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AP Calculus BC Practice Quiz: Finding the Area Between Curves Expressed as Functions of $x$

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

Which of the following definite integrals represents the area of the region in the plane enclosed by the graphs of y = e^x, y = x, x = 0, and x = 1?

All Questions (7)

Which of the following definite integrals represents the area of the region in the plane enclosed by the graphs of y = e^x, y = x, x = 0, and x = 1?

A) ∫[0, 1] (x - e^x) dx

B) ∫[0, 1] (e^x - x) dx

C) ∫[0, 1] (e^x + x) dx

D) ∫[0, e] (e^x - x) dx

Correct Answer: B

The area of the region between two curves f(x) and g(x) from x = a to x = b, where f(x) ≥ g(x) on [a, b], is calculated by the definite integral ∫[a, b] (f(x) - g(x)) dx. In the interval [0, 1], the value of e^x is always greater than or equal to the value of x. Therefore, the area is calculated by the integral of the top function (e^x) minus the bottom function (x) over the given interval [0, 1]. The correct setup is ∫[0, 1] (e^x - x) dx.

Which definite integral calculates the area of the region enclosed by the graphs of y = 4 and y = x^2?

A) ∫[-2, 2] (x^2 - 4) dx

B) ∫[0, 4] (4 - x^2) dx

C) ∫[-2, 2] (4 + x^2) dx

D) ∫[-2, 2] (4 - x^2) dx

Correct Answer: D

First, find the points of intersection by setting the functions equal: 4 = x^2, which gives x = ±2. These are the bounds of integration. In the interval [-2, 2], the graph of the constant function y = 4 is above the graph of the parabola y = x^2. The area is calculated by the definite integral of (top function - bottom function) over the interval. Thus, the integral is ∫[-2, 2] (4 - x^2) dx.

The figure shows the graphs of two functions, y = f(x) and y = g(x), which intersect at x = a, x = b, and x = c. Which of the following expressions represents the total area of the shaded regions enclosed between the curves?

A) ∫[a, c] (f(x) - g(x)) dx

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

C) ∫[a, c] (g(x) - f(x)) dx

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

Correct Answer: B

To find the total area between curves that cross, the definite integral must be split at each intersection point. Based on the described graph, on the interval [a, b], the graph of f(x) is above g(x), so the area is calculated by ∫[a, b] (f(x) - g(x)) dx. On the interval [b, c], the graph of g(x) is above f(x), so the area is calculated by ∫[b, c] (g(x) - f(x)) dx. The total area is the sum of the areas of these two regions.

Let f(x) and g(x) be continuous functions such that f(x) ≥ g(x) for all x in the interval [a, b]. The area of the region in the plane bounded by the graphs of y = f(x), y = g(x), x = a, and x = b is calculated by which definite integral?

A) ∫[a, b] (g(x) - f(x)) dx

B) ∫[a, b] (f(x) + g(x)) dx

C) ∫[a, b] (f(x) - g(x)) dx

D) ∫[g(a), f(b)] (f(x) - g(x)) dx

Correct Answer: C

The area of a region in the plane between two curves is calculated by the definite integral of the upper function minus the lower function over the specified interval. Since it is given that f(x) ≥ g(x) on the interval [a, b], f(x) is the upper function and g(x) is the lower function. Therefore, the correct definite integral is ∫[a, b] (f(x) - g(x)) dx.

The region in the plane is enclosed by the graphs of y = x^3 and y = x. Which of the following expressions calculates the total area of this region?

A) ∫[-1, 1] (x - x^3) dx

B) 2 ∫[0, 1] (x^3 - x) dx

C) ∫[-1, 1] (x^3 - x) dx

D) ∫[-1, 0] (x^3 - x) dx + ∫[0, 1] (x - x^3) dx

Correct Answer: D

First, find the intersection points by setting x^3 = x, which gives x^3 - x = 0, or x(x-1)(x+1) = 0. The intersections are at x = -1, x = 0, and x = 1. On the interval [-1, 0], x^3 ≥ x. For example, at x=-1/2, (-1/2)^3 = -1/8 and x = -1/2, so -1/8 > -1/2. The area for this portion is ∫[-1, 0] (x^3 - x) dx. On the interval [0, 1], x ≥ x^3. For example, at x=1/2, x=1/2 and (1/2)^3 = 1/8. The area for this portion is ∫[0, 1] (x - x^3) dx. The total area is the sum of these two definite integrals.

Let R be the region enclosed by the graphs of y = 8 - x^2 and y = x^2. Which definite integral calculates the area of the portion of R for which x ≥ 0?

A) ∫[0, 2] (x^2 - (8 - x^2)) dx

B) ∫[0, 4] ((8 - x^2) - x^2) dx

C) ∫[0, 2] ((8 - x^2) - x^2) dx

D) ∫[-2, 2] ((8 - x^2) - x^2) dx

Correct Answer: C

First, find the points of intersection by setting 8 - x^2 = x^2, which gives 2x^2 = 8, so x^2 = 4, and x = ±2. The problem specifies the region for x ≥ 0, so the interval of integration is from x = 0 to the positive intersection point, x = 2. On the interval [0, 2], the graph of y = 8 - x^2 is above the graph of y = x^2. The area is calculated by the definite integral of (top function - bottom function), which is ∫[0, 2] ((8 - x^2) - x^2) dx.

The area of the region in the plane enclosed by the graphs of two continuous functions y = f(x) and y = g(x) from x = a to x = b can always be calculated by which of the following definite integrals, regardless of whether f(x) ≥ g(x) or g(x) ≥ f(x) on the interval?

A) |∫[a, b] (f(x) - g(x)) dx|

B) ∫[a, b] (f(x) - g(x)) dx

C) ∫[a, b] |f(x) - g(x)| dx

D) ∫[a, b] (f(x) + g(x)) dx

Correct Answer: C

The definite integral ∫[a, b] (f(x) - g(x)) dx calculates the net signed area between the curves. If g(x) > f(x) on some subinterval, this integral would subtract that area, which is incorrect for finding total geometric area. To calculate the total area, one must integrate the absolute difference between the functions, |f(x) - g(x)|, which represents the positive vertical distance between the curves at every point x. Therefore, ∫[a, b] |f(x) - g(x)| dx correctly calculates the total area. Option A is incorrect because it calculates the absolute value of the net area, not the total area, which can be different if the functions cross.