Study Fundamental Theorem Of Calculus Definite Intervals in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards
Flashcard 1: Find the integral of f(x)=4x3−x2+2 from 0 to 1.
Answer: 413. Evaluate x4−3x3+2x at bounds 1 and 0.
Flashcard 2: Find the integral of f(x)=1+x from 0 to 2.
Answer:
- Antiderivative is x+2x2, evaluate at bounds.
Flashcard 3: Calculate the integral of f(x)=x−1 from 1 to 4.
Answer: ln(4). Same as ∫x1dx, antiderivative is ln(x).
Flashcard 4: Calculate the integral of f(x)=x21 from 1 to 2.
Answer: 21. Antiderivative is −x1, then −21−(−1)=21.
Flashcard 5: What is the integral of f(x)=4x from 1 to 3?
Answer:
- Antiderivative is 2x2, then 18−2=16.
Flashcard 6: What is the antiderivative of f(x)=2x3?
Answer: 2x4+C. Power rule integration: increase exponent, divide by new exponent.
Flashcard 7: Find the integral of f(x)=4x3−x2+2 from 0 to 1.
Answer: 413. Evaluate x4−3x3+2x at bounds 1 and 0.
Flashcard 8: What is the antiderivative of f(x)=2x3?
Answer: 2x4+C. Power rule integration: increase exponent, divide by new exponent.
Flashcard 9: Calculate the integral of f(x)=x21 from 1 to 2.
Answer: 21. Antiderivative is −x1, then −21−(−1)=21.
Flashcard 10: Find the integral of f(x)=1+x from 0 to 2.
Answer:
- Antiderivative is x+2x2, evaluate at bounds.
Flashcard 11: Determine the integral of f(x)=x4 from 0 to 1.
Answer: 51. Antiderivative is 5x5, then 51−0.
Flashcard 12: Find the value of the integral of f(x)=x2−x from 0 to 3.
Answer: 29. Evaluate 3x3−2x2 at bounds 3 and 0.
Flashcard 13: Identify the antiderivative given f(x)=2x.
Answer: F(x)=x2+C. The antiderivative of 2x is x2 plus a constant.
Flashcard 14: Find the integral of f(x)=3x2+2x from 0 to 2.
Answer:
- Evaluate x3+x2 at bounds 2 and 0.
Flashcard 15: Determine the integral of f(x)=x4 from 0 to 1.
Answer: 51. Antiderivative is 5x5, then 51−0.
Flashcard 16: Find the integral of f(x)=3x2+2x from 0 to 2.
Answer:
- Evaluate x3+x2 at bounds 2 and 0.
Flashcard 17: Evaluate the integral of f(x)=x3−3x2+2x from 0 to 1.
Answer:
- Evaluate 4x4−x3+x2 at bounds 1 and 0.
Flashcard 18: Evaluate the integral of f(x)=x3 from 0 to 2.
Answer:
- Antiderivative is 4x4, then 416−0=4.
Flashcard 19: Evaluate the integral of f(x)=x1 from 1 to 2.
Answer: ln(2). Antiderivative of x1 is ln(x), then ln(2)−ln(1).
Flashcard 20: Evaluate the integral of f(x)=x1 from 1 to 2.
Answer: ln(2). Antiderivative of x1 is ln(x), then ln(2)−ln(1).
Flashcard 21: What does the Fundamental Theorem of Calculus connect?
Answer: Differentiation and integration. Links the inverse operations of calculus.
Flashcard 22: Calculate the integral of f(x)=x−1 from 1 to 4.
Answer: ln(4). Same as ∫x1dx, antiderivative is ln(x).
Flashcard 23: What does the Fundamental Theorem of Calculus connect?
Answer: Differentiation and integration. Links the inverse operations of calculus.
Flashcard 24: What is the integral of f(x)=e2x from 0 to 1?
Answer: 2e2−1. Antiderivative is 2e2x, evaluate at bounds.
Flashcard 25: What is the integral of f(x)=1 from a to b?
Answer: b−a. Antiderivative of 1 is x, difference of bounds gives length.
Flashcard 26: Determine the integral of f(x)=ex from 0 to 1.
Answer: e−1. Antiderivative of ex is ex, then e−1.
Flashcard 27: What is the definite integral of f(x)=5 from 0 to 3?
Answer:
- Integral of constant function equals constant times interval length.
Flashcard 28: What is the definite integral of f(x)=5 from 0 to 3?
Answer:
- Integral of constant function equals constant times interval length.
Flashcard 29: Identify the antiderivative given f(x)=2x.
Answer: F(x)=x2+C. The antiderivative of 2x is x2 plus a constant.
Flashcard 30: Evaluate the integral of f(x)=x3−3x2+2x from 0 to 1.
Answer:
- Evaluate 4x4−x3+x2 at bounds 1 and 0.
Flashcard 31: Determine the integral of f(x)=ex from 0 to 1.
Answer: e−1. Antiderivative of ex is ex, then e−1.
Flashcard 32: Find the integral of f(x)=2x2−3x+1 from 1 to 2.
Answer: 37. Evaluate 32x3−23x2+x at bounds.
Flashcard 33: Calculate the integral of f(x)=x1 from 1 to e.
Answer:
- Antiderivative of x1 is ln(x), then ln(e)−ln(1)=1.
Flashcard 34: What is the integral of f(x)=1 from a to b?
Answer: b−a. Antiderivative of 1 is x, difference of bounds gives length.
Flashcard 35: What is the integral of f(x)=4x from 1 to 3?
Answer:
- Antiderivative is 2x2, then 18−2=16.
Flashcard 36: What is the integral of f(x)=e2x from 0 to 1?
Answer: 2e2−1. Antiderivative is 2e2x, evaluate at bounds.
Flashcard 37: Find the integral of f(x)=2x2−3x+1 from 1 to 2.
Answer: 37. Evaluate 32x3−23x2+x at bounds.
Flashcard 38: Find the value of the integral of f(x)=x2−x from 0 to 3.
Answer: 29. Evaluate 3x3−2x2 at bounds 3 and 0.
Flashcard 39: Calculate the integral of f(x)=x1 from 1 to e.
Answer:
- Antiderivative of x1 is ln(x), then ln(e)−ln(1)=1.
Flashcard 40: Evaluate the integral of f(x)=x3 from 0 to 2.
Answer:
- Antiderivative is 4x4, then 416−0=4.