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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 5 - IntegrationProblem 5.4.3b
Chapter 5, Problem 5.4.3b

Suppose ƒ is an even function and ∫⁸₋₈ ƒ(𝓍) d𝓍 = 18
(b) Evaluate ∫₋₈⁸ 𝓍ƒ(𝓍) d𝓍 .

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1
Step 1: Recall the definition of an even function. An even function satisfies ƒ(𝓍) = ƒ(-𝓍) for all 𝓍 in its domain. This symmetry will be important for solving the integral.
Step 2: Consider the integral ∫₋₈⁸ 𝓍ƒ(𝓍) d𝓍. Notice that the integrand 𝓍ƒ(𝓍) involves the product of 𝓍 and ƒ(𝓍).
Step 3: Analyze the symmetry of the integrand. Since ƒ(𝓍) is even, ƒ(𝓍) = ƒ(-𝓍). However, 𝓍 is an odd function because 𝓍 = -𝓍 when reflected about the origin. The product of an even function and an odd function is an odd function.
Step 4: Recall a key property of definite integrals: the integral of an odd function over a symmetric interval [−a, a] is always 0. This is because the positive and negative contributions cancel each other out.
Step 5: Conclude that ∫₋₈⁸ 𝓍ƒ(𝓍) d𝓍 = 0 based on the symmetry of the integrand and the properties of odd functions over symmetric intervals.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Even Functions

An even function is defined as a function f(x) that satisfies the condition f(-x) = f(x) for all x in its domain. This symmetry about the y-axis implies that the area under the curve from -a to 0 is equal to the area from 0 to a. This property is crucial for evaluating integrals involving even functions, as it simplifies calculations and allows for certain symmetries to be exploited.
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Properties of Definite Integrals

Definite integrals have several important properties, one of which is that the integral of an odd function over a symmetric interval around zero is zero. This is because the areas above and below the x-axis cancel each other out. Understanding these properties helps in evaluating integrals more efficiently, especially when dealing with functions that exhibit symmetry.
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Integration by Parts

Integration by parts is a technique used to integrate products of functions and is based on the product rule for differentiation. It is expressed as ∫u dv = uv - ∫v du, where u and v are differentiable functions. This method is particularly useful when integrating products of polynomials and other functions, allowing for the transformation of complex integrals into simpler forms.
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