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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
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.
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Properties of Functions

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.
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Definition of the Definite Integral

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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Related Practice
Textbook Question

Identifying definite integrals as limits of sums Consider the following limits of Riemann sums for a function ƒ on [a,b]. Identify ƒ and express the limit as a definite integral.                                

          n                                                                                                                                                                              

    lim   ∑ (𝓍ₖ*² + 1) ∆𝓍ₖ on [0,2]                                                                                                                                                                            

  ∆ → 0   k=1                                                                                                                                                                                                                      

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Textbook Question

Definite integrals from graphs The figure shows the areas of regions bounded by the graph of ƒ and the 𝓍-axis. Evaluate the following integrals.



∫₀ᵃ ƒ(𝓍) d𝓍

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Textbook Question

Explain the statement that a continuous function on an interval [a,b] equals its average value at some point on (a,b).

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Textbook Question

Average values Find the average value of the following functions on the given interval. Draw a graph of the function and indicate the average value.

ƒ(𝓍) = cos 𝓍 on [―π/2 , π/2]

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Textbook Question

Derivatives of integrals Simplify the following expressions.


d/d𝓍 ∫₃ˣ (t² + t + 1) dt

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Textbook Question

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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