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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 5.4.4a
Chapter 5, Problem 5.4.4a

Suppose ƒ is an odd function, ∫₀⁴ ƒ(𝓍) d𝓍 = 3 , and ∫₀⁸ ƒ(𝓍) d𝓍 = 9 .


(a) Evaluate ∫₋₈⁴ ƒ(𝓍) d𝓍 .

Verified step by step guidance
1
Step 1: Recall the property of odd functions. An odd function satisfies ƒ(−𝓍) = −ƒ(𝓍). This property implies that the integral of an odd function over a symmetric interval [−𝓎, 𝓎] is zero, i.e., ∫₋𝓎⁺𝓎 ƒ(𝓍) d𝓍 = 0.
Step 2: Break the integral ∫₋₈⁴ ƒ(𝓍) d𝓍 into two parts using the additive property of integrals: ∫₋₈⁴ ƒ(𝓍) d𝓍 = ∫₋₈⁰ ƒ(𝓍) d𝓍 + ∫₀⁴ ƒ(𝓍) d𝓍.
Step 3: Evaluate ∫₋₈⁰ ƒ(𝓍) d𝓍 using the property of odd functions. Since the interval [−8, 0] is symmetric about zero, ∫₋₈⁰ ƒ(𝓍) d𝓍 = −∫₀⁸ ƒ(𝓍) d𝓍. Substitute the given value ∫₀⁸ ƒ(𝓍) d𝓍 = 9 to find ∫₋₈⁰ ƒ(𝓍) d𝓍 = −9.
Step 4: Substitute the given value of ∫₀⁴ ƒ(𝓍) d𝓍 = 3 into the equation from Step 2: ∫₋₈⁴ ƒ(𝓍) d𝓍 = −9 + 3.
Step 5: Combine the results from Step 4 to express the final integral value. The result is ∫₋₈⁴ ƒ(𝓍) d𝓍 = −6.

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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 symmetry about the origin implies that the area under the curve from -a to 0 is the negative of the area from 0 to a. Understanding this property is crucial for evaluating integrals involving odd functions, especially when considering limits that span both positive and negative values.
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Definite Integrals

A definite integral, denoted as ∫ₐᵇ ƒ(x) dx, represents the signed area under the curve of the function ƒ(x) from x = a to x = b. The value of a definite integral can be interpreted as the accumulation of the function's values over the specified interval. In this problem, the given integrals provide specific areas that can be used to find other related integrals through properties of symmetry and linearity.
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Properties of Integrals

The properties of integrals, such as the linearity of integration and the relationship between integrals over symmetric intervals, are essential for solving problems involving definite integrals. For instance, the integral from -a to a of an odd function is zero, and the integral from a to b can be expressed in terms of integrals over other intervals. These properties allow for simplifications and transformations that facilitate the evaluation of complex integrals.
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Related Practice
Textbook Question

Area functions for linear functions Consider the following functions ƒ and real numbers a (see figure).                                                                                           

                                                                                                                                                                                     

 (a) Find and graph the area function A (𝓍) = ∫ₐˣ ƒ(t) dt .                                                                                                                               

                                                                                                                                                                               

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 ƒ(t) = 4t + 2 , a = 0

Textbook Question

Matching functions with area functions Match the functions ƒ, whose graphs are given in a― d, with the area functions A (𝓍) = ∫₀ˣ ƒ(t) dt, whose graphs are given in A–D.



Textbook Question

Working with area functions Consider the function ƒ and its graph.

(a) Estimate the zeros of the area function A(𝓍) = ∫₀ˣ ƒ(t) dt , for 0 ≤ 𝓍 ≤ 10 .


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Textbook Question

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals. 

I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1

(a) ∫₀^π/2 (2 sin θ ― cos θ) dθ

Textbook Question

Approximating displacement The velocity in ft/s of an object moving along a line is given by v = 3t² + 1 on the interval 0 ≤ t ≤ 4, where t is measured in seconds.

(a) Divide the interval [0,4] into n = 4 subintervals, [0,1] , [1.2] , [2,3] , and [3,4]. On each subinterval, assume the object moves at a constant velocity equal to v evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [0, 4] (see part (a) of the figure)

Textbook Question

Working with area functions Consider the function ƒ and the points a, b, and c.

(a) Find the area function A (𝓍) = ∫ₐˣ ƒ(t) dt using the Fundamental Theorem.

ƒ(𝓍) = sin 𝓍 ; a = 0 , b = π/2 , c = π