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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
Capitolo 5, Problema 5.2.55b

Properties of integrals Consider two functions ƒ and g on [1,6] such that ∫₁⁶ƒ(𝓍) d𝓍 = 10 and ∫₁⁶g(𝓍) d𝓍 = 5, ∫₄⁶ƒ(𝓍) d𝓍 = 5 , and ∫₁⁴g(𝓍) d𝓍 = 2. Evaluate the following integrals.


(b) ∫₁⁶ (f(𝓍) ― g(𝓍)) d𝓍

Guida verificata passo dopo passo
1
Step 1: Recall the property of integrals that states ∫ₐᵇ (ƒ(𝓍) ± g(𝓍)) d𝓍 = ∫ₐᵇ ƒ(𝓍) d𝓍 ± ∫ₐᵇ g(𝓍) d𝓍. This allows us to split the integral into two separate integrals.
Step 2: Apply the property to the given integral ∫₁⁶ (ƒ(𝓍) ― g(𝓍)) d𝓍. This becomes ∫₁⁶ ƒ(𝓍) d𝓍 ― ∫₁⁶ g(𝓍) d𝓍.
Step 3: Substitute the given values for the integrals: ∫₁⁶ ƒ(𝓍) d𝓍 = 10 and ∫₁⁶ g(𝓍) d𝓍 = 5.
Step 4: Perform the subtraction operation symbolically: 10 ― 5.
Step 5: The result of the subtraction gives the value of the integral ∫₁⁶ (ƒ(𝓍) ― g(𝓍)) d𝓍. This completes the evaluation process.

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Properties of Definite Integrals

Definite integrals have several key properties, including linearity, which states that the integral of a sum of functions is the sum of their integrals. This means that for any two functions f and g, ∫(f + g) dx = ∫f dx + ∫g dx. Additionally, the integral of a constant multiplied by a function can be factored out: ∫k * f dx = k * ∫f dx. Understanding these properties is essential for evaluating integrals involving combinations of functions.
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Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, stating that if F is an antiderivative of f on an interval [a, b], then ∫ₐᵇ f(x) dx = F(b) - F(a). This theorem allows us to evaluate definite integrals by finding the antiderivative of the integrand. It is crucial for solving problems involving definite integrals, as it provides a method to compute the area under a curve.
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Substitution in Integrals

Substitution is a technique used in integration to simplify the process of finding integrals. It involves changing the variable of integration to make the integral easier to evaluate. For example, if we let u = g(x), then the integral ∫f(g(x))g'(x)dx can be transformed into ∫f(u)du. This method is particularly useful when dealing with composite functions or when the integrand can be expressed in a simpler form.
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