Ch. 5 - Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 5 - Integration

Problem 5.4.1

Chapter 5, Problem 5.4.1

If ƒ is an odd function, why is ∫ᵃ₋ₐ ƒ(𝓍) d𝓍 = 0?

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Understand the definition of an odd function: A function ƒ is odd if ƒ(-𝓍) = -ƒ(𝓍) for all 𝓍 in its domain. This symmetry property will be key to solving the problem.

Visualize the graph of an odd function: Odd functions are symmetric about the origin, meaning that the part of the graph on the left side of the y-axis (negative 𝓍-values) is a mirror image of the part on the right side of the y-axis (positive 𝓍-values), but flipped vertically.

Set up the integral: The integral ∫ᵃ₋ₐ ƒ(𝓍) d𝓍 represents the total area under the curve of ƒ(𝓍) from -a to a. This includes contributions from both the interval [-a, 0] and [0, a].

Break the integral into two parts: Use the property of definite integrals to write ∫ᵃ₋ₐ ƒ(𝓍) d𝓍 = ∫⁰₋ₐ ƒ(𝓍) d𝓍 + ∫ᵃ₀ ƒ(𝓍) d𝓍. This separates the integral into the left and right halves of the interval.

Apply the odd function property: For an odd function, ƒ(-𝓍) = -ƒ(𝓍). Substituting this into the integral for the left half, ∫⁰₋ₐ ƒ(𝓍) d𝓍 becomes -∫ᵃ₀ ƒ(𝓍) d𝓍. Adding these two parts together results in ∫ᵃ₋ₐ ƒ(𝓍) d𝓍 = 0, because the areas cancel each other out.

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Key Concepts

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

Odd Functions

An odd function is defined by the property that ƒ(-x) = -ƒ(x) for all x in its domain. This means that the function is symmetric about the origin, leading to the cancellation of areas under the curve when integrated over symmetric intervals. For example, the function ƒ(x) = x³ is odd, as ƒ(-x) = -x³.

Definite Integrals

A definite integral, represented as ∫ᵃᵇ ƒ(x) dx, calculates the net area under the curve of the function ƒ(x) from x = a to x = b. If the interval is symmetric around zero, such as from -a to a, the contributions from positive and negative areas can cancel each other out, especially for odd functions.

Symmetry in Integration

Symmetry plays a crucial role in integration, particularly with odd and even functions. For odd functions integrated over symmetric limits, the area above the x-axis is equal in magnitude but opposite in sign to the area below the x-axis. Thus, when integrating an odd function from -a to a, the total area sums to zero, resulting in ∫ᵃ₋ₐ ƒ(x) dx = 0.

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Mean Value Theorem for Integrals Find or approximate all points at which the given function equals its average value on the given interval.

ƒ(𝓍) = 1 ― |𝓍| on [―1, 1]

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