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This deck focuses on Exploring Accumulations Of Change, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.
Study Exploring Accumulations Of Change 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 is the integral of f(x)=cos(x) from 0 to 2π?
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∫02πcos(x)dx=1. Antiderivative is sin(x), evaluated from 0 to 2π gives 1−0=1.
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This deck focuses on Exploring Accumulations Of Change, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.
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.
Answer: ∫02πcos(x)dx=1. Antiderivative is sin(x), evaluated from 0 to 2π gives 1−0=1.
Answer: 0. Linear function with zero net area due to symmetry about x=2.
Answer: The limit of Riemann sums: limn→∞∑i=1nf(xi∗)Δx. Formal definition using limit of approximating rectangular areas.
Answer: ∫01exdx=e−1. Antiderivative is ex, evaluated from 0 to 1 gives e−1.
Answer: ∫051dx=5. Integral of constant function equals constant times interval length.
Answer: 31∫03x2dx=3. Average value formula: b−a1∫abf(x)dx applied to x2 on [0,3].
Answer: ∫01exdx=e−1. Antiderivative is ex, evaluated from 0 to 1 gives e−1.
Answer: A function F(x) such that F′(x)=f(x) for all x in the domain. Function whose derivative equals the given function.
Answer: 37. Sum of power functions integrated using power rule.
Answer: Area = ∫abf(x)dx. Fundamental connection between integration and area under curves.
Answer: 2. Cubic polynomial integrated using power rule for each term.
Answer: A function F(x) such that F′(x)=f(x) for all x in the domain. Function whose derivative equals the given function.
Answer: 0. Linear function with zero net area due to symmetry about x=2.
Answer: F(x)=x5+C. Power rule for antiderivatives: increase exponent by 1, divide by new exponent.
Answer: ∫132xdx=8. Antiderivative is x2, evaluated from 1 to 3 gives 9−1=8.
Answer: ln(x). By FTC Part 1, derivative of integral equals integrand.
Answer: ∫0πsin(x)dx=2. Antiderivative is −cos(x), giving total area under sine curve.
Answer: 2. Antiderivative is −cos(x), evaluated from 0 to π gives −(−1)−(−1)=2.
Answer: The limit of Riemann sums: limn→∞∑i=1nf(xi∗)Δx. Formal definition using limit of approximating rectangular areas.
Answer: ∫143x2dx=63. Antiderivative is x3, so F(4)−F(1)=64−1=63.
Answer: 65. Sum of power functions integrated using standard power rule.
Answer: ex2. By FTC Part 1, derivative equals the integrand with x substituted.
Answer: ∫035xdx=22.5. Linear function 5x has antiderivative 25x2.
Answer: sin(x). By FTC Part 1, derivative of integral with variable upper limit equals integrand.
Answer: ∃c∈[a,b] such that f(c)=b−a1∫abf(x)dx. Guarantees existence of point where function equals its average value.
Answer: ∫051dx=5. Integral of constant function equals constant times interval length.
Answer: ∫014x3dx=1. Power rule: antiderivative of x3 is 4x4.
Answer: 316. Quadratic function forming parabolic region with positive area.
Answer: ∫abf(x)dx=F(b)−F(a), where F is an antiderivative of f. States that definite integral equals antiderivative evaluated at bounds.
Answer: 65. Quadratic function integrated using power rule for each term.
Answer: 2. Antiderivative is −cos(x), evaluated from 0 to π gives −(−1)−(−1)=2.
Answer: sin(x). By FTC Part 1, derivative of integral with variable upper limit equals integrand.
Answer: ∫132xdx=8. Antiderivative is x2, evaluated from 1 to 3 gives 9−1=8.
Answer: 31∫03x2dx=3. Average value formula: b−a1∫abf(x)dx applied to x2 on [0,3].
Answer: ∫143x2dx=63. Antiderivative is x3, so F(4)−F(1)=64−1=63.
Answer: F(x)=x5+C. Power rule for antiderivatives: increase exponent by 1, divide by new exponent.
Answer: ∫014x3dx=1. Power rule: antiderivative of x3 is 4x4.
Answer: 37. Sum of power functions integrated using power rule.
Answer: ∫02πcos(x)dx=1. Antiderivative is sin(x), evaluated from 0 to 2π gives 1−0=1.
Answer: ∫023dx=6. Integral of constant equals constant times interval width.
Answer: 65. Quadratic function integrated using power rule for each term.
Answer: 2. Cubic polynomial integrated using power rule for each term.
Answer: 12. Antiderivative is x3+x2, evaluated from 0 to 2 gives 8+4=12.
Answer: 12. Antiderivative is x3+x2, evaluated from 0 to 2 gives 8+4=12.
Answer: ex2. By FTC Part 1, derivative equals the integrand with x substituted.
Answer: ∫abf(x)dx=F(b)−F(a), where F is an antiderivative of f. States that definite integral equals antiderivative evaluated at bounds.
Answer: 0. Odd function over symmetric interval gives zero due to cancellation.
Answer: ∫0πsin(x)dx=2. Antiderivative is −cos(x), giving total area under sine curve.
Answer: ∫035xdx=22.5. Linear function 5x has antiderivative 25x2.
Answer: Area = ∫abf(x)dx. Fundamental connection between integration and area under curves.
Answer: 316. Quadratic function forming parabolic region with positive area.
Answer: ∫023dx=6. Integral of constant equals constant times interval width.
Answer: 0. Odd function over symmetric interval gives zero due to cancellation.
Answer: 65. Sum of power functions integrated using standard power rule.
Answer: ∃c∈[a,b] such that f(c)=b−a1∫abf(x)dx. Guarantees existence of point where function equals its average value.
Answer: ln(x). By FTC Part 1, derivative of integral equals integrand.