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AP Calculus BC Flashcards: Evaluating Improper Integrals

Study Evaluating Improper Integrals in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Evaluating Improper Integrals, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Calculus BC Flashcards: Evaluating Improper Integrals

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QUESTION

Evaluate: ∫1∞e−x dx\int_{1}^{\infty} e^{-x} \, dx∫1∞​e−xdx.

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ANSWER

Converges to 1e\frac{1}{e}e1​. lim⁡b→∞[−e−x]1b=0−(−e−1)=1e\lim_{b \to \infty} [-e^{-x}]_1^b = 0 - (-e^{-1}) = \frac{1}{e}limb→∞​[−e−x]1b​=0−(−e−1)=e1​

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Flashcard 1: Evaluate: ∫1∞e−x dx\int_{1}^{\infty} e^{-x} \, dx∫1∞​e−xdx.

Answer: Converges to 1e\frac{1}{e}e1​. lim⁡b→∞[−e−x]1b=0−(−e−1)=1e\lim_{b \to \infty} [-e^{-x}]_1^b = 0 - (-e^{-1}) = \frac{1}{e}limb→∞​[−e−x]1b​=0−(−e−1)=e1​

Flashcard 2: Find the limit: lim⁡b→∞∫1b1x3dx\textstyle \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^3} dxlimb→∞​∫1b​x31​dx.

Answer: Converges to 12\frac{1}{2}21​. lim⁡b→∞[−12x2]1b=0−(−12)=12\lim_{b \to \infty} [-\frac{1}{2x^2}]_1^b = 0 - (-\frac{1}{2}) = \frac{1}{2}limb→∞​[−2x21​]1b​=0−(−21​)=21​.

Flashcard 3: Evaluate limit: lim⁡a→0+∫a11xdx\textstyle \lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x} dxlima→0+​∫a1​x1​dx.

Answer: Diverges. lim⁡a→0+[ln⁡x]a1=0−(−∞)=∞\lim_{a \to 0^+} [\ln x]_a^1 = 0 - (-\infty) = \inftylima→0+​[lnx]a1​=0−(−∞)=∞.

Flashcard 4: Identify singularity: ∫1∞1(x−1)2dx\int_{1}^{\infty} \frac{1}{(x-1)^2} dx∫1∞​(x−1)21​dx.

Answer: Unbounded integrand at x=1x=1x=1. Function becomes infinite at the lower limit x=1x = 1x=1.

Flashcard 5: Evaluate: ∫0111−x2 dx\textstyle \int_{0}^{1} \frac{1}{\sqrt{1-x^2}} \, dx∫01​1−x2​1​dx.

Answer: Converges to π2\frac{\pi}{2}2π​. Arcsine function evaluation: arcsin⁡(1)−arcsin⁡(0)=π2\arcsin(1) - \arcsin(0) = \frac{\pi}{2}arcsin(1)−arcsin(0)=2π​.

Flashcard 6: Identify the singularity: ∫011x dx\textstyle \int_{0}^{1} \frac{1}{x} \, dx∫01​x1​dx.

Answer: Unbounded integrand at x=0x = 0x=0. Integrand has vertical asymptote at the lower limit.

Flashcard 7: Determine convergence: ∫0∞e−x2dx\textstyle \int_{0}^{\infty} e^{-x^2} \text{dx}∫0∞​e−x2dx

Answer: Converges. Gaussian integral converges due to exponential decay.

Flashcard 8: Evaluate: ∫1∞1x(ln⁡x)2 dx\textstyle \int_{1}^{\infty} \frac{1}{x (\ln x)^2} \, dx∫1∞​x(lnx)21​dx.

Answer: Converges. Higher power in denominator ensures convergence.

Flashcard 9: Identify the type of improper integral: ∫011xdx\int_{0}^{1} \frac{1}{\sqrt{x}} \text{dx}∫01​x​1​dx.

Answer: Unbounded integrand at x=0x = 0x=0. 1x\frac{1}{\sqrt{x}}x​1​ becomes infinite as x→0+.x \to 0^+.x→0+.

Flashcard 10: Identify convergence: ∫1∞1x dx\int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx∫1∞​x​1​dx.

Answer: Diverges. Since p=12≤1p = \frac{1}{2} \leq 1p=21​≤1, the integral diverges.

Flashcard 11: Convert to limit: ∫011x dx\int_{0}^{1} \frac{1}{x} \, \mathrm{d}x∫01​x1​dx.

Answer: lim⁡a→0+∫a11x dx\lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x} \, \mathrm{d}xlima→0+​∫a1​x1​dx. Integrand 1x\frac{1}{x}x1​ is unbounded at lower limit x=0x = 0x=0.

Flashcard 12: Determine convergence: ∫1∞1xdx\textstyle \int_{1}^{\infty} \frac{1}{x} \text{dx}∫1∞​x1​dx.

Answer: Diverges. ln⁡(x)\ln(x)ln(x) grows without bound, so the integral diverges.

Flashcard 13: Determine behavior: ∫011xdx\int_{0}^{1} \frac{1}{\sqrt{x}} \mathrm{d}x∫01​x​1​dx.

Answer: Converges. Power −12>−1-\frac{1}{2} > -1−21​>−1 ensures convergence at x=0x = 0x=0.

Flashcard 14: Convert to limit: ∫0∞xe−x dx\textstyle \int_{0}^{\infty} x e^{-x} \, dx∫0∞​xe−xdx.

Answer: lim⁡b→∞∫0bxe−x dx\textstyle \lim_{b \to \infty} \int_{0}^{b} x e^{-x} \, dxlimb→∞​∫0b​xe−xdx. Infinite upper limit requires limit evaluation.

Flashcard 15: State the condition for the convergence of ∫0a(a−x)−1/2 dx\textstyle \int_{0}^{a} (a-x)^{-1/2} \, dx∫0a​(a−x)−1/2dx.

Answer: Converges if a>0a > 0a>0. For a>0a > 0a>0, the singularity at x=ax = ax=a is integrable.

Flashcard 16: State the behavior of ∫01x−1dx\int_0^1 x^{-1} \text{dx}∫01​x−1dx

Answer: Diverges. Same as ∫011xdx\int_0^1 \frac{1}{x} dx∫01​x1​dx which diverges.

Flashcard 17: Evaluate: \textstyle \text{∫}_{0}^{1} \frac{1}{\text{√}x} \text{dx}.

Answer: Converges to 222. lim⁡a→0+[2x]a1=21−0=2\lim_{a \to 0^+} [2\sqrt{x}]_a^1 = 2\sqrt{1} - 0 = 2lima→0+​[2x​]a1​=21​−0=2.

Flashcard 18: What does it mean if an improper integral converges?

Answer: The limit defining the integral exists and is finite. The limit exists and equals a finite real number.

Flashcard 19: Explain divergence in the context of improper integrals.

Answer: The integral's limit is infinite or does not exist. The limit is infinite or undefined.

Flashcard 20: When does an integral ∫abf(x) dx\int_{a}^{b} f(x) \, dx∫ab​f(x)dx become improper?

Answer: If aaa, bbb, or f(x)f(x)f(x) are infinite or unbounded. Any infinite bound or unbounded integrand makes it improper.

Flashcard 21: What is the behavior of ∫−∞∞e−x2 dx\int_{-\infty}^{\infty} e^{-x^2} \, dx∫−∞∞​e−x2dx?

Answer: Converges to π\sqrt{\pi}π​. Famous Gaussian integral equals π\sqrt{\pi}π​.

Flashcard 22: What is an improper integral?

Answer: An integral with infinite limits or an unbounded integrand. Defined when limits of integration or integrand are infinite.

Flashcard 23: Convert to limit form: ∫−∞0ex dx\int_{-\infty}^{0} e^{x} \, dx∫−∞0​exdx.

Answer: lim⁡a→−∞∫a0ex dx\lim_{a \to -\infty} \int_{a}^{0} e^{x} \, dxlima→−∞​∫a0​exdx. Negative infinite lower limit requires limit form.

Flashcard 24: What is the behavior of \textstyle \text{∫}_{0}^{1} x^{-1/3} \text{dx}?

Answer: Converges. Power −13>−1-\frac{1}{3} > -1−31​>−1 ensures convergence.

Flashcard 25: Evaluate: \textstyle \text{∫}_{0}^{\text{∞}} e^{-x} \text{dx}.

Answer: Converges to 111. lim⁡b→∞[−e−x]0b=0−(−1)=1\lim_{b \to \infty} [-e^{-x}]_0^b = 0 - (-1) = 1limb→∞​[−e−x]0b​=0−(−1)=1.

Flashcard 26: What is convergence for ∫1∞1x4 dx\int_1^\infty \frac{1}{x^4} \, dx∫1∞​x41​dx?

Answer: Converges to 13\frac{1}{3}31​. lim⁡b→∞[−13x3]1b=0−(−13)=13\lim_{b \to \infty} [-\frac{1}{3x^3}]_1^b = 0 - (-\frac{1}{3}) = \frac{1}{3}limb→∞​[−3x31​]1b​=0−(−31​)=31​.

Flashcard 27: What is the behavior of \textstyle \text{∫}_{-\text{∞}}^{0} e^x \text{dx}?

Answer: Converges to 111. lim⁡a→−∞[ex]a0=1−0=1\lim_{a \to -\infty} [e^x]_a^0 = 1 - 0 = 1lima→−∞​[ex]a0​=1−0=1.

Flashcard 28: Evaluate: \textstyle \text{∫}_{1}^{\text{∞}} \frac{1}{x^2} \text{dx}.

Answer: Converges to 111. lim⁡b→∞[−1x]1b=lim⁡b→∞(0+1)=1\lim_{b \to \infty} [-\frac{1}{x}]_1^b = \lim_{b \to \infty} (0 + 1) = 1limb→∞​[−x1​]1b​=limb→∞​(0+1)=1.

Flashcard 29: Determine behavior: ∫011xpdx\textstyle \int_{0}^{1} \frac{1}{x^p} dx∫01​xp1​dx for p<1p<1p<1.

Answer: Converges. When p<1p < 1p<1, the singularity at x=0x = 0x=0 is integrable.

Flashcard 30: Identify convergence: ∫1∞1xdx\int_{1}^{\infty} \frac{1}{\sqrt{x}} dx∫1∞​x​1​dx.

Answer: Diverges. Since p=12≤1p = \frac{1}{2} \leq 1p=21​≤1, the integral diverges.