AP Calculus AB Quiz: Behavior Of Accumulation Functions Involving Area

20 questions · exam conditions

0:00

Behavior Of Accumulation Functions Involving AreaQuestion 1 of 20

Define $F(x)=\int_{0}^{x} f(t)\,dt$ . If $f(x)>0$ for $0<x<2$ and $f(2)=0$ with no sign change, what is true about $F$ at $x=2$ ?

F

has neither a local maximum nor local minimum at

2

F

has a local maximum at

2

F

has a local minimum at

2

F(2)=0

F

is constant on

(0,2)

← Back to quizzes

AP Calculus AB Quiz

AP Calculus AB Quiz: Behavior Of Accumulation Functions Involving Area

Practice Behavior Of Accumulation Functions Involving Area in AP Calculus AB 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 Behavior Of Accumulation Functions Involving Area, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

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

Define $F(x)=\int_{0}^{x} f(t)\,dt$ . If $f(x)>0$ for $0<x<2$ and $f(2)=0$ with no sign change, what is true about $F$ at $x=2$ ?

$F$ has neither a local maximum nor local minimum at $2$ (correct answer)
$F$ has a local maximum at $2$
$F$ has a local minimum at $2$
$F(2)=0$
$F$ is constant on $(0,2)$

Explanation: This problem tests understanding of accumulation function behavior when the integrand equals zero without changing sign. Since

F'(x) = f(x)

by the Fundamental Theorem, critical points occur where

f(x) = 0

. At

x = 2

, we have

f(2) = 0

with no sign change, meaning

F

has a critical point but no local extremum. The function continues its previous trend (increasing since

f(x) > 0

before

x = 2

) without creating a maximum or minimum. Choice B would require a sign change from positive to negative. When the integrand touches zero without changing sign, the accumulation function has neither a local maximum nor minimum.

Question 2

Define $G(x)=\int_{1}^{x} g(t)\,dt$ . If $g(x)>0$ for all $x$ and $g(3)=g(5)$ , what must be true about $G$ on $(1,6)$ ?

$G$ is increasing on $(1,6)$ (correct answer)
$G$ is constant on $(1,6)$
$G$ is decreasing on $(1,6)$
$G$ has a local maximum at $x=3$
$G(3)=G(5)$

Explanation: This problem tests understanding of accumulation function monotonicity when the integrand is always positive. Since

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

for all

x

, the accumulation function

G

has a positive derivative everywhere, making it increasing throughout its domain. The fact that

g(3) = g(5)

affects the concavity but not the monotonicity of

G

. Even when the integrand has equal values at different points, as long as it remains positive, the accumulation function continues increasing. Choice B incorrectly suggests constant behavior. As long as the integrand remains positive, accumulation functions are always increasing.

Question 3

Let $G(x)=\int_{0}^{x} q(t)\,dt$ . If $q$ has a local maximum at $x=5$ , what can be concluded about $G$ at $x=5$ ?

$G$ has a local maximum
$G$ has an inflection point (correct answer)
$G(5)=0$
$G$ is decreasing at $5$
$G$ is undefined at $5$

Explanation: This problem requires understanding how extrema in the integrand affect inflection points in accumulation functions. Since

G'(x) = q(x)

and

G''(x) = q'(x)

by differentiation, inflection points in

G

occur where

q'(x) = 0

with a sign change. At

x = 5

q

has a local maximum, which means

q'(5) = 0

and

q'(x)

changes from positive to negative. This creates a sign change in

G''(x)

, producing an inflection point in

G

x = 5

. Choice A incorrectly suggests a local maximum, but extrema in

G

require

q(x) = 0

, not extrema in

q

. When the integrand has extrema, the accumulation function has inflection points.

Question 4

Define $A(x)=\int_{2}^{x} g(t)\,dt$ . If $g$ is decreasing on $(0,5)$ , what is true about the concavity of $A$ on $(0,5)$ ?

$A$ is concave down on $(0,5)$ (correct answer)
$A$ is concave up on $(0,5)$
$A$ is linear on $(0,5)$
$A$ is increasing on $(0,5)$
$A$ is concave down only where $A(x)<0$

Explanation: This problem requires understanding how the integrand's monotonicity affects the accumulation function's concavity. Since

A'(x) = g(x)

and

A''(x) = g'(x)

by differentiation, the concavity of

A

depends on whether

g'(x)

is positive or negative. Given that

g

is decreasing on

(0,5)

, we have

g'(x) < 0

throughout this interval. This means

A''(x) < 0

, so

A

is concave down on

(0,5)

. Choice B incorrectly suggests concave up, which would require

g

to be increasing rather than decreasing. When analyzing accumulation function concavity, examine whether the integrand is increasing or decreasing.

Question 5

Define $F(x)=\int_{3}^{x} r(t)\,dt$ . If $r(x)>0$ for $x<1$ and $r(x)<0$ for $x>1$ , where does $F$ have a local extremum?

At $x=3$ only
At $x=1$ only (correct answer)
At every $x$ where $F(x)=0$
At $x=0$ only
Nowhere because the lower limit is $3$

Explanation: This problem tests our ability to locate extrema in accumulation functions with varying lower limits. Since

F'(x) = r(x)

by the Fundamental Theorem, critical points occur where

r(x) = 0

. Given that

r(x) > 0

for

x < 1

and

r(x) < 0

for

x > 1

, the integrand changes from positive to negative at

x = 1

. This means

F'(x)

changes from positive to negative, indicating

F

has a local maximum at

x = 1

. Choice A incorrectly identifies

x = 3

, but

x = 3

is the lower limit where

F(3) = 0

by definition. For accumulation functions, extrema occur where the integrand changes sign, regardless of the lower limit value.

Question 6

Let $A(x)=\int_{4}^{x} f(t)\,dt$ . If $f(x)<0$ and $f'(x)<0$ on $(0,7)$ , what describes $A$ on $(0,7)$ ?

Decreasing and concave down (correct answer)
Decreasing and concave up
Increasing and concave down
Increasing and concave up
Constant and concave up

Explanation: This problem involves analyzing accumulation functions when both the integrand and its derivative are negative. Since

A'(x) = f(x) < 0

(0,7)

, the function

A

is decreasing throughout this interval. For concavity,

A''(x) = f'(x) < 0

, so

A

is concave down on

(0,7)

. Both conditions being negative creates an accumulation function that decreases at an increasing rate. Choice B would be incorrect since concave up requires

f'(x) > 0

. When both the integrand and its derivative are negative, the accumulation function decreases with decreasing rate of change (accelerating downward).

Question 7

Let $G(x)=\int_{-3}^{x} p(t)\,dt$ . If $p$ is increasing on $(-3,1)$ and decreasing on $(1,5)$ , where can $G$ change concavity?

At $x=1$ only (correct answer)
At $x=-3$ only
Wherever $G(x)=0$
At all $x$ where $p(x)=0$
Nowhere because $G$ is an integral

Explanation: This problem requires understanding how monotonicity changes in the integrand create inflection points in accumulation functions. Since

G'(x) = p(x)

and

G''(x) = p'(x)

by differentiation, inflection points occur where

p'(x) = 0

with a sign change. Given that

p

increases on

(-3,1)

and decreases on

(1,5)

, we have

p'(x) > 0

before

x = 1

and

p'(x) < 0

after

x = 1

. This means

G''(x)

changes sign at

x = 1

, creating an inflection point. Choice D incorrectly suggests all zeros of

p

create inflection points, but inflection points require sign changes in

p'

. Inflection points in accumulation functions occur where the integrand changes from increasing to decreasing or vice versa.

Question 8

Define $A(x)=\int_{-2}^{x} f(t)\,dt$ . If $f(x)=3$ for all $x$ in $[-2,2]$ , what is true about $A$ on $(-2,2)$ ?

$A$ is increasing at a constant rate (correct answer)
$A$ is decreasing at a constant rate
$A$ is constant
$A$ is concave down
$A$ changes direction at $x=0$

Explanation: This problem involves analyzing accumulation functions with constant integrands. Since

A'(x) = f(x) = 3

for all

x

[-2,2]

, the accumulation function has a constant positive derivative. This means

A

is increasing at a constant rate of 3 units per unit of

x

, making

A

linear with slope 3. Choice B would be incorrect since the constant is positive, not negative. When the integrand is a positive constant, the accumulation function increases linearly at that constant rate.

Question 9

For $F(x)=\int_{1}^{x} f(t)\,dt$ , suppose $f$ is decreasing on $(1,5)$ . What can be concluded about the concavity of $F$ on $(1,5)$ ?

$F$ is concave up
$F$ is concave down (correct answer)
$F$ is linear
$F$ is increasing
$F$ is decreasing

Explanation: This problem tests understanding of how an integrand's monotonicity affects the accumulation function's concavity. Since F'(x) = f(x) and F''(x) = f'(x), the fact that f is decreasing on (1,5) means f'(x) < 0, which makes F''(x) < 0. Therefore, F is concave down on (1,5). Students often confuse the integrand being decreasing with the accumulation function being decreasing, but a decreasing integrand actually determines the concavity, not the monotonicity, of F. The key principle is that a decreasing integrand produces a concave down accumulation function.

Question 10

Let $H(x)=\int_{-1}^{x} h(t)\,dt$ . If $h(x)=0$ only at $x=3$ and changes from negative to positive there, what occurs at $x=3$ for $H$ ?

$H$ has a local maximum
$H$ has a point of inflection
$H$ has a local minimum (correct answer)
$H$ is undefined
$H(3)=0$

Explanation: This question examines critical points of accumulation functions and their relationship to sign changes in the integrand. Since H'(x) = h(x), when h(x) = 0 at x = 3, we have H'(3) = 0, making x = 3 a critical point of H. Because h changes from negative to positive at x = 3, H' changes from negative to positive there, indicating H transitions from decreasing to increasing. This means H has a local minimum at x = 3. Students might confuse this with a maximum by reversing the sign analysis. Remember that when the integrand changes from negative to positive, the accumulation function reaches a local minimum.

Question 11

Let $M(x)=\int_{-2}^{x} m(t)\,dt$ , where $m(x)<0$ and is increasing on $(-2,3)$ . Which describes $M$ on $(-2,3)$ ?

Decreasing and concave up (correct answer)
Increasing and concave up
Decreasing and concave down
Increasing and concave down
Constant and concave up

Explanation: This question requires analyzing both monotonicity and concavity of an accumulation function simultaneously. Since M'(x) = m(x) and m(x) < 0 on (-2,3), M is decreasing throughout the interval. Additionally, since M''(x) = m'(x) and m is increasing (so m'(x) > 0), M is concave up on (-2,3). Students might think a negative integrand implies concave down behavior, but the concavity depends on whether the integrand is increasing or decreasing, not its sign. The key insight is that an integrand that is negative but increasing produces an accumulation function that is decreasing and concave up.

Question 12

Define $A(x)=\int_{0}^{x} f(t)\,dt$ . If $f(2)=0$ and $f$ changes from positive to negative at $x=2$ , what happens to $A$ at $x=2$ ?

$A$ has a local minimum
$A$ has a local maximum (correct answer)
$A$ has a vertical asymptote
$A(2)=0$
$A$ is not differentiable

Explanation: This problem examines how sign changes in the integrand create extrema in the accumulation function. Since A'(x) = f(x), when f(2) = 0, we have A'(2) = 0, making x = 2 a critical point. Because f changes from positive to negative at x = 2, A' changes from positive to negative, meaning A transitions from increasing to decreasing. This indicates A has a local maximum at x = 2. Students might think A(2) = 0, confusing the value of the integrand with the value of the accumulation function. Remember that when the integrand changes from positive to negative, the accumulation function achieves a local maximum.

Question 13

Define $G(x)=\int_{0}^{x} g(t)\,dt$ . If $g$ is increasing on $(0,4)$ , what is true about $G$ on $(0,4)$ ?

$G$ is decreasing
$G$ is concave up (correct answer)
$G$ is concave down
$G$ is constant
$G$ has a local maximum at $x=2$

Explanation: This problem requires analyzing how properties of an integrand transfer to its accumulation function. Since G'(x) = g(x) by the Fundamental Theorem, and G''(x) = g'(x), the fact that g is increasing means g'(x) > 0, which makes G''(x) > 0. Therefore, G is concave up on (0,4). Students might mistakenly think that an increasing integrand makes the accumulation function increasing, but g being increasing tells us about G's concavity, not its monotonicity. The essential insight is that the derivative of the integrand determines the concavity of the accumulation function.

Question 14

Let $A(x)=\int_{2}^{x} f(t)\,dt$ , where $f(x)>0$ for $2<x<6$ and $f(x)<0$ for $x>6$ . Where is $A$ increasing?

Only on $(2,6)$ (correct answer)
Only on $(6,\infty)$
On $(2,\infty)$
On $(-\infty,6)$
Nowhere, because $f$ changes sign

Explanation: This question tests understanding of how the behavior of an integrand affects its accumulation function. Since A'(x) = f(x) by the Fundamental Theorem of Calculus, A is increasing when f(x) > 0 and decreasing when f(x) < 0. Given that f(x) > 0 only on (2,6) and f(x) < 0 for x > 6, A increases only on the interval (2,6). Students might incorrectly think A increases on the entire interval (2,∞) by forgetting that the sign of f determines whether A increases or decreases. The key strategy is to remember that an accumulation function increases when its integrand is positive and decreases when its integrand is negative.

Question 15

Let $P(x)=\int_{4}^{x} p(t)\,dt$ . If $p(x)>0$ for $x<1$ and $p(x)<0$ for $1<x<4$ , which is true about $P$ on $(1,4)$ ?

$P$ is increasing
$P$ is constant
$P$ is decreasing (correct answer)
$P(x)>0$ for all $x$ in $(1,4)$
$P$ has a critical point at every $x$ in $(1,4)$

Explanation: This question involves understanding accumulation functions with reversed limits of integration. Since P(x) = ∫₄ˣ p(t)dt = -∫ₓ⁴ p(t)dt, we have P'(x) = p(x). Given that p(x) < 0 for 1 < x < 4, this means P'(x) < 0 on (1,4), so P is decreasing on this interval. Students might think the reversed lower limit changes the analysis, but the Fundamental Theorem still gives P'(x) = p(x) regardless of which limit is variable. The strategy is to apply the Fundamental Theorem directly: when the integrand is negative, the accumulation function decreases.

Question 16

Let $A(x)=\int_{-1}^{x} f(t)\,dt$ . If $f(x)=0$ at $x=3$ and changes from positive to negative, what happens to $A$ at $x=3$ ?

$A$ has a local maximum (correct answer)
$A$ has a local minimum
$A$ has an inflection point
$A(3)=0$
$A$ is undefined at $3$

Explanation: This problem requires identifying local maxima in accumulation functions based on integrand sign changes. Since

A'(x) = f(x)

by the Fundamental Theorem, critical points occur where

f(x) = 0

. At

x = 3

, the integrand changes from positive to negative, meaning

A'(x)

changes from positive to negative. This indicates

A

changes from increasing to decreasing, creating a local maximum at

x = 3

. Choice B would be incorrect since a minimum requires the opposite sign change pattern. When the integrand changes from positive to negative, the accumulation function has a local maximum.

Question 17

Let $A(x)=\int_{0}^{x} f(t)\,dt$ . If $f$ is negative on $(0,1)$ and positive on $(1,2)$ , which best describes $A(2)$ ?

$A(2)$ could be positive, negative, or zero depending on areas (correct answer)
$A(2)$ must be positive because $f$ is positive near $2$
$A(2)$ must be negative because $f$ is negative near $0$
$A(2)=f(2)$
$A(2)=0$ because signs cancel

Explanation: This problem requires analyzing accumulation function values based on competing positive and negative areas. Since

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

, the value

A(2)

represents the net area from 0 to 2. With

f

negative on

(0,1)

and positive on

(1,2)

, we get

A(2) = \int_0^1 f(t) dt + \int_1^2 f(t) dt

, where the first integral is negative and the second is positive. The sign of

A(2)

depends on which area magnitude is larger, making it potentially positive, negative, or zero. Choice B incorrectly assumes only the final behavior matters. The value of an accumulation function depends on the net balance of all signed areas.

Question 18

Let $H(x)=\int_{-2}^{x} h(t)\,dt$ . If $h(x)>0$ on $(-2,0)$ and $h(x)=0$ on $(0,3)$ , how does $H$ behave on $(0,3)$ ?

$H$ is constant on $(0,3)$ (correct answer)
$H$ is increasing on $(0,3)$
$H$ is decreasing on $(0,3)$
$H$ has a local maximum at every point in $(0,3)$
$H(x)=0$ on $(0,3)$

Explanation: This problem tests understanding of accumulation function behavior when the integrand becomes zero on an interval. Since

H'(x) = h(x)

by the Fundamental Theorem, the behavior of

H

depends on the values of

h

. On

(0,3)

where

h(x) = 0

, we have

H'(x) = 0

, making the derivative zero throughout this interval. This means

H

has no rate of change and remains constant on

(0,3)

. The previous behavior on

(-2,0)

determines the constant value, but doesn't affect the constant behavior on

(0,3)

. Choice B would be incorrect since increasing requires

h(x) > 0

. When the integrand equals zero on an interval, the accumulation function is constant there.

Question 19

Define $G(x)=\int_{2}^{x} f(t)\,dt$ . If $f(x)=0$ at $x=7$ and $f$ changes from negative to positive there, what happens to $G$ at $x=7$ ?

$G$ has a local minimum (correct answer)
$G$ has a local maximum
$G$ has an inflection point
$G(7)=0$
$G$ is constant near $7$

Explanation: This problem tests identification of local minima based on integrand sign changes in accumulation functions. Since

G'(x) = f(x)

by the Fundamental Theorem, critical points occur where

f(x) = 0

. At

x = 7

, the integrand changes from negative to positive, meaning

G'(x)

changes from negative to positive. This indicates

G

changes from decreasing to increasing, creating a local minimum at

x = 7

. Choice B would be incorrect since a maximum requires the sign change from positive to negative. When the integrand changes from negative to positive, the accumulation function transitions from decreasing to increasing, producing a local minimum.

Question 20

Define $A(x)=\int_{0}^{x} f(t)\,dt$ . If $f(1)=0$ and $f$ is increasing through $x=1$ , what is true about $A$ at $x=1$ ?

$A$ has an inflection point at $x=1$ (correct answer)
$A$ has a local maximum at $x=1$
$A$ has a local minimum at $x=1$
$A(1)=0$
$A$ is undefined at $x=1$

Explanation: This problem involves analyzing accumulation functions when the integrand crosses zero while changing monotonicity. Since

A'(x) = f(x)

and

A''(x) = f'(x)

by differentiation, we examine both the value and derivative of

f

x = 1

. At this point,

f(1) = 0

means

A'(1) = 0

, creating a critical point. Since

f

is increasing through

x = 1

, we have

f'(1) > 0

, so

A''(1) > 0

, but the sign change in

f

creates an inflection point rather than an extremum. Choice B would be incorrect since extrema require definite sign changes in the derivative. When the integrand equals zero and is increasing, the accumulation function has an inflection point.