Symmetry properties Suppose β«ββ΄ Ζ(π) dπ = 10 and β«ββ΄ g(π) dπ = 20. Furthermore, suppose Ζ is an even function and g is an odd function. Evaluate the following integrals.
(e) β«ββΒ² 3πΖ(π)dπ
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Step 1: Understand the symmetry properties of the functions. An even function satisfies Ζ(π) = Ζ(-π), meaning it is symmetric about the y-axis. An odd function satisfies g(π) = -g(-π), meaning it is symmetric about the origin.
Step 2: Analyze the integral β«ββΒ² 3πΖ(π)dπ. Notice that the integrand 3πΖ(π) is a product of π (an odd function) and Ζ(π) (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: β«βπͺ^πͺ h(π)dπ = 0 if h(π) is an odd function. This property applies because the contributions from the interval [-πͺ, 0] and [0, πͺ] cancel each other out.
Step 4: Conclude that the integrand 3πΖ(π) is odd, and the integral β«ββΒ² 3πΖ(π)dπ 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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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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.
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.
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.