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  2. AP Calculus AB
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AP Calculus AB Flashcards: Fundamental Theorem Of Calculus Definite Intervals

Study Fundamental Theorem Of Calculus Definite Intervals 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 Fundamental Theorem Of Calculus Definite Intervals, 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: Fundamental Theorem Of Calculus Definite Intervals

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QUESTION

What is an antiderivative of f(x)=3x2f(x) = 3x^2f(x)=3x2?

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ANSWER

F(x)=x3+CF(x) = x^3 + CF(x)=x3+C. Power rule: increase exponent by 1, divide by new exponent.

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Flashcard 1: What is an antiderivative of f(x)=3x2f(x) = 3x^2f(x)=3x2?

Answer: F(x)=x3+CF(x) = x^3 + CF(x)=x3+C. Power rule: increase exponent by 1, divide by new exponent.

Flashcard 2: Calculate ∫13(2x2−x) dx\int_1^3 (2x^2 - x)\,dx∫13​(2x2−x)dx using the Fundamental Theorem.

Answer: 263\frac{26}{3}326​. Use antiderivative 2x33−x22\frac{2x^3}{3} - \frac{x^2}{2}32x3​−2x2​, evaluate at bounds.

Flashcard 3: Evaluate ∫−22(x3+2x) dx\int_{-2}^2 (x^3 + 2x)\,dx∫−22​(x3+2x)dx using symmetry properties.

Answer: 000. Both functions are odd, so integral over symmetric interval is zero.

Flashcard 4: Evaluate ∫04(3x2+2) dx\int_0^4 (3x^2 + 2)\,dx∫04​(3x2+2)dx using the Fundamental Theorem.

Answer:

  1. Use antiderivative x3+2xx^3 + 2xx3+2x, evaluate at bounds 444 and 000.

Flashcard 5: State the Fundamental Theorem of Calculus, Part 1.

Answer: If FFF is an antiderivative of fff on [a,b][a,b][a,b], then ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)∫ab​f(x)dx=F(b)−F(a). This allows calculation of definite integrals using antiderivatives.

Flashcard 6: Determine ∫141x dx\int_1^4 \frac{1}{x}\,dx∫14​x1​dx.

Answer: ln⁡(4)\ln(4)ln(4). Antiderivative of 1x\frac{1}{x}x1​ is ln⁡(x)\ln(x)ln(x).

Flashcard 7: For F(x)=∫0xet dtF(x) = \int_0^x e^t\,dtF(x)=∫0x​etdt, find F′(x)F'(x)F′(x).

Answer: exe^xex. By Part 2 of FTC, the derivative equals the integrand.

Flashcard 8: For F(x)=∫0x sin⁡(t) dtF(x) = \int_0^x \,\sin(t)\,dtF(x)=∫0x​sin(t)dt, find F′(x)F'(x)F′(x).

Answer: sin⁡(x)\sin(x)sin(x). By Part 2 of FTC, derivative equals the integrand.

Flashcard 9: What does the second part of the Fundamental Theorem allow us to compute?

Answer: The derivative of an integral function. It gives the rate of change of the area function.

Flashcard 10: Determine ∫02x2 dx\int_0^2 x^2\,dx∫02​x2dx using an antiderivative.

Answer: 83\frac{8}{3}38​. Use antiderivative x33\frac{x^3}{3}3x3​, evaluate at bounds.

Flashcard 11: Calculate ∫13(2x2−x) dx\int_1^3 (2x^2 - x)\,dx∫13​(2x2−x)dx using the Fundamental Theorem.

Answer: 263\frac{26}{3}326​. Use antiderivative 2x33−x22\frac{2x^3}{3} - \frac{x^2}{2}32x3​−2x2​, evaluate at bounds.

Flashcard 12: What is the integral of f(x)=exf(x) = e^xf(x)=ex over [0,1][0, 1][0,1]?

Answer: e−1e - 1e−1. Antiderivative of exe^xex is exe^xex; evaluate at bounds.

Flashcard 13: What is a definite integral's geometric interpretation?

Answer: Net area between the curve and the x-axis over [a,b][a, b][a,b]. Represents the signed area between curve and x-axis.

Flashcard 14: Find the derivative of F(x)=∫2xln⁡(t) dtF(x) = \int_2^x \ln(t)\,dtF(x)=∫2x​ln(t)dt.

Answer: ln⁡(x)\ln(x)ln(x). By Part 2 of FTC, the derivative equals the integrand.

Flashcard 15: Find ∫02(4x3+x) dx\int_0^2 (4x^3 + x)\,dx∫02​(4x3+x)dx using an antiderivative.

Answer:

  1. Use antiderivative x4+x22x^4 + \frac{x^2}{2}x4+2x2​, evaluate at bounds.

Flashcard 16: State the Fundamental Theorem of Calculus, Part 2.

Answer: If fff is continuous on [a,b][a,b][a,b], then F(x)=∫axf(t) dtF(x) = \int_a^x f(t)\,dtF(x)=∫ax​f(t)dt is differentiable and F′(x)=f(x)F'(x) = f(x)F′(x)=f(x). Shows that differentiation and integration are inverse operations.

Flashcard 17: What is the definite integral of a constant ccc over an interval [a,b][a, b][a,b]?

Answer: c(b−a)c(b-a)c(b−a). A constant function integrates to constant times interval length.

Flashcard 18: Compute ∫−11x3 dx\int_{-1}^1 x^3\,dx∫−11​x3dx using symmetry properties.

Answer:

  1. Odd function over symmetric interval gives zero area.

Flashcard 19: Evaluate ∫012x dx\int_0^1 2x\,dx∫01​2xdx using the Fundamental Theorem of Calculus.

Answer:

  1. Find antiderivative x2x^2x2, then evaluate 12−02=11^2 - 0^2 = 112−02=1.

Flashcard 20: Determine ∫13x2 dx\int_1^3 x^2\,dx∫13​x2dx using an antiderivative.

Answer: 26/326/326/3. Use antiderivative x33\frac{x^3}{3}3x3​, then evaluate at bounds.

Flashcard 21: What does the Fundamental Theorem of Calculus connect?

Answer: It connects differentiation and integration. They are inverse operations of each other.

Flashcard 22: Evaluate ∫03(x2+x+1) dx\int_0^3 (x^2 + x + 1)\,dx∫03​(x2+x+1)dx using the Fundamental Theorem.

Answer:

  1. Use antiderivative x33+x22+x\frac{x^3}{3} + \frac{x^2}{2} + x3x3​+2x2​+x, evaluate at bounds.

Flashcard 23: Find ∫04(x2−2) dx\int_0^4 (x^2 - 2)\,dx∫04​(x2−2)dx using the Fundamental Theorem.

Answer: 323\frac{32}{3}332​. Use antiderivative x33−2x\frac{x^3}{3} - 2x3x3​−2x, evaluate at bounds.

Flashcard 24: Determine ∫01(5x2−3x) dx\int_0^1 (5x^2 - 3x)\,dx∫01​(5x2−3x)dx using an antiderivative.

Answer: 13\frac{1}{3}31​. Use antiderivative 5x33−3x22\frac{5x^3}{3} - \frac{3x^2}{2}35x3​−23x2​, evaluate at bounds.

Flashcard 25: Evaluate ∫01(4x3−x) dx\int_0^1 (4x^3 - x)\,dx∫01​(4x3−x)dx using the Fundamental Theorem.

Answer: 34\frac{3}{4}43​. Use antiderivative x4−x22x^4 - \frac{x^2}{2}x4−2x2​, evaluate at bounds.

Flashcard 26: What is the result of ∫021 dx\int_0^2 1\,dx∫02​1dx?

Answer:

  1. Integral of constant 111 over interval length 222.

Flashcard 27: Find ∫01(x2+2x) dx\int_0^1 (x^2 + 2x)\,dx∫01​(x2+2x)dx using an antiderivative.

Answer: 73\frac{7}{3}37​. Use antiderivative x33+x2\frac{x^3}{3} + x^23x3​+x2, evaluate at bounds.

Flashcard 28: What is the integral of f(x)=cos⁡(x)f(x) = \cos(x)f(x)=cos(x) over [0,π][0, \pi][0,π]?

Answer:

  1. Antiderivative is sin⁡(x)\sin(x)sin(x); sin⁡(π)−sin⁡(0)=0\sin(\pi) - \sin(0) = 0sin(π)−sin(0)=0.

Flashcard 29: What is the integral of f(x)=1/xf(x) = 1/xf(x)=1/x over [1,e][1, e][1,e]?

Answer:

  1. Antiderivative of 1x\frac{1}{x}x1​ is ln⁡(x)\ln(x)ln(x); ln⁡(e)−ln⁡(1)=1\ln(e) - \ln(1) = 1ln(e)−ln(1)=1.

Flashcard 30: Calculate ∫01(x3+1) dx\int_0^1 (x^3 + 1)\,dx∫01​(x3+1)dx using the Fundamental Theorem.

Answer: 54\frac{5}{4}45​. Use antiderivative x44+x\frac{x^4}{4} + x4x4​+x, evaluate at bounds.