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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.RE.15e
Capitolo 5, Problema 5.RE.15e

Symmetry properties Suppose ∫₀⁴ ƒ(𝓍) d𝓍 = 10 and ∫₀⁴ g(𝓍) d𝓍 = 20. Furthermore, suppose ƒ is an even function and g is an odd function. Evaluate the following integrals.


(e) ∫₋₂² 3𝓍ƒ(𝓍)d𝓍

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Step 1: Understand the symmetry properties of the functions. An even function satisfies ƒ(𝓍) = ƒ(-𝓍), meaning it is symmetric about the y-axis. An odd function satisfies g(𝓍) = -g(-𝓍), meaning it is symmetric about the origin.
Step 2: Analyze the integral ∫₋₂² 3𝓍ƒ(𝓍)d𝓍. Notice that the integrand 3𝓍ƒ(𝓍) is a product of 𝓍 (an odd function) and ƒ(𝓍) (an even function). The product of an odd function and an even function is an odd function.
Step 3: Recall a key property of definite integrals for odd functions: ∫₋𝓪^𝓪 h(𝓍)d𝓍 = 0 if h(𝓍) is an odd function. This property applies because the contributions from the interval [-𝓪, 0] and [0, 𝓪] cancel each other out.
Step 4: Conclude that the integrand 3𝓍ƒ(𝓍) is odd, and the integral ∫₋₂² 3𝓍ƒ(𝓍)d𝓍 evaluates to 0 based on the symmetry property of odd functions.
Step 5: Summarize the reasoning: The integral evaluates to 0 because the integrand is an odd function and the limits of integration are symmetric about the origin.

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Even and Odd Functions

An even function is defined by the property f(-x) = f(x) for all x in its domain, which means its graph is symmetric about the y-axis. Conversely, an odd function satisfies g(-x) = -g(x), indicating that its graph is symmetric about the origin. Understanding these properties is crucial for evaluating integrals, as they can simplify calculations by exploiting symmetry.
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Properties of Functions

Properties of Definite Integrals

Definite integrals have specific properties that can simplify their evaluation. For instance, the integral of an even function over a symmetric interval [-a, a] can be expressed as twice the integral from 0 to a. In contrast, the integral of an odd function over a symmetric interval is zero. These properties are essential for solving integrals involving even and odd functions.
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Definition of the Definite Integral

Integration Techniques

Integration techniques involve various methods for calculating integrals, including substitution, integration by parts, and recognizing patterns in functions. In this context, recognizing the symmetry of the functions involved allows for the application of specific techniques that can simplify the evaluation of the integral, particularly when combined with the properties of even and odd functions.
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Integration by Parts for Definite Integrals
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Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

(a) If ƒ is symmetric about the line 𝓍 = 2 , then ∫₀⁴ ƒ(𝓍) d𝓍 = 2 ∫₀² ƒ(𝓍) d𝓍.

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Function defined by an integral Let ƒ(𝓍) = ∫₀ˣ (t ― 1)¹⁵ (t―2)⁹ dt .

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Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

(a) Consider the linear function ƒ(𝓍) = 2x + 5 and the region bounded by its graph and the x-axis on the interval [3,6]. Suppose the area of this region is approximated using midpoint Riemann sums. Then the approximations give the exact area of the region for any number of subintervals.

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Symmetry properties Suppose ∫₀⁴ ƒ(𝓍) d𝓍 = 10 and ∫₀⁴ g(𝓍) d𝓍 = 20. Furthermore, suppose ƒ is an even function and g is an odd function. Evaluate the following integrals.


(a) ∫₋₄⁴ ƒ(𝓍) d𝓍

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Symmetry properties Suppose ∫₀⁴ ƒ(𝓍) d𝓍 = 10 and ∫₀⁴ g(𝓍) d𝓍 = 20. Furthermore, suppose ƒ is an even function and g is an odd function. Evaluate the following integrals.


(c) ∫₋₄⁴ (4ƒ(𝓍) ― 3g(𝓍))d𝓍

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