AP Calculus BC Flashcards: Integrating Using Integration By Parts

Study Integrating Using Integration By Parts in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Calculus BC

Integrating Using Integration By Parts

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QUESTION
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What is ddx(x)\frac{d}{dx}(x) when using integration by parts?

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ANSWER

du=dxdu = dx. The derivative of xx is 11.

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What this deck covers

This deck focuses on Integrating Using Integration By Parts, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

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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.

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Flashcard 1: What is ddx(x)\frac{d}{dx}(x) when using integration by parts?

Answer: du=dxdu = dx. The derivative of xx is 11.

Flashcard 2: Find the integral of x2exx^2 e^x using the tabular method.

Answer: x2ex2xex+2ex+Cx^2 e^x - 2x e^x + 2e^x + C. Complete solution using repeated integration by parts or tabular method.

Flashcard 3: What is a key benefit of the tabular method?

Answer: Simplifies repeated integration by parts. Reduces calculation time and errors for polynomial-exponential products.

Flashcard 4: Find vv if dv=sin(x)dxdv = \sin(x) \, dx.

Answer: v=cos(x)v = -\cos(x). The antiderivative of sin(x)\sin(x) is cos(x)-\cos(x).

Flashcard 5: What is the derivative of sin(x)\sin(x) for integration by parts?

Answer: du=cos(x)dxdu = \cos(x) \, dx. The derivative of sin(x)\sin(x) is cos(x)\cos(x).

Flashcard 6: Determine dvdv for exsin(x)dx\int e^x \sin(x) \, dx using integration by parts.

Answer: dv=exdxdv = e^x \, dx. Choose exdxe^x dx as dvdv since exe^x is easy to integrate.

Flashcard 7: What is the integral of x3exx^3 e^x using integration by parts?

Answer: x3ex3x2exdxx^3 e^x - 3 \int x^2 e^x \, dx. First application of integration by parts, requires further iterations.

Flashcard 8: Identify uu for x2exdx\int x^2 e^x \, dx using integration by parts.

Answer: u=x2u = x^2. Choose x2x^2 as uu since it simplifies when differentiated.

Flashcard 9: Which method simplifies repeated parts integration?

Answer: Tabular integration. Alternative name for the tabular method of integration by parts.

Flashcard 10: What is ddx(x2)\frac{d}{dx}(x^2) when using integration by parts?

Answer: du=2xdxdu = 2x \, dx. The derivative of x2x^2 is 2x2x.

Flashcard 11: What is the integral of x3exx^3 e^x using integration by parts?

Answer: x3ex3x2exdxx^3 e^x - 3 \int x^2 e^x \, dx. First application of integration by parts, requires further iterations.

Flashcard 12: Find vv if dv=xdxdv = x \, dx.

Answer: v=x22v = \frac{x^2}{2}. The antiderivative of xx is x22\frac{x^2}{2}.

Flashcard 13: Identify uu for xexdx\int x e^x \, dx using integration by parts.

Answer: u=xu = x. Choose xx as uu since it simplifies when differentiated.

Flashcard 14: What is ddx(ln(x))\frac{d}{dx}(\ln(x)) when using integration by parts?

Answer: du=1xdxdu = \frac{1}{x} \, dx. The derivative of ln(x)\ln(x) is 1x\frac{1}{x}.

Flashcard 15: Which part is chosen as dvdv in integration by parts?

Answer: The part that is easy to integrate. Choose the function that can be integrated easily.

Flashcard 16: What is the integral of xcos(x)x \cos(x) using integration by parts?

Answer: xsin(x)+cos(x)+Cx \sin(x) + \cos(x) + C. Result after applying integration by parts to xcos(x)dx\int x \cos(x) dx.

Flashcard 17: Identify uu for x2exdx\int x^2 e^x \, dx using integration by parts.

Answer: u=x2u = x^2. Choose x2x^2 as uu since it simplifies when differentiated.

Flashcard 18: Determine uu for exsin(x)dx\int e^x \sin(x) \, dx using integration by parts.

Answer: u=sin(x)u = \sin(x). Choose sin(x)\sin(x) as uu since it cycles when differentiated.

Flashcard 19: Find vv if dv=xdxdv = x \, dx.

Answer: v=x22v = \frac{x^2}{2}. The antiderivative of xx is x22\frac{x^2}{2}.

Flashcard 20: What is ddx(x)\frac{d}{dx}(x) when using integration by parts?

Answer: du=dxdu = dx. The derivative of xx is 11.

Flashcard 21: Solve xe2xdx\int x e^{2x} \, dx using integration by parts.

Answer: x2e2x14e2x+C\frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} + C. Apply integration by parts with u=xu = x and dv=e2xdxdv = e^{2x} dx.

Flashcard 22: Find vv if dv=exdxdv = e^x \, dx.

Answer: v=exv = e^x. The antiderivative of exe^x is exe^x.

Flashcard 23: Find vv if dv=cos(x)dxdv = \cos(x) \, dx.

Answer: v=sin(x)v = \sin(x). The antiderivative of cos(x)\cos(x) is sin(x)\sin(x).

Flashcard 24: Determine uu for exsin(x)dx\int e^x \sin(x) \, dx using integration by parts.

Answer: u=sin(x)u = \sin(x). Choose sin(x)\sin(x) as uu since it cycles when differentiated.

Flashcard 25: Identify dvdv for xexdx\int x e^x \, dx using integration by parts.

Answer: dv=exdxdv = e^x \, dx. Choose exdxe^x dx as dvdv since exe^x is easy to integrate.

Flashcard 26: Find vv if dv=cos(x)dxdv = \cos(x) \, dx.

Answer: v=sin(x)v = \sin(x). The antiderivative of cos(x)\cos(x) is sin(x)\sin(x).

Flashcard 27: Identify dvdv for x2exdx\int x^2 e^x \, dx using integration by parts.

Answer: dv=exdxdv = e^x \, dx. Choose exdxe^x dx as dvdv since exe^x is easy to integrate.

Flashcard 28: What is the integration by parts formula?

Answer: udv=uvvdu\int u \, dv = uv - \int v \, du. The fundamental formula for integration by parts.

Flashcard 29: What is the result of ln(x)dx\int \ln(x) \, dx using integration by parts?

Answer: xln(x)x+Cx \ln(x) - x + C. Set u=ln(x)u = \ln(x) and dv=dxdv = dx, then apply the formula.

Flashcard 30: Identify dvdv for x2exdx\int x^2 e^x \, dx using integration by parts.

Answer: dv=exdxdv = e^x \, dx. Choose exdxe^x dx as dvdv since exe^x is easy to integrate.

Flashcard 31: Identify uu for xsin(x)dx\int x \sin(x) \, dx using integration by parts.

Answer: u=xu = x. Choose xx as uu since it simplifies when differentiated.

Flashcard 32: What is a key benefit of the tabular method?

Answer: Simplifies repeated integration by parts. Reduces calculation time and errors for polynomial-exponential products.

Flashcard 33: Calculate vv if dv=exdxdv = e^x \, dx.

Answer: v=exv = e^x. The antiderivative of exe^x is exe^x.

Flashcard 34: Identify dvdv for xsin(x)dx\int x \sin(x) \, dx using integration by parts.

Answer: dv=sin(x)dxdv = \sin(x) \, dx. Choose sin(x)dx\sin(x) dx as dvdv since sin(x)\sin(x) is easy to integrate.

Flashcard 35: Identify dvdv for xsin(x)dx\int x \sin(x) \, dx using integration by parts.

Answer: dv=sin(x)dxdv = \sin(x) \, dx. Choose sin(x)dx\sin(x) dx as dvdv since sin(x)\sin(x) is easy to integrate.

Flashcard 36: What is ddx(x)\frac{d}{dx}(x) when using integration by parts?

Answer: du=dxdu = dx. The derivative of xx is 11.

Flashcard 37: Solve xe2xdx\int x e^{2x} \, dx using integration by parts.

Answer: x2e2x14e2x+C\frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} + C. Apply integration by parts with u=xu = x and dv=e2xdxdv = e^{2x} dx.

Flashcard 38: Find vv if dv=exdxdv = e^x \, dx.

Answer: v=exv = e^x. The antiderivative of exe^x is exe^x.

Flashcard 39: What is ddx(x2)\frac{d}{dx}(x^2) when using integration by parts?

Answer: du=2xdxdu = 2x \, dx. The derivative of x2x^2 is 2x2x.

Flashcard 40: What is the first step in integration by parts for xln(x)dx\int x \ln(x) \, dx?

Answer: Choose u=ln(x)u = \ln(x) and dv=xdxdv = x \, dx. Choose ln(x)\ln(x) as uu since it simplifies when differentiated.

Flashcard 41: What rule can simplify repeated integration by parts?

Answer: Tabular method. Systematic approach for multiple applications of integration by parts.

Flashcard 42: What is the first step in integration by parts for xln(x)dx\int x \ln(x) \, dx?

Answer: Choose u=ln(x)u = \ln(x) and dv=xdxdv = x \, dx. Choose ln(x)\ln(x) as uu since it simplifies when differentiated.

Flashcard 43: What is the derivative of sin(x)\sin(x) for integration by parts?

Answer: du=cos(x)dxdu = \cos(x) \, dx. The derivative of sin(x)\sin(x) is cos(x)\cos(x).

Flashcard 44: Find vv if dv=sin(x)dxdv = \sin(x) \, dx.

Answer: v=cos(x)v = -\cos(x). The antiderivative of sin(x)\sin(x) is cos(x)-\cos(x).

Flashcard 45: Which method simplifies repeated parts integration?

Answer: Tabular integration. Alternative name for the tabular method of integration by parts.

Flashcard 46: Identify dvdv for xexdx\int x e^x \, dx using integration by parts.

Answer: dv=exdxdv = e^x \, dx. Choose exdxe^x dx as dvdv since exe^x is easy to integrate.

Flashcard 47: What is the result of ln(x)dx\int \ln(x) \, dx using integration by parts?

Answer: xln(x)x+Cx \ln(x) - x + C. Set u=ln(x)u = \ln(x) and dv=dxdv = dx, then apply the formula.

Flashcard 48: Find the integral of x2exx^2 e^x using the tabular method.

Answer: x2ex2xex+2ex+Cx^2 e^x - 2x e^x + 2e^x + C. Complete solution using repeated integration by parts or tabular method.

Flashcard 49: Find the integral of x2exx^2 e^x using integration by parts.

Answer: x2ex2xexdxx^2 e^x - 2 \int x e^x \, dx. First application of integration by parts, requires second iteration.

Flashcard 50: Which part is chosen as uu in integration by parts?

Answer: The part that simplifies when differentiated. Choose the function that becomes simpler when differentiated.

Flashcard 51: Determine dvdv for exsin(x)dx\int e^x \sin(x) \, dx using integration by parts.

Answer: dv=exdxdv = e^x \, dx. Choose exdxe^x dx as dvdv since exe^x is easy to integrate.

Flashcard 52: Which part is chosen as dvdv in integration by parts?

Answer: The part that is easy to integrate. Choose the function that can be integrated easily.

Flashcard 53: What is ddx(ln(x))\frac{d}{dx}(\ln(x)) when using integration by parts?

Answer: du=1xdxdu = \frac{1}{x} \, dx. The derivative of ln(x)\ln(x) is 1x\frac{1}{x}.

Flashcard 54: Calculate vv if dv=exdxdv = e^x \, dx.

Answer: v=exv = e^x. The antiderivative of exe^x is exe^x.

Flashcard 55: Which part is chosen as uu in integration by parts?

Answer: The part that simplifies when differentiated. Choose the function that becomes simpler when differentiated.

Flashcard 56: Find arctan(x)dx\int \arctan(x) \, dx using integration by parts.

Answer: xarctan(x)12ln(1+x2)+Cx \arctan(x) - \frac{1}{2} \ln(1 + x^2) + C. Set u=arctan(x)u = \arctan(x) and dv=dxdv = dx, then apply the formula.

Flashcard 57: Identify uu for xsin(x)dx\int x \sin(x) \, dx using integration by parts.

Answer: u=xu = x. Choose xx as uu since it simplifies when differentiated.

Flashcard 58: What is the integration by parts formula?

Answer: udv=uvvdu\int u \, dv = uv - \int v \, du. The fundamental formula for integration by parts.

Flashcard 59: What is the integral of xcos(x)x \cos(x) using integration by parts?

Answer: xsin(x)+cos(x)+Cx \sin(x) + \cos(x) + C. Result after applying integration by parts to xcos(x)dx\int x \cos(x) dx.

Flashcard 60: Find arctan(x)dx\int \arctan(x) \, dx using integration by parts.

Answer: xarctan(x)12ln(1+x2)+Cx \arctan(x) - \frac{1}{2} \ln(1 + x^2) + C. Set u=arctan(x)u = \arctan(x) and dv=dxdv = dx, then apply the formula.

Flashcard 61: What rule can simplify repeated integration by parts?

Answer: Tabular method. Systematic approach for multiple applications of integration by parts.

Flashcard 62: Find the integral of x2exx^2 e^x using integration by parts.

Answer: x2ex2xexdxx^2 e^x - 2 \int x e^x \, dx. First application of integration by parts, requires second iteration.

Flashcard 63: Identify uu for xexdx\int x e^x \, dx using integration by parts.

Answer: u=xu = x. Choose xx as uu since it simplifies when differentiated.