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  2. AP Calculus AB
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AP Calculus AB Flashcards: Applying Properties Of Definite Integrals

Study Applying Properties Of Definite Integrals 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 Applying Properties Of Definite Integrals, 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: Applying Properties Of Definite Integrals

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QUESTION

What is the antiderivative of exe^xex?

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ANSWER

ex+Ce^x + Cex+C. The exponential function is its own derivative and antiderivative.

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All flashcards

Flashcard 1: What is the antiderivative of exe^xex?

Answer: ex+Ce^x + Cex+C. The exponential function is its own derivative and antiderivative.

Flashcard 2: Evaluate ddx(1csc(x))\frac{d}{dx}\bigg(\frac{1}{\text{csc}(x)}\bigg)dxd​(csc(x)1​).

Answer: cot(x)csc(x)\text{cot}(x)\text{csc}(x)cot(x)csc(x). Chain rule on sin⁡(x)=(csc⁡(x))−1\sin(x) = (\csc(x))^{-1}sin(x)=(csc(x))−1 gives cos⁡(x)\cos(x)cos(x).

Flashcard 3: Evaluate ddx(1cos(x))\frac{d}{dx}\bigg(\frac{1}{\text{cos}(x)}\bigg)dxd​(cos(x)1​).

Answer: sin(x)cos2(x)\frac{\text{sin}(x)}{\text{cos}^2(x)}cos2(x)sin(x)​. Chain rule on sec⁡(x)=(cos⁡(x))−1\sec(x) = (\cos(x))^{-1}sec(x)=(cos(x))−1 gives sec⁡(x)tan⁡(x)\sec(x)\tan(x)sec(x)tan(x).

Flashcard 4: What is the antiderivative of sin(x)\text{sin}(x)sin(x)?

Answer: −cos(x)+C-\text{cos}(x) + C−cos(x)+C. Standard antiderivative: derivative of negative cosine is sine.

Flashcard 5: Evaluate ddx(1sin(x))\frac{d}{dx}\bigg(\frac{1}{\text{sin}(x)}\bigg)dxd​(sin(x)1​).

Answer: −cos(x)sin2(x)-\frac{\text{cos}(x)}{\text{sin}^2(x)}−sin2(x)cos(x)​. Chain rule on csc⁡(x)=(sin⁡(x))−1\csc(x) = (\sin(x))^{-1}csc(x)=(sin(x))−1 gives −csc⁡(x)cot⁡(x)-\csc(x)\cot(x)−csc(x)cot(x).

Flashcard 6: Evaluate ddx(1cot(x))\frac{d}{dx}\bigg(\frac{1}{\text{cot}(x)}\bigg)dxd​(cot(x)1​).

Answer: 1sin2(x)\frac{1}{\text{sin}^2(x)}sin2(x)1​. Chain rule on tan⁡(x)=(cot⁡(x))−1\tan(x) = (\cot(x))^{-1}tan(x)=(cot(x))−1 gives sec⁡2(x)\sec^2(x)sec2(x).

Flashcard 7: Evaluate ddx(1x)\frac{d}{dx} \bigg(\frac{1}{x}\bigg)dxd​(x1​).

Answer: −1x2-\frac{1}{x^2}−x21​. Power rule: ddx(x−1)=−1⋅x−2\frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2}dxd​(x−1)=−1⋅x−2.

Flashcard 8: Evaluate ddx(1x2+1)\frac{d}{dx}\bigg(\frac{1}{x^2 + 1}\bigg)dxd​(x2+11​).

Answer: −2x(x2+1)2-\frac{2x}{(x^2 + 1)^2}−(x2+1)22x​. Chain rule on (x2+1)−1(x^2 + 1)^{-1}(x2+1)−1 gives −1⋅(x2+1)−2⋅2x-1 \cdot (x^2 + 1)^{-2} \cdot 2x−1⋅(x2+1)−2⋅2x.

Flashcard 9: What is the integral of cos(x)\text{cos}(x)cos(x)?

Answer: sin(x)+C\text{sin}(x) + Csin(x)+C. Standard antiderivative: derivative of sine is cosine.

Flashcard 10: Find the derivative of e2xe^{2x}e2x.

Answer: 2e2x2e^{2x}2e2x. Chain rule: ddx(e2x)=e2x⋅2\frac{d}{dx}(e^{2x}) = e^{2x} \cdot 2dxd​(e2x)=e2x⋅2.

Flashcard 11: Evaluate ddx(1eax)\frac{d}{dx}\bigg(\frac{1}{\text{e}^{ax}}\bigg)dxd​(eax1​).

Answer: −ae−ax-a\text{e}^{-ax}−ae−ax. Chain rule on e−axe^{-ax}e−ax gives −a⋅e−ax-a \cdot e^{-ax}−a⋅e−ax.

Flashcard 12: Evaluate ddx(1tan(x))\frac{d}{dx}\bigg(\frac{1}{\text{tan}(x)}\bigg)dxd​(tan(x)1​).

Answer: −1sin2(x)-\frac{1}{\text{sin}^2(x)}−sin2(x)1​. Chain rule on cot⁡(x)=(tan⁡(x))−1\cot(x) = (\tan(x))^{-1}cot(x)=(tan(x))−1 gives −csc⁡2(x)-\csc^2(x)−csc2(x).

Flashcard 13: What is the value of ddx(1x)\frac{d}{dx}\bigg(\frac{1}{x}\bigg)dxd​(x1​)?

Answer: −1x2-\frac{1}{x^2}−x21​. Power rule applied to x−1x^{-1}x−1.

Flashcard 14: Find ddx(ln∣x∣)\frac{d}{dx} \bigg( \text{ln}|x| \bigg)dxd​(ln∣x∣).

Answer: 1x\frac{1}{x}x1​. Derivative of natural logarithm is 1x\frac{1}{x}x1​.

Flashcard 15: State the integral of 1x\frac{1}{x}x1​ with respect to xxx.

Answer: ln∣x∣+C\text{ln}|x| + Cln∣x∣+C. Standard antiderivative formula for 1x\frac{1}{x}x1​.

Flashcard 16: What is ddx(1x)\frac{d}{dx}\bigg(\frac{1}{x}\bigg)dxd​(x1​)?

Answer: −1x2-\frac{1}{x^2}−x21​. Derivative of x−1x^{-1}x−1 using power rule.

Flashcard 17: What is the integral of xnx^nxn with respect to xxx?

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.

Flashcard 18: What is the integral of eax\text{e}^{ax}eax?

Answer: 1aeax+C\frac{1}{a}\text{e}^{ax} + Ca1​eax+C. Standard exponential integral with constant aaa.

Flashcard 19: State the integral of 1x\frac{1}{x}x1​ with respect to xxx.

Answer: ln⁡∣x∣+C\ln |x| + Cln∣x∣+C. Standard antiderivative formula for 1x\frac{1}{x}x1​.

Flashcard 20: What is the integral of eax\text{e}^{ax}eax?

Answer: 1aeax+C\frac{1}{a}\text{e}^{ax} + Ca1​eax+C. Standard exponential integral with constant aaa.

Flashcard 21: What is the value of ddx(1x)\frac{d}{dx}\bigg(\frac{1}{x}\bigg)dxd​(x1​)?

Answer: −1x2-\frac{1}{x^2}−x21​. Power rule applied to x−1x^{-1}x−1.

Flashcard 22: Find ddx(ln∣x∣)\frac{d}{dx} \bigg( \text{ln}|x| \bigg)dxd​(ln∣x∣).

Answer: 1x\frac{1}{x}x1​. Derivative of natural logarithm is 1x\frac{1}{x}x1​.

Flashcard 23: Evaluate ddx(1sin(x))\frac{d}{dx}\bigg(\frac{1}{\text{sin}(x)}\bigg)dxd​(sin(x)1​).

Answer: −cos(x)sin2(x)-\frac{\text{cos}(x)}{\text{sin}^2(x)}−sin2(x)cos(x)​. Chain rule on csc⁡(x)=(sin⁡(x))−1\csc(x) = (\sin(x))^{-1}csc(x)=(sin(x))−1 gives −csc⁡(x)cot⁡(x)-\csc(x)\cot(x)−csc(x)cot(x).

Flashcard 24: Evaluate ddx(1cot(x))\frac{d}{dx}\bigg(\frac{1}{\text{cot}(x)}\bigg)dxd​(cot(x)1​).

Answer: 1sin2(x)\frac{1}{\text{sin}^2(x)}sin2(x)1​. Chain rule on tan⁡(x)=(cot⁡(x))−1\tan(x) = (\cot(x))^{-1}tan(x)=(cot(x))−1 gives sec⁡2(x)\sec^2(x)sec2(x).

Flashcard 25: Evaluate ddx(1csc(x))\frac{d}{dx}\bigg(\frac{1}{\text{csc}(x)}\bigg)dxd​(csc(x)1​).

Answer: cot(x)csc(x)\text{cot}(x)\text{csc}(x)cot(x)csc(x). Chain rule on sin⁡(x)=(csc⁡(x))−1\sin(x) = (\csc(x))^{-1}sin(x)=(csc(x))−1 gives cos⁡(x)\cos(x)cos(x).

Flashcard 26: Evaluate ddx(1cos(x))\frac{d}{dx}\bigg(\frac{1}{\text{cos}(x)}\bigg)dxd​(cos(x)1​).

Answer: sin(x)cos2(x)\frac{\text{sin}(x)}{\text{cos}^2(x)}cos2(x)sin(x)​. Chain rule on sec⁡(x)=(cos⁡(x))−1\sec(x) = (\cos(x))^{-1}sec(x)=(cos(x))−1 gives sec⁡(x)tan⁡(x)\sec(x)\tan(x)sec(x)tan(x).

Flashcard 27: Find the derivative of e2xe^{2x}e2x.

Answer: 2e2x2e^{2x}2e2x. Chain rule: ddx(e2x)=e2x⋅2\frac{d}{dx}(e^{2x}) = e^{2x} \cdot 2dxd​(e2x)=e2x⋅2.

Flashcard 28: Evaluate ddx(1eax)\frac{d}{dx}\bigg(\frac{1}{\text{e}^{ax}}\bigg)dxd​(eax1​).

Answer: −ae−ax-a\text{e}^{-ax}−ae−ax. Chain rule on e−axe^{-ax}e−ax gives −a⋅e−ax-a \cdot e^{-ax}−a⋅e−ax.

Flashcard 29: What is ddx(1x)\frac{d}{dx}\bigg(\frac{1}{x}\bigg)dxd​(x1​)?

Answer: −1x2-\frac{1}{x^2}−x21​. Derivative of x−1x^{-1}x−1 using power rule.

Flashcard 30: What is the integral of xnx^nxn with respect to xxx?

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.