AP Calculus BC 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
All Questions (7)
A) 4
B) 8
C) 12
D) 8 - 2π
Correct Answer: B
The value of g(2) is given by the definite integral ∫[-2, 2] f(t) dt, which represents the net area under the curve of f from t = -2 to t = 2. The region is a triangle with a base of length 4 (from -2 to 2) and a height of 4. The area is calculated as (1/2) * base * height = (1/2) * 4 * 4 = 8.
A) (-2, 0)
B) (0, 2)
C) (2, 6)
D) (-2, 2)
Correct Answer: C
The function g is decreasing when its derivative, g'(x), is negative. By the Fundamental Theorem of Calculus, g'(x) = f(x). Therefore, g is decreasing on the interval where the graph of f(x) is below the x-axis. Based on the provided figure, f(x) < 0 on the interval (2, 6).
A) 0
B) 8 - 2π
C) 8
D) 12
Correct Answer: C
To find the absolute maximum of g, we must test the endpoints and the critical points where g'(x) = f(x) = 0. The critical points are x=2 and x=6. We evaluate g at x=-2, x=2, and x=6. g(-2) = ∫[-2, -2] f(t) dt = 0. g(2) = ∫[-2, 2] f(t) dt = Area of the triangle above the x-axis = 8. g(6) = ∫[-2, 6] f(t) dt = g(2) + ∫[2, 6] f(t) dt = 8 + (Area of semicircle below x-axis) = 8 - (1/2)π(2)^2 = 8 - 2π. Comparing the values {0, 8, 8-2π}, the largest value is 8.
A) (-2, 0) and (4, 6)
B) (0, 4)
C) (2, 6)
D) (-2, 2)
Correct Answer: A
The graph of g is concave up when its second derivative, g''(x), is positive. Since g'(x) = f(x), we have g''(x) = f'(x). Thus, g is concave up where the slope of f is positive, which is equivalent to where f is increasing. From the graph, f is increasing on the intervals (-2, 0) and (4, 6).
A) x = 0 only
B) x = 2 only
C) x = 0 and x = 4
D) x = 2 and x = 6
Correct Answer: C
A point of inflection on the graph of g occurs where g''(x) changes sign. Since g''(x) = f'(x), this corresponds to a point where the slope of f changes sign. This occurs at local extrema of f. From the graph, f has a local maximum at x = 0 (where the slope of f changes from positive to negative) and a local minimum at x = 4 (where the slope of f changes from negative to positive). Therefore, g has points of inflection at x = 0 and x = 4.
A) The graph of g is increasing and concave up.
B) The graph of g is increasing and concave down.
C) The graph of g is decreasing and concave up.
D) The graph of g is decreasing and concave down.
Correct Answer: B
By the Fundamental Theorem of Calculus, g'(x) = f(x) and g''(x) = f'(x). We are given that f(t) > 0, which implies g'(x) > 0, so g is increasing. We are also given that f'(t) < 0, which implies g''(x) < 0, so g is concave down. Therefore, the graph of g is increasing and concave down.
A) y = 4x
B) y = 2x + 4
C) y = 4x + 8
D) y = 4x + 4
Correct Answer: D
The equation of the tangent line is y - g(0) = g'(0)(x - 0). First, we find the y-coordinate of the point of tangency, g(0) = ∫[-2, 0] f(t) dt. This is the area of the triangle from x=-2 to x=0, which is (1/2) * base * height = (1/2) * 2 * 4 = 4. Next, we find the slope, g'(0). By the Fundamental Theorem of Calculus, g'(x) = f(x), so the slope is g'(0) = f(0). From the graph, f(0) = 4. The equation is y - 4 = 4(x - 0), which simplifies to y = 4x + 4.