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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 8 - Techniques of IntegrationProblema 8.5.14
Capitolo 8, Problema 8.5.14

In Exercises 9–16, express the integrand as a sum of partial fractions and evaluate the integrals.
∫ (y + 4) / (y² + y) dy from 1/2 to 1

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1
Start by factoring the denominator of the integrand. The denominator is \(y^{2} + y\), which can be factored as \(y(y + 1)\).
Express the integrand \(\frac{y + 4}{y(y + 1)}\) as a sum of partial fractions in the form \(\frac{A}{y} + \frac{B}{y + 1}\), where \(A\) and \(B\) are constants to be determined.
Multiply both sides of the equation by the common denominator \(y(y + 1)\) to clear the fractions: $y + 4 = A(y + 1) + By$.
Expand the right side and collect like terms: $y + 4 = Ay + A + By = (A + B)y + A$. Then, equate the coefficients of like terms from both sides to form a system of equations: for \(y\) terms, \(1 = A + B\); for constant terms, \(4 = A\).
Solve the system for \(A\) and \(B\), substitute these values back into the partial fractions, and then integrate each term separately over the interval from \(\frac{1}{2}\) to \(1\).

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Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down a complex rational function into simpler fractions that are easier to integrate. It involves expressing the integrand as a sum of fractions with simpler denominators, typically linear or irreducible quadratic factors.
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Partial Fraction Decomposition: Distinct Linear Factors

Integration of Rational Functions

Integrating rational functions often requires rewriting the integrand into simpler parts, such as partial fractions. Once decomposed, each term can be integrated using basic integral formulas, including logarithmic and inverse trigonometric functions depending on the denominator.
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Intro to Rational Functions

Definite Integrals and Limits of Integration

A definite integral calculates the net area under a curve between two points. After finding the antiderivative, the Fundamental Theorem of Calculus is applied by evaluating the antiderivative at the upper and lower limits and subtracting to find the integral's value.
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Definition of the Definite Integral
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