All flashcards
Flashcard 1: What is the integral of f(x)=sin(x) from 0 to π?
Answer: ∫0πsin(x)dx=2. Antiderivative is −cos(x), giving total area under sine curve.
Flashcard 2: What is the integral of f(x)=5x from 0 to 3?
Answer: ∫035xdx=22.5. Linear function 5x has antiderivative 25x2.
Flashcard 3: What is the integral of f(x)=4x3 from 0 to 1?
Answer: ∫014x3dx=1. Power rule: antiderivative of x3 is 4x4.
Flashcard 4: What is the integral of f(x)=2x from 1 to 3?
Answer: ∫132xdx=8. Antiderivative is x2, evaluated from 1 to 3 gives 9−1=8.
Flashcard 5: Find the antiderivative of f(x)=5x4.
Answer: F(x)=x5+C. Power rule for antiderivatives: increase exponent by 1, divide by new exponent.
Flashcard 6: State the formula for the area under a curve using integration.
Answer: Area = ∫abf(x)dx. Fundamental connection between integration and area under curves.
Flashcard 7: What is the integral of f(x)=ex from 0 to 1?
Answer: ∫01exdx=e−1. Antiderivative is ex, evaluated from 0 to 1 gives e−1.
Flashcard 8: What is the integral of f(x)=cos(x) from 0 to 2π?
Answer: ∫02πcos(x)dx=1. Antiderivative is sin(x), evaluated from 0 to 2π gives 1−0=1.
Flashcard 9: What is the integral of f(x)=1 from 0 to 5?
Answer: ∫051dx=5. Integral of constant function equals constant times interval length.
Flashcard 10: State the definition of an antiderivative.
Answer: A function F(x) such that F′(x)=f(x) for all x in the domain. Function whose derivative equals the given function.
Flashcard 11: Compute ∫02(4x−x2)dx.
Answer: 316. Quadratic function forming parabolic region with positive area.
Flashcard 12: Find dxd∫0xet2dt.
Answer: ex2. By FTC Part 1, derivative equals the integrand with x substituted.
Flashcard 13: Compute ∫01(x2−x+1)dx.
Answer: 65. Quadratic function integrated using power rule for each term.
Flashcard 14: Find ∫02(x3−3x2+3x)dx.
Answer: 2. Cubic polynomial integrated using power rule for each term.
Flashcard 15: Evaluate ∫01(x4−x2+1)dx.
Answer: 65. Sum of power functions integrated using standard power rule.
Flashcard 16: Compute ∫04(x−2)dx.
Answer: 0. Linear function with zero net area due to symmetry about x=2.
Flashcard 17: Find dxd∫0xln(t)dt.
Answer: ln(x). By FTC Part 1, derivative of integral equals integrand.
Flashcard 18: Evaluate ∫12(x2+x)dx.
Answer: 37. Sum of power functions integrated using power rule.
Flashcard 19: State the Mean Value Theorem for Integrals.
Answer: ∃c∈[a,b] such that f(c)=b−a1∫abf(x)dx. Guarantees existence of point where function equals its average value.
Flashcard 20: Compute ∫02(3x2+2x)dx.
Answer: 12. Antiderivative is x3+x2, evaluated from 0 to 2 gives 8+4=12.
Flashcard 21: State the definition of a definite integral.
Answer: The limit of Riemann sums: limn→∞∑i=1nf(xi∗)Δx. Formal definition using limit of approximating rectangular areas.
Flashcard 22: What is the integral of f(x)=3x2 from 1 to 4?
Answer: ∫143x2dx=63. Antiderivative is x3, so F(4)−F(1)=64−1=63.
Flashcard 23: Find dxd∫0xsin(t)dt.
Answer: sin(x). By FTC Part 1, derivative of integral with variable upper limit equals integrand.
Flashcard 24: What is the Fundamental Theorem of Calculus, Part 2?
Answer: ∫abf(x)dx=F(b)−F(a), where F is an antiderivative of f. States that definite integral equals antiderivative evaluated at bounds.
Flashcard 25: What is the integral of f(x)=x3 from −1 to 1?
Answer: 0. Odd function over symmetric interval gives zero due to cancellation.
Flashcard 26: What is the average value of f(x)=x2 on [0,3]?
Answer: 31∫03x2dx=3. Average value formula: b−a1∫abf(x)dx applied to x2 on [0,3].
Flashcard 27: Evaluate ∫0πsin(x)dx.
Answer: 2. Antiderivative is −cos(x), evaluated from 0 to π gives −(−1)−(−1)=2.
Flashcard 28: What is the integral of f(x)=3 from 0 to 2?
Answer: ∫023dx=6. Integral of constant equals constant times interval width.
Flashcard 29: What is the integral of f(x)=ex from 0 to 1?
Answer: ∫01exdx=e−1. Antiderivative is ex, evaluated from 0 to 1 gives e−1.
Flashcard 30: State the formula for the area under a curve using integration.
Answer: Area = ∫abf(x)dx. Fundamental connection between integration and area under curves.