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AP Calculus AB Practice Quiz: Finding the Area Between Curves That Intersect at More Than Two Points

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(x) and g(x) be continuous functions that intersect at x = a, x = b, and x = c, where a < b < c. If f(x) ≥ g(x) on the interval [a, b] and g(x) ≥ f(x) on the interval [b, c], which expression represents the total area enclosed between the curves from x = a to x = c?

All Questions (7)

Let f(x) and g(x) be continuous functions that intersect at x = a, x = b, and x = c, where a < b < c. If f(x) ≥ g(x) on the interval [a, b] and g(x) ≥ f(x) on the interval [b, c], which expression represents the total area enclosed between the curves from x = a to x = c?

A) ∫[a,b] (f(x) - g(x)) dx + ∫[b,c] (g(x) - f(x)) dx

B) ∫[a,c] (f(x) - g(x)) dx

C) ∫[a,c] |f(x) - g(x)| dx

D) |∫[a,c] (f(x) - g(x)) dx|

Correct Answer: A

To find the total area between curves, we integrate the 'top function minus the bottom function'. Since the function on top changes at x = b, the area must be calculated as a sum of two separate definite integrals. The first integral from a to b uses (f(x) - g(x)), and the second from b to c uses (g(x) - f(x)). Option C is also a valid mathematical representation, but option A explicitly shows the calculation required by splitting the integral, which is a key skill.

The area of the region enclosed by the graphs of y = x³ - 3x² + 2x and y = 0 is to be calculated. The curves intersect at x = 0, x = 1, and x = 2. Which of the following definite integrals correctly represents this area?

A) ∫[0,2] (x³ - 3x² + 2x) dx

B) ∫[0,2] |x³ - 3x² + 2x| dx

C) ∫[0,1] (-x³ + 3x² - 2x) dx + ∫[1,2] (-x³ + 3x² - 2x) dx

D) |∫[0,2] (x³ - 3x² + 2x) dx|

Correct Answer: B

The function f(x) = x³ - 3x² + 2x is positive on [0, 1] and negative on [1, 2]. A single definite integral from 0 to 2 would calculate the net area, where the area below the x-axis is subtracted. To find the total geometric area, one must evaluate the definite integral of the absolute value of the function over the entire interval, which ensures all contributions are positive.

Which of the following gives the area of the region enclosed by the graphs of f(x) = sin(x) and g(x) = cos(x) between x = 0 and x = π?

A) ∫[0,π] (sin(x) - cos(x)) dx

B) ∫[0,π/4] (cos(x) - sin(x)) dx + ∫[π/4,π] (sin(x) - cos(x)) dx

C) ∫[0,π/4] (sin(x) - cos(x)) dx + ∫[π/4,π] (cos(x) - sin(x)) dx

D) ∫[0,π] (cos(x) - sin(x)) dx

Correct Answer: B

The graphs of sin(x) and cos(x) intersect at x = π/4. On the interval [0, π/4], cos(x) ≥ sin(x). On the interval [π/4, π], sin(x) ≥ cos(x). Therefore, the total area must be calculated by splitting the integral at the intersection point, with the correct 'top minus bottom' function for each interval.

What is the total area of the regions enclosed by the graphs of y = x³ and y = 4x?

A) 0

B) 4

C) 8

D) 16

Correct Answer: C

First, find the intersection points by setting x³ = 4x, which gives x³ - 4x = 0, so x(x² - 4) = 0. The intersections are at x = -2, 0, and 2. On [-2, 0], x³ ≥ 4x. On [0, 2], 4x ≥ x³. The total area is the sum of two integrals: ∫[-2,0] (x³ - 4x) dx + ∫[0,2] (4x - x³) dx. The first integral evaluates to [x⁴/4 - 2x²] from -2 to 0, which is 0 - (4 - 8) = 4. The second integral evaluates to [2x² - x⁴/4] from 0 to 2, which is (8 - 4) - 0 = 4. The total area is 4 + 4 = 8.

The graph shows the functions f and g, which intersect at x=a, x=b, and x=c. Which of the following expressions represents the total shaded area?

A) ∫[a,c] (g(x) - f(x)) dx

B) ∫[a,b] (f(x) - g(x)) dx + ∫[b,c] (g(x) - f(x)) dx

C) ∫[a,b] (g(x) - f(x)) dx + ∫[b,c] (f(x) - g(x)) dx

D) ∫[a,c] (f(x) + g(x)) dx

Correct Answer: C

The area between two curves is found by integrating the upper function minus the lower function. In the interval [a, b], the graph shows g(x) is above f(x), so the area is ∫[a,b] (g(x) - f(x)) dx. In the interval [b, c], the graph shows f(x) is above g(x), so the area is ∫[b,c] (f(x) - g(x)) dx. The total area is the sum of these two integrals.

Which integral expression represents the area of the region completely enclosed by the graphs of f(x) = 3x³ - x² - 10x and g(x) = -x² + 2x?

A) ∫[-2,2] (3x³ - 12x) dx

B) ∫[-2,0] (-3x³ + 12x) dx + ∫[0,2] (3x³ - 12x) dx

C) 2 ∫[0,2] (-3x³ + 12x) dx

D) ∫[-2,0] (3x³ - 12x) dx + ∫[0,2] (-3x³ + 12x) dx

Correct Answer: D

Find intersections: 3x³ - x² - 10x = -x² + 2x => 3x³ - 12x = 0 => 3x(x² - 4) = 0. Intersections are at x = -2, 0, 2. Let h(x) = f(x) - g(x) = 3x³ - 12x. On [-2, 0], test x = -1: h(-1) = 3(-1) - 12(-1) = 9 > 0, so f(x) is on top. The integral is ∫[-2,0] (3x³ - 12x) dx. On [0, 2], test x = 1: h(1) = 3(1) - 12(1) = -9 < 0, so g(x) is on top. The integral is ∫[0,2] (-3x³ + 12x) dx. The total area is the sum of these two integrals.

A student is asked to find the area between the curves y = f(x) and y = g(x) from x = a to x = c, where the curves intersect at x = b (a < b < c). The student calculates ∫[a,c] (f(x) - g(x)) dx and gets a result of 0. Assuming the calculations are correct, what must be true?

A) The area of the region from a to b is equal to the area of the region from b to c.

B) The functions f(x) and g(x) are identical.

C) The region does not exist, so the area is 0.

D) The student calculated the net signed area, not the total geometric area.

Correct Answer: D

The definite integral of the difference of two functions, ∫(f(x) - g(x))dx, calculates the net signed area. Where f(x) > g(x), the area is counted as positive, and where g(x) > f(x), it's counted as negative. A result of 0 indicates that these positive and negative areas canceled each other out. To find the total geometric area, the student should have used a sum of integrals with the correct upper and lower functions for each subinterval, or integrated the absolute value of the difference.