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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 5 - IntegrationProblema 5.5.84
Capitolo 5, Problema 5.5.84

Variations on the substitution method Evaluate the following integrals.                                                                                                        
                                                                                                                                                                    
 ∫ (𝒵 + 1) √(3𝒵 + 2) d𝒵

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1
Identify the integral to solve: \(\int (Z + 1) \sqrt{3Z + 2} \, dZ\).
Choose a substitution to simplify the integral. Let \(u = 3Z + 2\), so that the expression under the square root becomes \(\sqrt{u}\).
Compute the differential \(du\) in terms of \(dZ\): since \(u = 3Z + 2\), then \(\frac{du}{dZ} = 3\), which implies \(dZ = \frac{du}{3}\).
Express \(Z\) in terms of \(u\) to rewrite the factor \((Z + 1)\): from \(u = 3Z + 2\), solve for \(Z\) to get \(Z = \frac{u - 2}{3}\), so \(Z + 1 = \frac{u - 2}{3} + 1 = \frac{u + 1}{3}\).
Rewrite the integral entirely in terms of \(u\) and \(du\): substitute \((Z + 1)\) and \(\sqrt{3Z + 2}\) with their expressions in \(u\), and replace \(dZ\) with \(\frac{du}{3}\). Then simplify the integrand before integrating with respect to \(u\).

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Substitution Method in Integration

The substitution method simplifies integrals by changing variables to transform a complicated integral into a basic form. It involves choosing a substitution u = g(z) such that the integral becomes easier to evaluate. This method is especially useful when the integral contains a composite function and its derivative.
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Euler's Method

Chain Rule and Its Role in Integration

The chain rule in differentiation helps identify the inner function and its derivative, which guides the substitution choice in integration. Recognizing the derivative of the inner function within the integral allows for an effective substitution, turning the integral into a simpler polynomial or standard form.
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Integration of Polynomial and Root Functions

Integrals involving polynomials and roots often require rewriting the root as a fractional exponent. After substitution, the integral can be expressed as a sum of powers of the variable, which can be integrated using the power rule. Understanding how to manipulate and integrate these expressions is essential for solving such integrals.
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