AP Calculus BC Practice Quiz: Evaluating Improper Integrals (BC ONLY)

Correct Answer: C

The integral ∫ from 0 to 2 of (1/(x-2)) dx is improper because the integrand f(x) = 1/(x-2) has an infinite discontinuity (is unbounded) at x=2, which is an endpoint of the interval of integration. Option A is improper due to an infinite limit, not an unbounded integrand. Option B's integrand is continuous and bounded on [-1, 1]. Option D is a standard definite integral.

Correct Answer: B

According to the definition for determining an improper integral with an infinite upper limit, the integral from a to ∞ of f(x) dx is found by taking the limit of the definite integral from a to b as b approaches ∞. Therefore, ∫ from 1 to ∞ of (1/x³) dx is determined by lim (b→∞) ∫ from 1 to b of (1/x³) dx.

Correct Answer: A

First, set up the limit: lim (b→∞) ∫ from 2 to b of x⁻² dx. The antiderivative of x⁻² is -x⁻¹ or -1/x. Evaluating the definite integral gives lim (b→∞) [-1/x] from 2 to b = lim (b→∞) [-1/b - (-1/2)] = lim (b→∞) [-1/b + 1/2]. As b→∞, the term -1/b approaches 0. Thus, the limit is 0 + 1/2 = 1/2.

Correct Answer: B

The integrand is unbounded at x=0. We evaluate this using a limit: lim (a→0⁺) ∫ from a to 9 of x⁻¹/² dx. The antiderivative of x⁻¹/² is 2x¹/². Evaluating the definite integral gives lim (a→0⁺) [2√x] from a to 9 = lim (a→0⁺) [2√9 - 2√a] = 2(3) - 2(0) = 6.

Correct Answer: D

To evaluate the integral, we use a limit: lim (b→∞) ∫ from 1 to b of (1/x) dx. The antiderivative of 1/x is ln|x|. So we have lim (b→∞) [ln|x|] from 1 to b = lim (b→∞) [ln(b) - ln(1)]. Since ln(1) = 0, this simplifies to lim (b→∞) ln(b). As b approaches ∞, ln(b) also approaches ∞. Because the limit is infinite, the integral diverges.

Correct Answer: D

The integrand f(x) = 1/x² is unbounded at x=0. We must use a limit: lim (b→0⁻) ∫ from -2 to b of x⁻² dx. The antiderivative is -1/x. The limit becomes lim (b→0⁻) [-1/x] from -2 to b = lim (b→0⁻) [-1/b - (-1/-2)] = lim (b→0⁻) [-1/b - 1/2]. As b approaches 0 from the left, -1/b approaches +∞. Since the limit is infinite, the integral diverges.

Correct Answer: B

This integral has two infinite limits and must be split into two separate improper integrals, for example at x=0: ∫ from -∞ to 0 of (1/(x²+1)) dx + ∫ from 0 to ∞ of (1/(x²+1)) dx. Let's evaluate the second integral: lim (b→∞) ∫ from 0 to b of (1/(x²+1)) dx = lim (b→∞) [arctan(x)] from 0 to b = lim (b→∞) [arctan(b) - arctan(0)] = π/2 - 0 = π/2. Since the integrand is an even function, the first integral is also π/2. The total value is π/2 + π/2 = π.

Correct Answer: A

The integrand f(x) = x⁻¹/³ is unbounded at x=0, which is within the interval [-1, 8]. The integral must be split: ∫ from -1 to 0 of x⁻¹/³ dx + ∫ from 0 to 8 of x⁻¹/³ dx. The antiderivative is (3/2)x²/³. For the first part: lim (b→0⁻) [(3/2)x²/³] from -1 to b = lim (b→0⁻) [(3/2)b²/³ - (3/2)(-1)²/³] = 0 - 3/2 = -3/2. For the second part: lim (a→0⁺) [(3/2)x²/³] from a to 8 = lim (a→0⁺) [(3/2)(8)²/³ - (3/2)a²/³] = (3/2)(4) - 0 = 6. The total value is -3/2 + 6 = 9/2.

Correct Answer: B

First, find the antiderivative of x*e⁻ˣ using integration by parts. Let u=x and dv=e⁻ˣ dx. Then du=dx and v=-e⁻ˣ. The antiderivative is uv - ∫v du = -x*e⁻ˣ - ∫(-e⁻ˣ) dx = -x*e⁻ˣ - e⁻ˣ. Now, evaluate the improper integral using a limit: lim (b→∞) [-x*e⁻ˣ - e⁻ˣ] from 0 to b = lim (b→∞) [(-b*e⁻ᵇ - e⁻ᵇ) - (-0*e⁰ - e⁰)] = lim (b→∞) [-b/eᵇ - 1/eᵇ + 1]. The limits of b/eᵇ and 1/eᵇ as b→∞ are both 0 (the first can be found with L'Hôpital's Rule). The result is 0 - 0 + 1 = 1.