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

Correct Answer: B

First, find the points of intersection by setting f(x) = g(x): x^3 - 3x = x, which simplifies to x^3 - 4x = 0, or x(x-2)(x+2) = 0. The points of intersection are x = -2, 0, and 2. On the interval [-2, 0], f(x) ≥ g(x), so the area is ∫[-2, 0] ((x^3 - 3x) - x) dx = ∫[-2, 0] (x^3 - 4x) dx. On the interval [0, 2], g(x) ≥ f(x), so the area is ∫[0, 2] (x - (x^3 - 3x)) dx = ∫[0, 2] (4x - x^3) dx. The total area is the sum of these two integrals.

Correct Answer: B

The total area must be calculated as a sum of integrals over the intervals where the upper and lower functions do not change. On [a, b], the area is ∫[a, b] (f(x) - g(x)) dx because f(x) is the upper function. On [b, c], the area is ∫[b, c] (g(x) - f(x)) dx because g(x) is the upper function. The total area is the sum of these two definite integrals.

Correct Answer: A

The intersection points are found by solving x^3 = 4x, which gives x^3 - 4x = 0, or x(x-2)(x+2) = 0. The intersections are at x = -2, 0, and 2. The total area between two curves f(x) and g(x) from x=a to x=b can be expressed as a single definite integral of the absolute value of their difference: ∫[a, b] |f(x) - g(x)| dx. Therefore, the total area is correctly represented by ∫[-2, 2] |4x - x^3| dx.

Correct Answer: A

First, find the intersections: sin(x) = sin^2(x) implies sin(x) - sin^2(x) = 0, or sin(x)(1 - sin(x)) = 0. This occurs when sin(x) = 0 (at x=0, π) or sin(x) = 1 (at x=π/2). However, on the interval (0, π), the value of sin(x) is between 0 and 1. For any number z between 0 and 1, z ≥ z^2. Therefore, sin(x) ≥ sin^2(x) for all x in [0, π]. Because the upper function does not change, the area can be calculated with a single integral: ∫[0, π] (sin(x) - sin^2(x)) dx.

Correct Answer: C

We need the area between y = 4x - x^3 and y = 3x in the first quadrant. First, find the intersection points: 4x - x^3 = 3x, which simplifies to x - x^3 = 0, or x(1 - x^2) = 0. The intersections are at x = 0, x = 1, and x = -1. Since we are in the first quadrant, we consider the interval [0, 1]. On this interval, test a point like x=0.5: 4(0.5) - (0.5)^3 = 2 - 0.125 = 1.875, while 3(0.5) = 1.5. Thus, y = 4x - x^3 is the upper curve. The area is the integral of the upper curve minus the lower curve over the interval [0, 1], which is ∫[0, 1] ((4x - x^3) - 3x) dx.

Correct Answer: C

To find the intersections, set f(x) = g(x): x^3 - 2x^2 - x + 3 = x^2 - x - 1, which simplifies to x^3 - 3x^2 + 4 = 0. By testing integer factors of 4, we find x=-1, x=2 are roots. (x+1)(x-2)^2 = 0. The functions intersect at x=-1 and are tangent at x=2. On the interval (-1, 2), f(x) > g(x). However, the question asks why a single integral might be incorrect in a general case of multiple intersections. The reason a single integral fails is when the curves cross within the interval of integration, requiring the area to be calculated as a sum of two or more definite integrals. In this specific case, the single integral is actually correct because f(x) ≥ g(x) on [-1, 2], but the principle being tested is the need to check for interior intersection points where the top and bottom functions might switch.

Correct Answer: C

First, find the intersections: x^4 - 4x^2 = x^2 - 4, which gives x^4 - 5x^2 + 4 = 0. This factors as (x^2 - 1)(x^2 - 4) = 0, so the intersections are at x = ±1 and x = ±2. Let h(x) = (x^4 - 4x^2) - (x^2 - 4) = x^4 - 5x^2 + 4. On [-2, -1], h(x) ≥ 0. On [-1, 1], h(x) ≤ 0. On [1, 2], h(x) ≥ 0. Therefore, the area is ∫[-2, -1] h(x) dx + ∫[-1, 1] (-h(x)) dx + ∫[1, 2] h(x) dx. This corresponds to option C.