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AP Calculus AB Practice Quiz: Finding the Area Between Curves Expressed as Functions of $y$

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

A region in the xy-plane is bounded on the right by the curve x = f(y) and on the left by the curve x = g(y) over the interval c ≤ y ≤ d, where f(y) ≥ g(y). Which definite integral represents the area of this region?

All Questions (7)

A region in the xy-plane is bounded on the right by the curve x = f(y) and on the left by the curve x = g(y) over the interval c ≤ y ≤ d, where f(y) ≥ g(y). Which definite integral represents the area of this region?

A) ∫[c, d] (f(x) - g(x)) dx

B) ∫[c, d] (f(y) - g(y)) dy

C) ∫[c, d] (g(y) - f(y)) dy

D) ∫[c, d] (f(y) + g(y)) dy

Correct Answer: B

To find the area between curves expressed as functions of y, we integrate the difference between the right-hand curve and the left-hand curve with respect to y. The integral is set up as ∫[right curve - left curve] dy over the interval of y-values that bounds the region.

To calculate the area of the region bounded by the curves x = y² and x = y + 2, one must set up a definite integral with respect to y. What is the first step in determining the limits of integration?

A) Find the y-intercepts of both curves.

B) Set the two expressions for x equal to each other and solve for y.

C) Differentiate both functions to find their critical points.

D) Integrate both functions from y = 0 to y = 1.

Correct Answer: B

The limits of integration for the area between two curves are the y-values of their intersection points. To find these points, we set the functions equal to each other (y² = y + 2) and solve for y. This directly applies the principle of using a definite integral to calculate the area between curves.

Consider the region bounded by the curve y = √x, the y-axis (x=0), and the horizontal line y = 3. Which statement best explains why calculating the area using an integral with respect to y is often more direct than using an integral with respect to x for this specific region?

A) Integrating with respect to y is always easier than integrating with respect to x.

B) The region is bounded by a single continuous function on the right (x = y²) and a single continuous function on the left (x = 0) over the entire interval in y.

C) The limits of integration are integers when using y, but not when using x.

D) The function x = y² is simpler to integrate than the function y = √x.

Correct Answer: B

Areas can be calculated using functions of either x or y. For this region, if we integrate with respect to y, the right boundary is always x = y² and the left boundary is always x = 0, from y=0 to y=3. This requires only one integral. Integrating with respect to x would require splitting the region or using a different geometric formula, as the top boundary is not a single function of x.

When using a definite integral to find the area of a region bounded by curves expressed as functions of y, the area is conceptualized as a sum of the areas of infinitesimally thin rectangles. What is the orientation of these representative rectangles?

A) Vertical, with a width of Δx.

B) Horizontal, with a height of Δy.

C) Vertical, with a width of Δy.

D) Horizontal, with a height of Δx.

Correct Answer: B

When integrating with respect to y (using dy), the definite integral sums the areas of horizontal rectangles. Each rectangle has a height (or thickness) of Δy (which becomes dy in the integral) and a length determined by the difference between the right and left functions of y.

A student attempts to find the area of the region enclosed by x = f(y) and x = g(y) from y = c to y = d by calculating ∫[c, d] (f(y) - g(y)) dy. The result is negative. What is the most likely reason for this result?

A) The area of a region cannot be calculated using an integral with respect to y.

B) The student incorrectly integrated the functions f(y) and g(y).

C) Throughout the interval [c, d], the curve x = g(y) was to the right of the curve x = f(y).

D) The functions f(y) and g(y) are not continuous on the interval [c, d].

Correct Answer: C

The definite integral ∫[right curve - left curve] dy calculates the area. If the student calculated ∫[f(y) - g(y)] dy and got a negative answer, it implies that for most or all of the interval, g(y) > f(y), meaning the 'left' curve was subtracted from the 'right' curve, resulting in the negative of the area.

To find the area of the region bounded by the y-axis and the curve x = 4y - y², which integral should be set up?

A) ∫[0, 4] (4y - y²) dy

B) ∫[0, 2] (4y - y²) dy

C) ∫[0, 4] (4x - x²) dx

D) ∫[0, 2] (4x - x²) dx

Correct Answer: A

The region is bounded by x = 4y - y² and the y-axis (x = 0). First, find the intersection points by setting 4y - y² = 0, which gives y(4 - y) = 0, so y = 0 and y = 4. These are the limits of integration. Since 4y - y² is positive for y between 0 and 4, this curve is to the right of the y-axis. The area is the definite integral of the 'right curve' (4y - y²) minus the 'left curve' (0) with respect to y from 0 to 4.

The area of a region can be expressed as two separate integrals with respect to x, ∫[a, c] (f(x) - g(x)) dx + ∫[c, b] (h(x) - g(x)) dx. If this same region can be described by a single definite integral with respect to y, ∫[d, e] (p(y) - q(y)) dy, what does this imply about the region's boundaries?

A) The region has a top boundary defined by two different functions of x, but a right boundary defined by a single function of y.

B) The region is a simple rectangle.

C) It is impossible for an area requiring two integrals in x to require only one integral in y.

D) The functions f(x), g(x), and h(x) must be inverse functions of p(y) and q(y).

Correct Answer: A

The fact that the area calculation requires splitting the integral with respect to x implies that the 'top' or 'bottom' boundary function changes at x=c. The fact that it can be calculated with a single integral with respect to y implies that for the entire vertical span of the region (from y=d to y=e), the 'right' boundary and 'left' boundary are each described by a single, continuous function of y. This highlights the strategic choice of which variable to integrate with respect to, a key concept in calculating areas.