AP Calculus AB Practice Quiz: Interpreting the Behavior of Accumulation Functions Involving Area

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

Let f be the function whose graph is shown above, and let g be the function defined by g(x) = ∫(from 0 to x) f(t) dt. On which of the following intervals is g increasing?

A)(0, 2)B)(2, 6)C)(4, 8)D)(6, 10)

All Questions (7)

Let f be the function whose graph is shown above, and let g be the function defined by g(x) = ∫(from 0 to x) f(t) dt. On which of the following intervals is g increasing?

A) (0, 2)

B) (2, 6)

C) (4, 8)

D) (6, 10)

Correct Answer: B

The function g(x) is increasing on intervals where its derivative, g'(x), is positive. According to the Fundamental Theorem of Calculus, g'(x) = f(x). Therefore, g is increasing where the graph of f(x) is above the x-axis. Based on the provided graph, f(x) > 0 on the interval (2, 6).

The graph of a differentiable function f is shown above for the interval [0, 8]. If g(x) = ∫(from 1 to x) f(t) dt, at what value of x does g(x) have a local maximum?

A) x = 1

B) x = 3

C) x = 5

D) x = 7

Correct Answer: C

The function g(x) has a local maximum where its derivative, g'(x), changes from positive to negative. By the Fundamental Theorem of Calculus, g'(x) = f(x). We need to find where the graph of f(x) crosses the x-axis from above to below. This occurs at x = 5.

Let f be a continuous function, and let g be the function defined by g(x) = ∫(from -2 to x) f(t) dt. The graph of f is shown above. On which of the following intervals is the graph of g concave down?

A) (-4, -1)

B) (-1, 2)

C) (2, 4)

D) (-4, 2)

Correct Answer: B

The graph of g(x) is concave down when its second derivative, g''(x), is negative. Since g'(x) = f(x), we have g''(x) = f'(x). Therefore, g is concave down where f'(x) < 0, which means the function f is decreasing. Based on the graph, f is decreasing on the interval (-1, 2).

The graph of the function f is shown above. Let g be the function defined by g(x) = ∫(from 0 to x) f(t) dt. For what value of x in the open interval (0, 6) does the graph of g have a point of inflection?

A) x = 1

B) x = 2

C) x = 3

D) x = 4

Correct Answer: C

The graph of g has a point of inflection where its second derivative, g''(x), changes sign. Since g'(x) = f(x), we have g''(x) = f'(x). The sign of f'(x) changes where the function f has a local extremum (maximum or minimum). Looking at the graph of f, it has a local maximum at x = 3. Therefore, g has a point of inflection at x = 3.

Let g be a function defined by g(x) = ∫(from 1 to x) (6t^2 - 2) dt. What is the value of g(2)?

A) 8

B) 10

C) 12

D) 14

Correct Answer: C

To find g(2), we evaluate the definite integral from 1 to 2. The antiderivative of f(t) = 6t^2 - 2 is F(t) = 2t^3 - 2t. Using the Fundamental Theorem of Calculus, g(2) = F(2) - F(1) = (2(2)^3 - 2(2)) - (2(1)^3 - 2(1)) = (16 - 4) - (2 - 2) = 12 - 0 = 12.

A function f is continuous on the interval [0, ∞). It is known that f(x) is always positive and its rate of change is always negative. If g(x) = ∫(from 2 to x) f(t) dt, which of the following statements must be true about g for x > 2?

A) g is increasing and concave up.

B) g is increasing and concave down.

C) g is decreasing and concave up.

D) g is decreasing and concave down.

Correct Answer: B

We are given g(x) = ∫(from 2 to x) f(t) dt. By the Fundamental Theorem of Calculus, g'(x) = f(x) and g''(x) = f'(x). We are told f(x) is always positive, so g'(x) > 0, which means g is increasing. We are also told that the rate of change of f is always negative, which means f'(x) < 0. Therefore, g''(x) < 0, which means g is concave down. Thus, g is increasing and concave down.

The graph of f, a continuous function on the interval [0, 7], is shown above. Let g be the function defined by g(x) = ∫(from 3 to x) f(t) dt. What is the absolute minimum value of g on the interval [0, 7]?

A) -4

B) -2

C) 0

D) 3

Correct Answer: A

To find the absolute minimum of g on [0, 7], we must test the endpoints and the critical points. Critical points occur where g'(x) = f(x) = 0, which is at x=1, x=3, and x=5. The candidates are x=0, 1, 3, 5, 7. We calculate the value of g at each candidate by finding the net signed area from 3 to x. g(0) = ∫(from 3 to 0) f(t) dt = -∫(from 0 to 3) f(t) dt = -(-2) = 2. g(1) = ∫(from 3 to 1) f(t) dt = -∫(from 1 to 3) f(t) dt = -(-4) = 4. g(3) = ∫(from 3 to 3) f(t) dt = 0. g(5) = ∫(from 3 to 5) f(t) dt = -4. g(7) = ∫(from 3 to 7) f(t) dt = ∫(from 3 to 5) f(t) dt + ∫(from 5 to 7) f(t) dt = -4 + 2 = -2. Comparing the values {2, 4, 0, -4, -2}, the absolute minimum value is -4.