Ch. 5 - Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 5 - Integration

Problem 5.5.106

Chapter 5, Problem 5.5.106

General results Evaluate the following integrals in which the function ƒ is unspecified. Note that ƒ⁽ᵖ⁾ is the pth derivative of ƒ and ƒᵖ is the pth power of ƒ. Assume ƒ and its derivatives are continuous for all real numbers.
∫ (5 ƒ³ (𝓍) + 7ƒ² (𝓍) + ƒ (𝓍 )) ƒ'(𝓍) d𝓍

Verified step by step guidance

Recognize that the integral is of the form \( \int (5 f^3(x) + 7 f^2(x) + f(x)) f'(x) \, dx \), where \( f(x) \) is an unspecified function and \( f'(x) \) is its derivative.

Use the substitution method by letting \( u = f(x) \). Then, the differential \( du = f'(x) \, dx \). This transforms the integral into \( \int (5 u^3 + 7 u^2 + u) \, du \).

Rewrite the integral in terms of \( u \): \( \int (5 u^3 + 7 u^2 + u) \, du = \int 5 u^3 \, du + \int 7 u^2 \, du + \int u \, du \).

Integrate each term separately using the power rule for integration: \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). So, compute \( \int 5 u^3 \, du \), \( \int 7 u^2 \, du \), and \( \int u \, du \) accordingly.

After integrating, substitute back \( u = f(x) \) to express the result in terms of \( f(x) \). Don't forget to add the constant of integration \( + C \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration by Substitution

Integration by substitution is a method used to simplify integrals by changing variables. When an integral contains a function and its derivative, substituting the inner function with a new variable can transform the integral into a basic form, making it easier to evaluate.

Notation and Interpretation of Derivatives and Powers

Understanding the notation is crucial: ƒ⁽ᵖ⁾ denotes the pth derivative of ƒ, while ƒᵖ means the pth power of ƒ. Distinguishing between these helps correctly interpret the integrand and apply appropriate integration techniques.

Properties of Continuous Functions and Their Derivatives

Assuming ƒ and its derivatives are continuous ensures the validity of applying substitution and other integration rules. Continuity guarantees no abrupt changes, allowing the integral to be evaluated smoothly over the real numbers.

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