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AP Calculus AB Flashcards: Selecting Techniques For Antidifferentiation

Study Selecting Techniques For Antidifferentiation in AP Calculus AB 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 Selecting Techniques For Antidifferentiation, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.

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 AB Flashcards: Selecting Techniques For Antidifferentiation

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QUESTION

Find the antiderivative of xexxe^{x}xex.

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ANSWER

Integration by parts. Product of polynomial and exponential requires integration by parts.

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Flashcard 1: Find the antiderivative of xexxe^{x}xex.

Answer: Integration by parts. Product of polynomial and exponential requires integration by parts.

Flashcard 2: What is the antiderivative of 1x\frac{1}{x}x1​?

Answer: ln∣x∣+C\text{ln}|x| + Cln∣x∣+C. The derivative of ln∣x∣\text{ln}|x|ln∣x∣ is 1x\frac{1}{x}x1​.

Flashcard 3: What is the antiderivative of xnx^{n}xn?

Answer: xn+1n+1+C\frac{x^{n+1}}{n+1} + Cn+1xn+1​+C for n≠−1n \neq -1n=−1. Power rule for integration increases exponent by 1 and divides.

Flashcard 4: Identify the technique for 1sqrt(x2+a2)\frac{1}{\text{sqrt}(x^2 + a^2)}sqrt(x2+a2)1​?

Answer: Trigonometric substitution. Use x=atan(θ)x = a\text{tan}(\theta)x=atan(θ) to simplify the square root.

Flashcard 5: What is the antiderivative of sec⁡(x)tan⁡(x)\sec(x)\tan(x)sec(x)tan(x)?

Answer: sec⁡(x)+C\sec(x) + Csec(x)+C. The derivative of sec⁡(x)\sec(x)sec(x) is sec⁡(x)tan⁡(x)\sec(x)\tan(x)sec(x)tan(x).

Flashcard 6: What substitution is used for sqrt(x2+a2)\text{sqrt}(x^2 + a^2)sqrt(x2+a2)?

Answer: Trigonometric substitution. Use x=atan(θ)x = a\text{tan}(\theta)x=atan(θ) to handle the square root expression.

Flashcard 7: Choose the technique for ln(ax)\text{ln}(a^x)ln(ax).

Answer: Substitution. Simplify ln(ax)=xln(a)\text{ln}(a^x) = x\text{ln}(a)ln(ax)=xln(a) before integrating.

Flashcard 8: Find the antiderivative of sec2(x)\text{sec}^2(x)sec2(x).

Answer: tan(x)+C\text{tan}(x) + Ctan(x)+C. The derivative of tan(x)\text{tan}(x)tan(x) is sec2(x)\text{sec}^2(x)sec2(x).

Flashcard 9: What is the antiderivative of axa^xax?

Answer: axln(a)+C\frac{a^x}{\text{ln}(a)} + Cln(a)ax​+C. General exponential function antiderivative formula with base aaa.

Flashcard 10: Select the technique for ex2e^{x^2}ex2.

Answer: No elementary antiderivative. Cannot be expressed using elementary functions.

Flashcard 11: Identify the integration technique for ln(x)\text{ln}(x)ln(x).

Answer: Integration by parts. Use u=ln(x)u = \text{ln}(x)u=ln(x) and dv=dxdv = dxdv=dx for integration by parts.

Flashcard 12: What is the antiderivative of sin2(x)\text{sin}^2(x)sin2(x)?

Answer: x2−14sin(2x)+C\frac{x}{2} - \frac{1}{4}\text{sin}(2x) + C2x​−41​sin(2x)+C. Use double angle identity: sin2(x)=1−cos(2x)2\text{sin}^2(x) = \frac{1 - \text{cos}(2x)}{2}sin2(x)=21−cos(2x)​.

Flashcard 13: What is the antiderivative of cos2(x)\text{cos}^2(x)cos2(x)?

Answer: x2+14sin(2x)+C\frac{x}{2} + \frac{1}{4}\text{sin}(2x) + C2x​+41​sin(2x)+C. Use double angle identity: cos2(x)=1+cos(2x)2\text{cos}^2(x) = \frac{1 + \text{cos}(2x)}{2}cos2(x)=21+cos(2x)​.

Flashcard 14: Identify the technique for x2ln(x)x^2 \text{ln}(x)x2ln(x).

Answer: Integration by parts. Polynomial times logarithm requires integration by parts.

Flashcard 15: Which substitution is used for 1sqrt(a2−x2)\frac{1}{\text{sqrt}(a^2 - x^2)}sqrt(a2−x2)1​?

Answer: Trigonometric substitution. Use x=asin(θ)x = a\text{sin}(\theta)x=asin(θ) to eliminate the square root.

Flashcard 16: What is the antiderivative of csc2(x)\text{csc}^2(x)csc2(x)?

Answer: −cot(x)+C-\text{cot}(x) + C−cot(x)+C. The derivative of −cot(x)-\text{cot}(x)−cot(x) is csc2(x)\text{csc}^2(x)csc2(x).

Flashcard 17: Identify the technique for arccos(x)\text{arccos}(x)arccos(x).

Answer: Integration by parts. Inverse trig functions require integration by parts technique.

Flashcard 18: Which technique is suitable for xx2+1\frac{x}{x^2 + 1}x2+1x​?

Answer: Substitution. Use u=x2+1u = x^2 + 1u=x2+1 to simplify the integrand.

Flashcard 19: What is the antiderivative of tan(x)\text{tan}(x)tan(x)?

Answer: −ln∣cos(x)∣+C-\text{ln}|\text{cos}(x)| + C−ln∣cos(x)∣+C. Rewrite tan(x)=sin(x)cos(x)\text{tan}(x) = \frac{\text{sin}(x)}{\text{cos}(x)}tan(x)=cos(x)sin(x)​ and use substitution.

Flashcard 20: What is the antiderivative of 1x2\frac{1}{x^2}x21​?

Answer: −1x+C-\frac{1}{x} + C−x1​+C. Rewrite as x−2x^{-2}x−2 and apply power rule.

Flashcard 21: Identify the technique for e3xsin(x)e^{3x}\text{sin}(x)e3xsin(x).

Answer: Integration by parts. Product of exponential and sine requires integration by parts.

Flashcard 22: What is the antiderivative of 1x2+a2\frac{1}{x^2 + a^2}x2+a21​?

Answer: 1aarctan⁡(xa)+C\frac{1}{a}\arctan(\frac{x}{a}) + Ca1​arctan(ax​)+C. Standard arctangent formula with scaling factor 1a\frac{1}{a}a1​.

Flashcard 23: Which method is suitable for xcos(x)x\text{cos}(x)xcos(x)?

Answer: Integration by parts. Product of polynomial and cosine requires integration by parts.

Flashcard 24: What is the antiderivative of 1sqrt(1−x2)\frac{1}{\text{sqrt}(1-x^2)}sqrt(1−x2)1​?

Answer: arcsin(x)+C\text{arcsin}(x) + Carcsin(x)+C. Standard inverse sine antiderivative formula.

Flashcard 25: Which technique is appropriate for e2xe^{2x}e2x?

Answer: Substitution. Use u=2xu = 2xu=2x substitution to handle the coefficient.

Flashcard 26: What is the antiderivative of exe^xex?

Answer: ex+Ce^x + Cex+C. The derivative of exe^xex is itself, so the antiderivative reverses this.

Flashcard 27: Identify the integration technique for xsin(x)x\text{sin}(x)xsin(x).

Answer: Integration by parts. Product of xxx and trig function requires integration by parts.

Flashcard 28: State the antiderivative of sin(x)\text{sin}(x)sin(x).

Answer: −cos(x)+C-\text{cos}(x) + C−cos(x)+C. The derivative of −cos(x)-\text{cos}(x)−cos(x) is sin(x)\text{sin}(x)sin(x).

Flashcard 29: What is the antiderivative of arcsin(x)\text{arcsin}(x)arcsin(x)?

Answer: xarcsin(x)+sqrt(1−x2)+Cx \text{arcsin}(x) + \text{sqrt}(1 - x^2) + Cxarcsin(x)+sqrt(1−x2)+C. Standard result from integration by parts formula.

Flashcard 30: Choose the integration technique for arctan(x)\text{arctan}(x)arctan(x).

Answer: Integration by parts. Inverse trig functions require integration by parts technique.