Book cover for Calculus: Early Transcendentals

Calculus: Early Transcendentals

James Stewart

ISBN #9781285741550

8th Edition

6,422 Questions

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2,819,387 Students Helped

Homework Questions

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

Integration by parts is an essential technique in calculus that transforms complex integrals into simpler ones by applying the formula ? u dv = uv - ? v du. Proper selection of u and dv is crucial; a good choice results in a simpler derivative for u and an easily integrable dv. This method not only handles standard integrals but also extends to complex applications such as reduction formulas and trigonometric integrals, making it invaluable in both pure and applied mathematics.

Learning Objectives

1

Explain the integration by parts formula and its relationship with the differentiation Product Rule.

2

Identify appropriate choices for u and dv to simplify the integrals using integration by parts.

3

Apply integration by parts iteratively to solve complex integrals, including trigonometric integrals and those requiring reduction formulas.

4

Demonstrate the process of checking solutions by differentiating the result.

5

Develop strategies to use integration by parts in real-world applications such as calculating areas and volumes.

Key Concepts

CONCEPT

DEFINITION

Integration by Parts

A technique based on the product rule for differentiation. It states that ∫ u dv = uv - ∫ v du, and is used to integrate products of functions.

Product Rule

A differentiation rule stating that d(uv)/dx = u dv/dx + v du/dx. Integration by parts is its counterpart in integration.

Substitution Rule

A method for integration that involves replacing a function with a new variable to simplify the integrand.

Reduction Formula

An equation that expresses an integral in terms of another integral with a lower exponent, allowing repeated use to eventually reach a basic integral form.

Trigonometric Integrals

Integrals involving trigonometric functions that often require the use of identities or techniques like integration by parts and substitution to evaluate.

Example Problems

Example 1

Evaluate the integral using integration by parts with the indicated choices of $ u $ and $ dv $. $ \displaystyle \int xe^{2x} $ ; $ u = x $ , $ dv = e^{2x} dx $

Example 2

Evaluate the integral using integration by parts with the indicated choices of $ u $ and $ dv $. $ \displaystyle \int \sqrt{x} \ln x dx $ ; $ u = \displaystyle \ln x $ , $ dv = \sqrt{x} dx $

Example 3

Evaluate the integral. $ \displaystyle \int x \cos 5x dx $

Example 4

Evaluate the integral. $ \displaystyle \int ye^{0.2y} dy $

Example 5

Evaluate the integral. $ \displaystyle \int te^{-3t} dt $
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Step-by-Step Explanations

QUESTION

How do we evaluate the integral ∫ x sin x dx using integration by parts?

STEP-BY-STEP ANSWER:

Step 1: Choose u and dv. Let u = x (a function that simplifies when differentiated) and dv = sin x dx (whose antiderivative is known).
Step 2: Differentiate and integrate. Compute du = dx and v = -cos x (since ∫ sin x dx = -cos x).
Step 3: Apply the integration by parts formula: ∫ u dv = u*v - ∫ v du. Thus, substituting, we get: ∫ x sin x dx = x*(-cos x) - ∫ (-cos x)*dx.
Step 4: Simplify the expression: This becomes -x cos x + ∫ cos x dx.
Step 5: Evaluate the remaining integral: ∫ cos x dx = sin x.
Step 6: Write the final answer: -x cos x + sin x + C.
Final Answer:

Example: ∫ x sin x dx (Using Integration by Parts)

QUESTION

How do we evaluate the integral ∫ ln x dx using integration by parts?

STEP-BY-STEP ANSWER:

Step 1: Choose u and dv. Let u = ln x (which simplifies upon differentiation) and dv = dx.
Step 2: Differentiate u and integrate dv: du = (1/x) dx and v = x.
Step 3: Apply integration by parts: ∫ ln x dx = u*v - ∫ v du = x ln x - ∫ x*(1/x) dx.
Step 4: Simplify the integral: ∫ 1 dx = x.
Step 5: Write the final answer: x ln x - x + C.
Final Answer:

Example: ∫ ln x dx (Using Integration by Parts)

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Common Mistakes

  • Choosing u and dv poorly, leading to a more complicated integral rather than a simpler one.
  • Sign errors in differentiating or integrating, particularly missing negative signs (for example, when integrating sin x to get -cos x).
  • Forgetting to include the constant of integration (C) in indefinite integrals.
  • Assuming integration by parts can simplify every product without checking if the derivative of u actually simplifies the problem.
  • Not verifying the final answer by differentiating it to see if the original integrand is recovered.