amount of contaminant is decreasing for all $t \in [0,4]$ because the rate function $r(t)$ is a decreasing function.

Explanation: The function

A(t) = \int_0^t r(x) dx

represents the total accumulated amount of contaminant at time

t

. The rate of change of this amount is

A'(t) = r(t)

. The amount of contaminant increases as long as the rate

r(t)

is positive. We set

r(t) = 2 - \sqrt{t} > 0

, which implies

2 > \sqrt{t}

, or

t < 4

. The rate is zero at

t=4

. Since the rate is positive for

0 \le t < 4

and zero at

t=4

, the accumulated amount

A(t)

is increasing on the entire interval

[0, 4]

. Therefore, the maximum amount of contaminant occurs at the end of the interval, at

t=4

Question 6

If $G(x) = \int_0^x f(t) dt$ where $f$ is continuous, and $G(2) = 8$ , $G(5) = 20$ , what is the average value of $f$ on the interval $[2, 5]$ ?

$\frac{12}{3}$
$4$ (correct answer)
$6$
$\frac{20}{5}$

Explanation: This question tests your understanding of the Fundamental Theorem of Calculus and the formula for average value of a function. When you see a function defined as an integral with a variable upper limit, think about how changes in that integral relate to the original function. The key insight is recognizing what

G(5) - G(2)

represents. Since

G(x) = \int_0^x f(t) dt

, we have:

G(5) - G(2) = \int_0^5 f(t) dt - \int_0^2 f(t) dt = \int_2^5 f(t) dt

This gives us

\int_2^5 f(t) dt = 20 - 8 = 12

. The average value of a continuous function

f

on interval

[a,b]

\frac{1}{b-a}\int_a^b f(t) dt

. Therefore, the average value on

[2,5]

is:

\frac{1}{5-2}\int_2^5 f(t) dt = \frac{1}{3} \cdot 12 = 4

Choice A (

\frac{12}{3}

) shows the correct setup but leaves the fraction unreduced. While mathematically equivalent to 4, this suggests incomplete work. Choice C (6) likely comes from incorrectly using the interval length as 2 instead of 3, giving

\frac{12}{2} = 6

. Choice D (

\frac{20}{5}

) represents a common error: dividing

G(5)

by the right endpoint rather than using the difference

G(5) - G(2)

and the correct interval length. Remember: when finding average value over

[a,b]

, you need the integral specifically over that interval, not from 0. Use properties of definite integrals to extract the piece you need.

Question 7

Consider the accumulation function $F(x) = \int_1^x \frac{dt}{\sqrt{1+t^2}}$ . Which of the following best describes the relationship between $F(x)$ and $\sinh^{-1}(x)$ (the inverse hyperbolic sine function)?

$F(x) = \sinh^{-1}(x)$ for all $x \geq 1$
$F(x) = \sinh^{-1}(x) - \sinh^{-1}(1)$ for all $x \geq 1$
$F'(x) = \frac{d}{dx}[\sinh^{-1}(x)]$ but $F(x) \neq \sinh^{-1}(x)$ (correct answer)
$F(x)$ and $\sinh^{-1}(x)$ have different derivatives

Explanation: We know that

\frac{d}{dx}[\sinh^{-1}(x)] = \frac{1}{\sqrt{1+x^2}}

. By the Fundamental Theorem of Calculus,

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

. So

F'(x) = \frac{d}{dx}[\sinh^{-1}(x)]

, meaning they have the same derivative. However,

F(x)

and

\sinh^{-1}(x)

differ by a constant since

F(1) = 0

while

\sinh^{-1}(1) = \ln(1 + \sqrt{2}) \neq 0

. Therefore,

F(x) = \sinh^{-1}(x) - \sinh^{-1}(1)

, but the question asks about the relationship in general terms. Choice C correctly identifies that they have the same derivative but are not equal functions. Choice A is false because they differ by a constant. Choice B gives the exact relationship but is more specific than what the question seems to be testing. Choice D is incorrect since they do have the same derivative.

Question 8

What is the value of the limit $\lim_{x \to 0} \frac{1}{x^3} \int_0^x \sin(t^2) dt$ ?

0
$\infty$
$1/3$ (correct answer)
1

Explanation: The limit is of the form

\frac{0}{0}

because

\int_0^0 \sin(t^2) dt = 0

and

0^3 = 0

. We can apply L'Hôpital's Rule. Let the numerator be

N(x) = \int_0^x \sin(t^2) dt

and the denominator be

D(x) = x^3

. By FTC Part 1,

N'(x) = \sin(x^2)

. The derivative of the denominator is

D'(x) = 3x^2

. Applying L'Hôpital's Rule, the limit becomes

\lim_{x \to 0} \frac{\sin(x^2)}{3x^2}

. This limit can be rewritten as

\frac{1}{3} \lim_{x \to 0} \frac{\sin(x^2)}{x^2}

. Using the known limit

\lim_{u \to 0} \frac{\sin u}{u} = 1

, with

u = x^2

, the expression becomes

\frac{1}{3} \cdot 1 = \frac{1}{3}

Question 9

Consider $g(x) = \int_{x}^{2x} e^t dt$ . Which expression represents $g'(x)$ ?

$e^{2x} \cdot 2 - e^x$
$2e^{2x} - e^x$ (correct answer)
$e^{2x} - e^x$
$e^{2x} + e^x$

Explanation: Since both limits of integration contain

x

, we use:

\frac{d}{dx}\int_{u(x)}^{v(x)} f(t) dt = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x)

. Here,

u(x) = x

v(x) = 2x

, and

f(t) = e^t

. So

u'(x) = 1

v'(x) = 2

f(v(x)) = e^{2x}

, and

f(u(x)) = e^x

. Therefore:

g'(x) = e^{2x} \cdot 2 - e^x \cdot 1 = 2e^{2x} - e^x

. Choice A forgets to multiply

e^{2x}

by the derivative of

2x

. Choice C treats both limits as if their derivatives were 1. Choice D incorrectly adds instead of subtracting.

Question 10

Let $h(x) = \int_1^{x^2} \sqrt{9+t^2} dt$ . What is the equation of the tangent line to the graph of $y=h(x)$ at $x=1$ ?

$y = \sqrt{10}(x-1)$
$y = 2\sqrt{10}(x-1)$ (correct answer)
$y = 2\sqrt{10}x + (1-2\sqrt{10})$
$y = 0$

Explanation: To find the equation of the tangent line, we need a point and a slope. The point is

(1, h(1))

. We calculate

h(1) = \int_1^{1^2} \sqrt{9+t^2} dt = \int_1^1 \sqrt{9+t^2} dt = 0

. So the point is

(1, 0)

. The slope is

h'(1)

. Using the FTC Part 1 and the Chain Rule,

h'(x) = \sqrt{9+(x^2)^2} \cdot \frac{d}{dx}(x^2) = \sqrt{9+x^4} \cdot 2x

. The slope at

x=1

h'(1) = \sqrt{9+1^4} \cdot 2(1) = 2\sqrt{10}

. Using the point-slope form

y - y_1 = m(x - x_1)

, the equation of the tangent line is

y - 0 = 2\sqrt{10}(x - 1)

, which simplifies to

y = 2\sqrt{10}(x - 1)

Question 11

Let $G(x) = \int_{\sin(x)}^{\cos(x)} e^{t^2} dt$ . What is the value of $G'(\pi/4)$ ?

0
$\sqrt{e}$
$-\sqrt{2e}$ (correct answer)
$-2\sqrt{e}$

Explanation: Using the Fundamental Theorem of Calculus with the Chain Rule for an integral of the form

\int_{h(x)}^{g(x)} f(t) dt

, the derivative is

f(g(x))g'(x) - f(h(x))h'(x)

. Here,

f(t) = e^{t^2}

g(x) = \cos(x)

, and

h(x) = \sin(x)

. The derivatives are

g'(x) = -\sin(x)

and

h'(x) = \cos(x)

. So,

G'(x) = e^{(\cos x)^2} \cdot (-\sin x) - e^{(\sin x)^2} \cdot (\cos x)

. Evaluating at

x = \pi/4

, we have

\sin(\pi/4) = \cos(\pi/4) = \frac{\sqrt{2}}{2}

. Thus,

G'(\pi/4) = e^{(\sqrt{2}/2)^2} \cdot (-\frac{\sqrt{2}}{2}) - e^{(\sqrt{2}/2)^2} \cdot (\frac{\sqrt{2}}{2}) = e^{1/2} \cdot (-\frac{\sqrt{2}}{2}) - e^{1/2} \cdot (\frac{\sqrt{2}}{2}) = -2 e^{1/2} \frac{\sqrt{2}}{2} = -\sqrt{2}e^{1/2} = -\sqrt{2e}

Question 12

Let $F(x) = \int_0^{\sin(x)} t^2 dt$ . Find $F'(\pi/2)$ .

$\sin^2(\pi/2) \cdot \cos(\pi/2)$
$\sin^2(\pi/2)$
$\cos(\pi/2)$
$0$ (correct answer)

Explanation: Using the Fundamental Theorem with chain rule:

F'(x) = (\sin(x))^2 \cdot \frac{d}{dx}[\sin(x)] = \sin^2(x) \cdot \cos(x)

. At

x = \pi/2

F'(\pi/2) = \sin^2(\pi/2) \cdot \cos(\pi/2) = 1^2 \cdot 0 = 0

. The key insight is that

\cos(\pi/2) = 0

, making the entire expression equal to zero. Choice A shows the correct formula but doesn't evaluate it. Choice B forgets the chain rule factor

\cos(x)

. Choice C gives only the chain rule factor without the integrand evaluation.

Question 13

Let $F(x) = \int_x^{x+2} (4t+1) dt$ . What is the value of $F'(1)$ ?

8 (correct answer)
10
14
18

Explanation: We can solve this problem in two ways. Method 1: Use FTC Part 1 with variable limits. We can write

F(x) = \int_x^c (4t+1) dt + \int_c^{x+2} (4t+1) dt = -\int_c^x (4t+1) dt + \int_c^{x+2} (4t+1) dt

. Taking the derivative,

F'(x) = -(4x+1) + (4(x+2)+1) \cdot (1) = -4x-1 + 4x+8+1 = 8

. Since

F'(x)=8

for all

x

F'(1)=8

. Method 2: Evaluate the integral first.

F(x) = [2t^2+t]_x^{x+2} = (2(x+2)^2 + (x+2)) - (2x^2+x) = (2(x^2+4x+4) + x+2) - 2x^2-x = (2x^2+8x+8+x+2) - 2x^2-x = 8x+10

. Then,

F'(x) = 8

. So,

F'(1)=8

Question 14

Let $g(x) = \frac{d}{dx} \int_0^x (t^2 - 3t) dt$ . What is the instantaneous rate of change of $g(x)$ at $x=2$ ?

-2
1 (correct answer)
2
4

Explanation: This is a multi-step derivative problem. First, we must find the function

g(x)

. By the Fundamental Theorem of Calculus Part 1,

g(x) = \frac{d}{dx} \int_0^x (t^2 - 3t) dt = x^2 - 3x

. The question asks for the instantaneous rate of change of

g(x)

x=2

, which means we need to find

g'(2)

. First, we find the derivative of

g(x)

g'(x) = \frac{d}{dx}(x^2 - 3x) = 2x - 3

. Then, we evaluate

g'(x)

x=2

g'(2) = 2(2) - 3 = 4 - 3 = 1

Question 15

Let $f(x) = \int_{x^2}^{x^3} \frac{1}{\ln t} dt$ for $x>1$ . Find $f'(x)$ .

$\frac{x}{\ln x}$
$\frac{x^2-x}{\ln x}$
$\frac{3x^2}{\ln(x^3)} - \frac{2x}{\ln(x^2)}$ (correct answer)
$\frac{1}{\ln(x^3)} - \frac{1}{\ln(x^2)}$

Explanation: We use the Fundamental Theorem of Calculus for variable limits of integration:

\frac{d}{dx}\int_{h(x)}^{g(x)} F(t) dt = F(g(x))g'(x) - F(h(x))h'(x)

. In this problem,

F(t) = \frac{1}{\ln t}

g(x) = x^3

, and

h(x) = x^2

. The derivatives are

g'(x) = 3x^2

and

h'(x) = 2x

. Applying the formula:

f'(x) = \frac{1}{\ln(x^3)} \cdot (3x^2) - \frac{1}{\ln(x^2)} \cdot (2x) = \frac{3x^2}{\ln(x^3)} - \frac{2x}{\ln(x^2)}

Question 16

Let $F(x) = \int_1^x (t^2 - 4t + 3) \ln(t) dt$ for $x > 0$ . At which of the following values of $x$ does $F(x)$ have a local minimum?

$x=1$
$x=3$ (correct answer)
$x=4$
Both $x=1$ and $x=3$

Explanation: To find local extrema of

F(x)

, we must find its critical points by taking the derivative and setting it to zero. By the Fundamental Theorem of Calculus Part 1,

F'(x) = (x^2 - 4x + 3)\ln(x)

. We set

F'(x) = 0

. This occurs if

x^2 - 4x + 3 = 0

\ln(x) = 0

. Factoring gives

(x-1)(x-3) = 0

, so

x=1

x=3

. Also,

\ln(x)=0

gives

x=1

. The critical points are

x=1

and

x=3

. We use the first derivative test. For

0 < x < 1

F'(x) = (+)(-) = -

. For

1 < x < 3

F'(x) = (-)(+) = -

. For

x > 3

F'(x) = (+)(+) = +

. Since

F'(x)

changes from negative to positive at

x=3

F(x)

has a local minimum at

x=3

. At

x=1

, the derivative does not change sign, so it is not a local extremum.

Question 17

Let $f(x)$ be a continuous function and $F(x)$ be an antiderivative of $f(x)$ . If $\int_2^5 f(x) dx = -3$ and $F(2) = 7$ , what is the value of $F(5)$ ?

-10
10
4 (correct answer)
-4

Explanation: The Fundamental Theorem of Calculus Part 2 states that

\int_a^b f(x) dx = F(b) - F(a)

, where

F'(x) = f(x)

. Applying this to the given problem, we have

\int_2^5 f(x) dx = F(5) - F(2)

. We are given

\int_2^5 f(x) dx = -3

and

F(2) = 7

. Substituting these values into the equation gives

-3 = F(5) - 7

. Solving for

F(5)

, we add 7 to both sides:

F(5) = 7 - 3 = 4

Question 18

Let $f$ be a continuous function. If $G(x) = \int_1^x f(t) dt$ , $G(1)=0$ and $G(3)=4$ , what is the average value of $f$ on the interval $[1, 3]$ ?

1
2 (correct answer)
3
4

Explanation: The average value of a function

f(t)

on an interval

[a, b]

is given by the formula

\frac{1}{b-a} \int_a^b f(t) dt

. In this case, the interval is

[1, 3]

, so the average value is

\frac{1}{3-1} \int_1^3 f(t) dt

. We are given that

G(x) = \int_1^x f(t) dt

. Therefore, the integral

\int_1^3 f(t) dt

is equal to

G(3)

. We are given that

G(3)=4

. Substituting this into the average value formula, we get

\frac{1}{2} \cdot G(3) = \frac{1}{2} \cdot 4 = 2

. The information

G(1)=0

is consistent with the definition of

G(x)

but is not needed for the final calculation.

Question 19

Let $G(x) = \int_{-2}^x |t+1| dt$ . What is the value of $G(3)$ ?

8
9
7.5
8.5 (correct answer)

Explanation: When you encounter an integral with an absolute value function, you need to identify where the expression inside changes sign, then split the integral at those critical points. For

G(x) = \int_{-2}^x |t+1| dt

, the expression

t+1

equals zero when

t = -1

. This means

|t+1|

behaves differently on either side of

t = -1

When $t < -1$ : $t+1 < 0$ , so $|t+1| = -(t+1) = -t-1$
When $t > -1$ : $t+1 > 0$ , so $|t+1| = t+1$

To find

G(3)

, split the integral at

t = -1

G(3) = \int_{-2}^{-1} |t+1| dt + \int_{-1}^{3} |t+1| dt

= \int_{-2}^{-1} (-t-1) dt + \int_{-1}^{3} (t+1) dt

For the first integral:

\int_{-2}^{-1} (-t-1) dt = \left[-\frac{t^2}{2} - t\right]_{-2}^{-1} = \left(-\frac{1}{2} + 1\right) - \left(-2 + 2\right) = \frac{1}{2}

For the second integral:

\int_{-1}^{3} (t+1) dt = \left[\frac{t^2}{2} + t\right]_{-1}^{3} = \left(\frac{9}{2} + 3\right) - \left(\frac{1}{2} - 1\right) = 8

Therefore,

G(3) = \frac{1}{2} + 8 = 8.5

. Choice A (8) likely comes from forgetting the first piece of the integral. Choice B (9) might result from sign errors in the absolute value regions. Choice C (7.5) could stem from computational mistakes in the antiderivatives. Always sketch the absolute value function first to visualize where it changes behavior—this prevents sign errors and missed boundary points.

Question 20

Suppose $f(x)$ is a continuous function such that $f(x) > 0$ for all $x$ . Let $H(x) = \int_1^{x^2} f(t) dt$ . Which of the following statements about $H(x)$ must be true?

$H(x)$ is increasing for $x > 0$ and decreasing for $x < 0$ . (correct answer)
$H(x)$ is increasing for all $x$ .
$H(x)$ is decreasing for $x > 0$ and increasing for $x < 0$ .
$H(x)$ has a local maximum at $x = 0$ .

Explanation: To determine where

H(x)

is increasing or decreasing, we analyze its derivative,

H'(x)

. Using the FTC Part 1 and the Chain Rule,

H'(x) = f(x^2) \cdot \frac{d}{dx}(x^2) = f(x^2) \cdot 2x

. We are given that

f(x) > 0

for all

x

, so

f(x^2)

must be positive for all real

x

. Therefore, the sign of

H'(x)

is determined by the sign of

2x

. For

x > 0

2x > 0

, so

H'(x) > 0

, which means

H(x)

is increasing. For

x < 0

2x < 0

, so

H'(x) < 0

, which means

H(x)

is decreasing. At

x=0

H'(0)=0

, and since the derivative changes from negative to positive,

H(x)

has a local minimum at

x=0