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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 5.2.53c
Capitolo 5, Problema 5.2.53c

Properties of integrals Suppose ∫₀³ƒ(𝓍) d𝓍 = 2 , ∫₃⁶ƒ(𝓍) d𝓍 = ―5 , and ∫₃⁶g(𝓍) d𝓍 = 1. Evaluate the following integrals.
(c) ∫₃⁶ (3ƒ(𝓍) ― g(𝓍)) d𝓍

Guida verificata passo dopo passo
1
Step 1: Recognize that the integral ∫₃⁶ (3ƒ(𝓍) ― g(𝓍)) d𝓍 can be split into separate integrals using the linearity property of integrals. This property states that ∫ₐᵇ [c₁f(𝓍) + c₂g(𝓍)] d𝓍 = c₁∫ₐᵇ f(𝓍) d𝓍 + c₂∫ₐᵇ g(𝓍) d𝓍.
Step 2: Apply the linearity property to rewrite the integral as ∫₃⁶ (3ƒ(𝓍)) d𝓍 ― ∫₃⁶ g(𝓍) d𝓍.
Step 3: Factor out the constant 3 from the first integral using the constant multiple rule, which states that ∫ₐᵇ c·f(𝓍) d𝓍 = c·∫ₐᵇ f(𝓍) d𝓍. This gives 3∫₃⁶ ƒ(𝓍) d𝓍 ― ∫₃⁶ g(𝓍) d𝓍.
Step 4: Substitute the given values for the integrals. From the problem, ∫₃⁶ ƒ(𝓍) d𝓍 = ―5 and ∫₃⁶ g(𝓍) d𝓍 = 1.
Step 5: Combine the results algebraically to evaluate the expression. The final result will be 3(―5) ― 1.

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