Textbook Question
Suppose F is an antiderivative of Ę and A is an area function of Ę. What is the relationship between F and A?
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Ch. 5 - Integration
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.4.5
Use symmetry to explain why.
ā«ā“āā (5šā“ + 3šĀ³ + 2šĀ² + š + 1) dš = 2 ā«āā“ (5šā“ + 2šĀ² + š + 1) dš .
Verified step by step guidance1
Step 1: Recognize that the integral is over a symmetric interval [-4, 4]. Symmetry in integrals often simplifies calculations, especially when the integrand has specific properties like even or odd functions.
Step 2: Break down the integrand into individual terms: 5šā“, 3šĀ³, 2šĀ², š, and 1. Analyze each term to determine whether it is an even function or an odd function. Recall that even functions satisfy f(-š) = f(š), while odd functions satisfy f(-š) = -f(š).
Step 3: Identify the symmetry of each term: 5šā“ and 2šĀ² are even functions, while 3šĀ³ and š are odd functions. The constant term 1 is also even because it does not depend on š.
Step 4: Use the property of integrals over symmetric intervals: The integral of an odd function over [-a, a] is zero because the positive and negative contributions cancel out. Therefore, the terms 3šĀ³ and š do not contribute to the integral over [-4, 4].
Step 5: Rewrite the original integral by excluding the odd terms and focusing only on the even terms. This simplifies the integral to ā«ā“āā (5šā“ + 2šĀ² + š + 1) dš = 2 ā«āā“ (5šā“ + 2šĀ² + š + 1) dš, leveraging the symmetry of the even functions over the interval.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Symmetry in Functions
Symmetry in functions refers to the property where a function exhibits identical behavior on either side of a central point, typically the y-axis for even functions or the origin for odd functions. For example, a function f(x) is even if f(-x) = f(x), and odd if f(-x) = -f(x). This property can simplify the evaluation of integrals, particularly over symmetric intervals.
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Definite Integrals
A definite integral calculates the net area under a curve defined by a function over a specific interval [a, b]. It is represented as ā«_a^b f(x) dx and provides a numerical value that represents this area. Understanding how to manipulate definite integrals, especially with respect to symmetry, is crucial for simplifying calculations.
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Properties of Integrals
Properties of integrals include various rules that allow for the manipulation and evaluation of integrals. One important property is that the integral of an even function over a symmetric interval [-a, a] can be expressed as twice the integral from 0 to a. This property is essential for simplifying integrals involving symmetric functions, as seen in the given equation.
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Related Practice
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Evaluate ā«āĀ² 3šĀ² dš and ā«āāĀ² 3šĀ² dš.
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ā« 2 / (šā4šĀ² ā1) dš , š > Ā½
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Variations on the substitution method Evaluate the following integrals.
ā« (šµ + 1) ā(3šµ + 2) dšµ
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ā«āā“ (šĀ²ā1) dš
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Textbook Question
Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals.
ā«ā/āā āāā^Ā²/āµ dš/ xā(25šĀ²ā 1)
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