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

Correct Answer: A

The area between two curves f(x) and g(x) from x=a to x=b is given by the definite integral ∫ₐᵇ (top function - bottom function) dx. In the interval [0, 2], the graph of y = x² + 1 is always above the graph of y = x. Therefore, the area is calculated by the integral of ((x² + 1) - x) over the interval [0, 2], which simplifies to ∫₀² (x² - x + 1) dx.

Correct Answer: B

First, find the points of intersection by setting the functions equal: 4 - x² = 3, which gives x² = 1, so x = ±1. Since the question specifies the first quadrant, the region is bounded by x=0 and x=1. The area is calculated using a definite integral of the top function minus the bottom function. In the interval [0, 1], y = 4 - x² is above y = 3. The integral is ∫₀¹ ((4 - x²) - 3) dx = ∫₀¹ (1 - x²) dx. Evaluating this definite integral gives [x - (x³/3)] from 0 to 1, which is (1 - 1/3) - 0 = 2/3.

Correct Answer: C

To find the bounds of integration, set the two functions equal to find their intersection points: x² = 2x, which leads to x² - 2x = 0, or x(x - 2) = 0. The points of intersection are at x = 0 and x = 2. In the interval [0, 2], the line y = 2x is above the parabola y = x². The area is given by the definite integral ∫₀² (2x - x²) dx. The antiderivative is x² - (x³/3). Evaluating from 0 to 2 gives (2² - 2³/3) - (0² - 0³/3) = 4 - 8/3 = 4/3.

Correct Answer: D

First, find the intersection points of y = x³ and y = x by solving x³ = x, which gives x = -1, 0, and 1. In the first quadrant, the region is considered from x=0 to x=2. The top and bottom functions switch at the intersection point x = 1. For the interval [0, 1], y = x is above y = x³. For the interval [1, 2], y = x³ is above y = x. Therefore, the total area must be calculated as the sum of two definite integrals: ∫₀¹ (x - x³) dx + ∫₁² (x³ - x) dx.

Correct Answer: C

The area between two curves f(x) and g(x) over an interval [a, b] is given by the definite integral ∫ₐᵇ |f(x) - g(x)| dx. In the interval [0, π/2], the functions y = sin(x) and y = cos(x) cross at x = π/4. From x = 0 to x = π/4, cos(x) ≥ sin(x), and from x = π/4 to x = π/2, sin(x) ≥ cos(x). Using the absolute value correctly accounts for which function is on top throughout the entire interval without needing to split the integral. Option D incorrectly switches the integrands for the two intervals.

Correct Answer: C

The region is bounded by the curves y = eˣ, y = 1, and the vertical line x = 2. The intersection of y = eˣ and y = 1 occurs when eˣ = 1, which is at x = 0. This serves as the left bound of integration. The right bound is given as x = 2. Over the interval [0, 2], the function y = eˣ is always above the line y = 1. The area is calculated by the definite integral ∫₀² (eˣ - 1) dx. The antiderivative is eˣ - x. Evaluating this from 0 to 2 gives (e² - 2) - (e⁰ - 0) = (e² - 2) - 1 = e² - 3.

Correct Answer: D

The area between curves is calculated by integrating (top function - bottom function). The problem states that from x = a to x = c, f(x) is the top function, so the area on this interval is ∫ₐᶜ (f(x) - g(x)) dx. From x = c to x = b, g(x) is the top function, so the area on this interval is ∫ᶜᵇ (g(x) - f(x)) dx. The total area is the sum of these two definite integrals.