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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 5.2.58b
Capitolo 5, Problema 5.2.58b

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals. 
I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1
(b) ∫₀^π/2 (4 cos θ ― 8 sin θ) dθ

Guida verificata passo dopo passo
1
Step 1: Recognize that the given integral (b) ∫₀^π/2 (4 cos θ ― 8 sin θ) dθ can be expressed in terms of the original integral I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1.
Step 2: Factor out the constant multiplier from the integrand in (b). Using the property of integrals, ∫ₐᵇ k * f(x) dx = k * ∫ₐᵇ f(x) dx, rewrite the integral as ∫₀^π/2 (4 cos θ ― 8 sin θ) dθ = 4 * ∫₀^π/2 (cos θ ― 2 sin θ) dθ.
Step 3: Substitute the value of the original integral I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1 into the expression derived in Step 2.
Step 4: Multiply the constant factor (4) by the value of the original integral (―1) to simplify the expression.
Step 5: The result of the multiplication gives the value of the integral (b).

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Properties of Integrals

The properties of integrals, such as linearity, allow us to manipulate integrals in useful ways. For instance, the integral of a sum can be expressed as the sum of the integrals, and constants can be factored out. This is crucial for evaluating integrals efficiently, especially when they can be expressed in terms of known integrals.
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Properties of Functions

Definite Integrals

Definite integrals represent the signed area under a curve between two limits. In this case, the limits are from 0 to π/2. Understanding how to compute definite integrals and their properties is essential for evaluating integrals like the one given in the question, as it provides the numerical value of the area.
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Definition of the Definite Integral

Substitution in Integrals

Substitution is a technique used to simplify the evaluation of integrals by changing the variable of integration. This method can transform a complex integral into a simpler form, making it easier to compute. Recognizing when and how to apply substitution is key to solving integrals effectively.
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Substitution With an Extra Variable
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